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#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Re: Integration lessons continued ....

4.4.4. integrals involve zeta computations

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Prove that

$$\displaystyle \int^{\frac{\pi}{2}}_0 \,x \sin(x)\cos(x) \log(\sin x) \log(\cos x) \, dx = \frac{\pi}{16}-\frac{\pi^3}{192}$$

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PROOF

Start by the change of variable, $$\displaystyle t=\frac{\pi}{2}-x$$

$$\displaystyle \int^{\frac{\pi}{2}}_0 \,x \sin(x)\cos(x) \log(\sin x) \log(\cos x) \, dx\,=\frac{\pi}{4}\int^{\frac{\pi}{2}}_0 \,\sin(x)\cos(x) \log(\sin x) \log(\cos x) \, dx \text{ }\cdots (1)$$

We need to find :

$$\displaystyle \int^{\frac{\pi}{2}}_0 \,\sin(x)\cos(x) \log(\sin x) \log(\cos x) \, dx$$

Let us start by the following :

$$\displaystyle F(a,b)=2\int^{\frac{\pi}{2}}_0 \,\sin^{2a-1}(x)\cos^{2b-1}(x) \, dx\,=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$

Now let us differentiate with respect to $a$

$$\displaystyle \frac{\partial}{\partial a}(F(a,b))=4\int^{\frac{\pi}{2}}_0 \,\sin^{2a-1}(x)\cos^{2b-1}(x) \log(\sin x)\, dx\,=\frac{\Gamma(a)\Gamma(b) \left(\psi_0(a)-\psi_0(a+b)\right)}{\Gamma(a+b)}$$

Differentiate again but this time with respect to $b$

$$\displaystyle \frac{\partial}{\partial b}\left(F_a(a,b) \right)=8\int^{\frac{\pi}{2}}_0 \,\sin^{2a-1}(x)\cos^{2b-1}(x) \log(\sin x)\, \log( \cos x)dx\,$$

$$\displaystyle =\frac{\Gamma(a) \Gamma(b) \left( \psi_0^2(a+b) +\psi_0(a)\psi_0(b)-\psi_0(a)\psi_0(a+b)-\psi_0(b)\psi_0(a+b) +\psi_1(a+b)\right)}{\Gamma(a+b)}$$

putting $$\displaystyle a=b=1$$ we have the following

$$\displaystyle \int^{\frac{\pi}{2}}_0 \,\sin (x)\cos(x) \log(\sin x)\, \log( \cos x)dx\,= \frac{ \psi_0^2(2) +\psi_0^2(1)-\psi_0(1)\psi_0(2)-\psi_0(1)\psi(2) -\psi_1(2)}{8}$$

By simple algebra we arrive to

$$\displaystyle \int^{\frac{\pi}{2}}_0 \,\sin (x)\cos(x) \log(\sin x)\, \log( \cos x)dx\,= \frac{ (\psi_0(2) -\psi_0(1))^2-\psi_1(2)}{8}$$

We can easily see that

• $$\displaystyle \psi_0(1) = -\gamma$$
• $$\displaystyle \psi_0(2) =1 -\gamma$$

Now to evaluate $$\displaystyle \psi_1(2)$$ , we have to use the zeta function

we have already established the following relation :

$$\displaystyle \psi_1(z) = \sum_{k\geq 0} \frac{1}{(n+z)^2}$$

Now putting $$\displaystyle z =2$$ we have the following

$$\displaystyle \psi_1(2) = \sum_{k\geq 0} \frac{1}{(k+2)^2}$$

Let us write the first few terms in the expansion

$$\displaystyle \sum_{k\geq 0} \frac{1}{(k+2)^2} = \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+ \cdots$$

we see this is similar to $$\displaystyle \zeta(2)$$ but we are missing the first term

$$\displaystyle \psi_1(2) = \zeta(2)-1 = \frac{\pi^2}{6} -1$$

Collecting all these information together we have

$$\displaystyle \int^{\frac{\pi}{2}}_0 \,\sin (x)\cos(x) \log(\sin x)\, \log( \cos x)dx\,= \frac{1}{4}-\frac{\pi^2}{48}$$

substituting in (1) we get :

$$\displaystyle \int^{\frac{\pi}{2}}_0 \,x\sin(x)\cos(x) \log(\sin x) \log(\cos x) \, dx =\frac{\pi}{16}-\frac{\pi^3}{192}$$

To Be Continued ...

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Re: Integration lessons continued ....

4.4.5 Dirichlet eta function( an alternating form )

Dirichlet eta function is the alternating form of the zeta function

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DEFINITION :

$$\displaystyle \eta(s) = \sum_{n\geq 1} \frac{(-1)^{n-1}}{n^s}$$​

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The alternating form of the zeta function is easier to compute once we have established the main results of the zeta function because the alternating form is related to the zeta function through the relation

$$\displaystyle \eta(s) = \left( 1-2^{1-s} \right) \zeta(s)$$​

PROOF

We will start by the RHS

$$\displaystyle \left( 1-2^{1-s} \right) \zeta(s) = \zeta(s) - 2^{1-s} \zeta(s)$$

which can be written as sums of series

$$\displaystyle \sum_{n\geq 1} \frac{1}{n^s} - \frac{1}{2^{s-1}}\sum_{n\geq 1} \frac{1}{n^s}$$

$$\displaystyle \sum_{n\geq 1} \frac{1}{n^s} - 2\sum_{n\geq 1} \frac{1}{(2n)^s}$$

Clearly we can see that we are subtracting even terms twice , this is equivalent to

$$\displaystyle \sum_{n\geq 1} \frac{1}{(2n+1)^s} - \sum_{n\geq 1} \frac{1}{(2n)^s}$$

This looks easier to understand if we write the terms

$$\displaystyle \left(1+\frac{1}{3^s}+\frac{1}{5^s}+ \cdots\right) - \left( \frac{1}{2^s}+\frac{1}{4^s} + \frac{1}{6^s}+\cdots \right)$$

Rearranging the terms we establish the alternating form

$$\displaystyle 1-\frac{1}{2^s}+\frac{1}{3^s}-\frac{1}{4^s}+\cdots = \sum_{n\geq 1}\frac{(-1)^{n-1}}{n^s} =\eta(s) \, \square$$

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We know feel tempted to evaluate some values

$$\displaystyle \eta(2) = \left(1-\frac{1}{2} \right) \zeta(2) = \frac{\pi^2}{12}$$

Actually there is a nice integration formula similar to that we had for zeta

$$\displaystyle \eta(s) \, \Gamma(s) = \int^{\infty}_{0} \frac{t^{s-1}}{e^t+1} \, dt$$​

PROOF :

Start by the RHS

$$\displaystyle \int^{\infty}_{0} \frac{t^{s-1}}{e^t+1} \, dt = \int^{\infty}_0 \frac{e^{-t} t^{s-1}}{1+e^{-t}} \, dt$$

Now using the power expansion we arrive to

$$\displaystyle \int^{\infty}_0 e^{-t} t^{s-1}dt \left(\sum_{n\geq 0} (-1)^n e^{-nt} \right)$$

$$\displaystyle \sum_{n\geq 0}(-1)^n\int^{\infty}_0 e^{-(n+1) t} t^{s-1}dt$$

Using Laplace transform we can solve the inner integral

$$\displaystyle \int^{\infty}_0 e^{-(n+1) t} t^{s-1}dt = \frac{\Gamma(s)}{(n+1)^s}$$

Hence we have the following

$$\displaystyle \Gamma(s) \sum_{n\geq 0}\frac{(-1)^n}{(n+1)^s}=\Gamma(s) \sum_{n\geq 1}\frac{(-1)^{n-1}}{n^s}=\Gamma(s)\, \eta(s) \square$$

An easy result of the above integral

$$\displaystyle \int^{\infty}_{0}\frac{t}{e^t+1} \, dt = \Gamma(2)\, \eta(2) = \frac{\pi^2}{12}$$

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We shall look at the polylogarithms in the next thread

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#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Re: Integration lessons continued ....

4.Integration using special functions (continued)

4.5. Polylogarithm

As the name suggests, this special function is closely related to the logarithms .

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DEFINITION

$$\displaystyle \operatorname{Li}_{n}(z) = \sum_{k\geq 1} \frac{z^k}{k^n}$$​

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Note : As we see , we use $$\displaystyle \operatorname{Li}_{n}(z)$$ to denote the polylogarithm . The name contains two parts , (poly) because we can choose different $n$ and produce many functions . (logarithm) because we can express it integrals of logarithms .

Through that representation we can see how closely it is related to the zeta function

$$\displaystyle \operatorname{Li}_{n}(1) = \sum_{k \geq 1} \frac{1}{k^n}= \zeta(n)$$

In particular we have for $n=2$

$$\displaystyle \operatorname{Li}_{2}(1) = \zeta(2) = \frac{\pi^2}{6}$$

Also we can relate it to the eta function though $z=-1$

$$\displaystyle \operatorname{Li}_{n}(-1) = \sum_{k\geq 1} \frac{(-1)^k}{k^n}=-\eta(n)$$

Also we shall see how to relate this function to the logarithm

putting $$\displaystyle n=1$$ we have the following

$$\displaystyle \operatorname{Li}_{\, 1}(z) = \sum_{n\geq 1} \frac{z^k}{k}$$

The power expansion on the left is quire famous

$$\displaystyle \sum_{n\geq 1} \frac{z^k}{k} = - \log(1-z)$$

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There is an interesting recursive representation of this function

$$\displaystyle \operatorname{Li}_{\, n+1}(z) = \int^z_0 \frac{\operatorname{Li}_{\,n}(t) }{t}\, dt$$​

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PROOF

Using the series representation we have

$$\displaystyle \int^z_0 \frac{1}{t} \left( \sum_{k\geq 1} \frac{t^k}{k^n}\, \right) dt$$

$$\displaystyle \sum_{k\geq 1}\frac{1}{k^n} \int^z_0 t^{k-1} \, dt$$

Integrating term by term we have

$$\displaystyle \sum_{k\geq 1}\frac{z^{k}}{k^{n+1}} \, = \operatorname{Li}_{\, n+1}(z) \,\, \square$$

We can differentiate the result to obtain

$$\displaystyle \frac{\partial}{\partial z}\operatorname{Li}_{\, n+1}(z) = \frac{1}{z} \operatorname{Li}_{\,n}(z)$$

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Square formula

$$\displaystyle \operatorname{Li}_{\,n}(-z) + \operatorname{Li}_{\,n}(z) = 2^{1-n} \,\operatorname{Li}_{\,n}(z^2)$$​

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PROOF

As usual we write the series representation of the LHS

$$\displaystyle \sum_{k\geq 1} \frac{z^k}{k^n}+\sum_{k\geq 1} \frac{(-z)^k}{k^n}$$

Listing the first few terms

$$\displaystyle z+\frac{z^2}{2^n}+\frac{z^3}{3^n}+\cdots +\left(-z+\frac{z^2}{2^n}-\frac{z^3}{3^n}+\cdots \right)$$

Clearly the odd terms will cancel so we are left with

$$\displaystyle 2\frac{z^2}{2^n}+2\frac{z^4}{4^n}+2\frac{z^6}{6^n}+\cdots$$

By simple manipulation

$$\displaystyle 2^{1-n} \left( z^{2}+\frac{(z^2)^2}{2^n}+\frac{(z^2)^3}{3^n}+ \cdots \right)$$

$$\displaystyle 2^{1-n} \sum_{k \geq 1} \frac{(z^2)^{k}}{k^n} = 2^{1-n} \, \operatorname{Li}_{\, n}(z^2) \, \square$$

[HW] prove that $$\displaystyle \operatorname{Li}_{\, 2}(z) = -\int^z_0 \frac{\log(1-t)}{t}\, dt$$ Dilogarithm

To Be Continued ...

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Re: Integration lessons continued ....

4.5.1.Dilogarithms

Of all polylogarithms $$\displaystyle \operatorname{Li}_2(z)$$ is the most interesting one , in this section we will see why.

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DEFINITION

$$\displaystyle \operatorname{Li}_2(z) = \sum_{k\geq 1} \frac{z^k}{k^2} = - \int^z_0 \frac{\log(1-t)}{t}\, dt$$​

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The curious reader should try to prove the integral representation using the recursive definition we introduced in the previous section .

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Some Functional equations

$$\displaystyle \operatorname{Li}_2\left(\frac{-1}{z}\right) + \operatorname{Li}_2(-z) = - \frac{1}{2}\log^2(z)-\frac{\pi^2}{6}$$

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PROOF

We will start by the following

$$\displaystyle \operatorname{Li}_2\left(\frac{-1}{z}\right) = -\int^{\frac{-1}{z}}_0 \frac{\log(1-t)}{t}\, dt$$

Differentiate with respect to $z$

$$\displaystyle \frac{d}{dz}\operatorname{Li}_2\left(\frac{-1}{z}\right) = \frac{1}{z^2} \left(-\frac{\log \left(1+\frac{1}{z} \right)}{\frac{-1}{z}} \right) = \frac{\log\left( 1+ \frac{1}{z} \right)}{z} = \frac{\log(1+z) - \log(z)}{z}$$

Now integrate with respect to $z$

$$\displaystyle \operatorname{Li}_2\left(\frac{-1}{z}\right) = \int^{-z}_0 \frac{\log(1-t)}{t} \, dt - \frac{1}{2} \log^2(z) \,+ C$$

$$\displaystyle \operatorname{Li}_2\left(\frac{-1}{z}\right) = -\operatorname{Li}_2 {-z} - \frac{1}{2} \log^2(z) \,+ C$$

To find the constant $C$ let $z = 1$

$$\displaystyle C= 2\operatorname{Li}_2\left(-1\right)$$

Now we must be aware that

$$\displaystyle \displaystyle \operatorname{Li}_{2}(-1) =-\eta(2) = \frac{-\pi^2}{12}$$

Hence we have

$$\displaystyle C= \frac{-\pi^2}{6}$$

which proves the result by simple rearrangement

$$\displaystyle \operatorname{Li}_2\left(\frac{-1}{z}\right)+\operatorname{Li}_2(-z) = -\frac{1}{2} \log^2(z) -\frac{\pi^2}{6} \, \square$$

We can let $$\displaystyle z=-1$$

$$\displaystyle 2\operatorname{Li}_2\left(1\right)=-\frac{1}{2} \log^2(-1) -\frac{\pi^2}{6} \,$$

knowing that $$\displaystyle \log(-1) = i \pi$$

$$\displaystyle 2\operatorname{Li}_2\left(1\right)=\frac{\pi^2}{2} -\frac{\pi^2}{6} \, = \frac{\pi^2}{3}$$

Hence we have $$\displaystyle \operatorname{Li}_2\left(1\right) = \frac{\pi^2}{6}$$ as expected .

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Another function equation

$$\displaystyle \operatorname{Li}_2(z) + \operatorname{Li}_{2}(1-z) = \frac{\pi^2}{6}-\log(z) \log(1-z) \,\,\,\, 0<z<1$$

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PROOF

Start by the following

$$\displaystyle \operatorname{Li}_2\left(z\right) = -\int^{z}_0 \frac{\log(1-t)}{t} \, dt$$

Now integrate by parts to obtain

$$\displaystyle \operatorname{Li}_2\left(z\right)= -\int^z_0 \frac{\log(t)}{1-t} \, dt -\log(z) \log(1-z)$$

To solve $$\displaystyle \int^z_0 \frac{\log(t)}{1-t} \, dt \,$$ let $$\displaystyle t = 1-x$$

$$\displaystyle -\int^{1-z}_{1} \frac{\log(1-x)}{x} \, dx$$

For $$\displaystyle 0<z <1$$

$$\displaystyle \int^{1}_{1-z} \frac{\log(1-x)}{x} \, dx = \int^1_0 \frac{\log(1-x)}{x}\, dx - \int_0^{1-z} \frac{\log(1-x)}{x}\, dx$$

Now it is easy to see that

$$\displaystyle \int^{1}_{1-z} \frac{\log(1-x)}{x} \, dx =-\operatorname{Li}_2(1)+\operatorname{Li}_2(1-z)$$

$$\displaystyle \operatorname{Li}_2\left(z\right)= \operatorname{Li}_2(1)-\operatorname{Li}_2(1-z) -\log(z) \log(1-z)$$

$$\displaystyle \operatorname{Li}_2\left(z\right)+\operatorname{Li}_2(1-z)\, = \, \operatorname{Li}_2(1)-\log(z) \log(1-z)$$

Now since $$\displaystyle \operatorname{Li}_2(1) = \frac{\pi^2}{6}$$

$$\displaystyle \operatorname{Li}_2\left(z\right)+\operatorname{Li}_2(1-z)\, = \frac{\pi^2}{6}-\log(z) \log(1-z) \,\, \square$$

We can easily deduce that for $$\displaystyle z=\frac{1}{2}$$

$$\displaystyle 2\operatorname{Li}_2\left(\frac{1}{2}\right)= \frac{\pi^2}{6}-\log^2\left(\frac{1}{2}\right) \,\,$$

$$\displaystyle \operatorname{Li}_2\left(\frac{1}{2}\right)= \frac{\pi^2}{12}-\frac{1}{2}\log^2 \left(\frac{1}{2}\right)$$

To be continued ...

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#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Re: Integration lessons continued ....

4.5.1.Dilogarithms (continued)

In this section we shall continue looking at some amazing results related to the dilogarithm.

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Yet another functional equation

$$\displaystyle \operatorname{Li}_2(z) + \operatorname{Li}_2 \left(\frac{z}{z-1} \right) = - \frac{1}{2} \log^2 (1-z) \,\,\,\, \, z<1$$​

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PROOF

Start by the following

$$\displaystyle \operatorname{Li}_2 \left(\frac{z}{z-1} \right) = -\int^{\frac{z}{z-1}}_0 \frac{ \log(1-t)}{t}\, dt$$

Differentiate both sides with respect to $z$

$$\displaystyle \frac{d}{dz}\operatorname{Li}_2 \left(\frac{z}{z-1} \right) = \frac{1}{(z-1)^2}\left( \frac{ \log \left(1-\frac{z}{z-1}\right)}{\frac{z}{z-1}} \right)$$

Upon simplification we obtain

$$\displaystyle \frac{d}{dz}\operatorname{Li}_2 \left(\frac{z}{z-1} \right) = \frac{- \log(1-z)}{z(z-1)}$$

Using partial fractions decomposition

$$\displaystyle \frac{d}{dz}\operatorname{Li}_2 \left(\frac{z}{z-1} \right) = \frac{\log(1-z)}{1-z}+ \frac{\log(1-z)}{z}$$

Integrate both sides with respect to $z$

$$\displaystyle \operatorname{Li}_2 \left(\frac{z}{z-1} \right) = -\frac{1}{2} \log^2(1-z) - \operatorname{Li}_2(z) +C$$

put $z=-1$ to find the constant

$$\displaystyle \operatorname{Li}_2 \left(\frac{1}{2} \right) = -\frac{1}{2} \log^2(2) - \operatorname{Li}_2(-1) +C$$

Remember that

• $$\displaystyle \operatorname{Li}_2 \left(\frac{1}{2} \right) = \frac{\pi^2}{12}-\frac{1}{2} \log^2\left(\frac{1}{2} \right)$$
• $$\displaystyle \operatorname{Li}_2(-1) = -\frac{\pi^2}{12}$$
• $$\displaystyle \log^2 \left( \frac{1}{2} \right) = \log^2 \left(2 \right)$$

Hence we deduce that $C=0$ , so

$$\displaystyle \operatorname{Li}_2 \left(\frac{z}{z-1} \right) = -\frac{1}{2} \log^2(1-z) - \operatorname{Li}_2(z)$$

which can be written as

$$\displaystyle \operatorname{Li}_2 \left(\frac{z}{z-1} \right)+\operatorname{Li}_2(z) = -\frac{1}{2} \log^2(1-z) \, \,\square$$

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Well, that doesn't end here , prove

$$\displaystyle \frac{1}{2} \operatorname{Li}_2 (z^2) = \operatorname{Li}_2 (z)+\operatorname{Li}_2 (-z)$$​

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PROOF

We will continue with the same manner

$$\displaystyle \frac{d}{dz}\operatorname{Li}_2 (z^2) = -\int^{z^2}_0 \frac{\log(1-t)}{t}\, dt$$

$$\displaystyle \frac{d}{dz}\operatorname{Li}_2 (z^2) = -2 \frac{\log(1-z^2)}{z}$$

$$\displaystyle \frac{1}{2} \frac{d}{dz} \operatorname{Li}_2 (z^2) = -\frac{\log(1-z)}{z}- \frac{\log(1+z)}{z}$$

$$\displaystyle \frac{1}{2}\operatorname{Li}_2 (z^2) = \operatorname{Li}_2(z)+\operatorname{Li}_2(-z) \,+ C$$

putting $$\displaystyle z=1$$ we get $$\displaystyle C=0$$ , hence the result

$$\displaystyle \frac{1}{2}\operatorname{Li}_2 (z^2) = \operatorname{Li}_2(z)+\operatorname{Li}_2(-z) \, \square$$

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Prove

$$\displaystyle \operatorname{Li}_2\left( \frac{\sqrt{5}-1}{2} \right) = \frac{\pi^2}{10} - \log^2 \left( \frac{\sqrt{5}-1}{2}\right)$$​

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PROOF

First we add the two functional equations of this section to obtain

$$\displaystyle \operatorname{Li}_2 \left( \frac{z}{z-1} \right) + \frac{1}{2} \operatorname{Li}_2 (z^2) - \operatorname{Li}_2(-z) = -\frac{1}{2} \log^2 (1-z)$$

Now let $$\displaystyle z = \frac{1-\sqrt{5}}{2}$$

• $$\displaystyle z^2 = \frac{1-2\sqrt{5}+5}{4}= \frac{3-\sqrt{5}}{2}$$

• $$\displaystyle \frac{z}{z-1} = \frac{\sqrt{5}-1}{1+\sqrt{5}} = \frac{3-\sqrt{5}}{2}$$

$$\displaystyle \frac{3}{2} \operatorname{Li}_2 \left( \frac{3-\sqrt{5}}{2} \right) - \operatorname{Li}_2 \left( \frac{\sqrt{5}-1}{2} \right) = -\frac{1}{2} \log^2 \left( \frac{\sqrt{5}+1}{2}\right)$$ ----(1)

We already established the following functional equation

$$\displaystyle \operatorname{Li}_2 (z) + \operatorname{Li}_2 (1-z) = \frac{\pi^2}{6} - \log(z) \log(1-z)$$

putting $$\displaystyle z = \frac{3-\sqrt{5}}{2}$$

$$\displaystyle \operatorname{Li}_2 \left( \frac{3-\sqrt{5}}{2}\right) + \operatorname{Li}_2 \left(\frac{\sqrt{5}-1}{2}\right) = \frac{\pi^2}{6} - \log\left(\frac{3-\sqrt{5}}{2} \right) \log \left(\frac{\sqrt{5}-1}{2}\right)$$ -----(2)

Solving (1) , (2) for $$\displaystyle \operatorname{Li}_2 \left(\frac{\sqrt{5}-1}{2}\right)$$ we get our result $$\displaystyle \square$$

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[HW] prove that $$\displaystyle \operatorname{Li}_2 \left( \frac{3-\sqrt{5}}{2} \right) = \frac{\pi^2}{15} - \frac{1}{4}\log^2 \left( \frac{3-\sqrt{5}}{2}\right)$$ .

To Be continued ...

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Integration lessons continued ....

4.5.2.Exercises

In this section I will give some exercises that involve Polylogarithms

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Find the following integral

$$\displaystyle I=\int^1_0 \, \frac{\log(1-x) \log(x) }{x} \, dx$$​

where $$\displaystyle \log$$ represents the natural logarithm

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Integrate by parts by Integrating

$$\displaystyle \frac{\log(1-x)}{x} = - \text{Li}_2(x)$$

Differentiating

$$\displaystyle \log(x) = \frac{1}{x}$$

So we have

$$\displaystyle I=-\log(x) \text{Li}_2(x) |^1_0+\int^1_0 \frac{\text{Li}_2(t)}{t}\, dt$$

$$\displaystyle I=\text{Li}_3(1) = \sum_{k\geq 1} \frac{1}{k^3} = \zeta(3)$$

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Evaluate the following integral

$$\displaystyle \int^x_0 \frac{\log^2(1-t)}{t}\, dt \,\,\,\,\, 0<x<1$$​

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Integrating by parts we get the following

$$\displaystyle \int^x_0 \frac{\log^2(1-t)}{t}\, dt = - \log(1-x) \text{Li}_2(x) -\int^x_0\frac{\text{Li}_2 (t)}{1-t} \, dt$$

Now we are left with the following integral

$$\displaystyle \int^x_0\frac{\text{Li}_2 (t)}{1-t} \, dt = \int^{1}_{1-x} \frac{\text{Li}_2 (1-t)}{t} \, dt$$

Using a result we obtained earlier

$$\displaystyle \int^{1}_{1-x} \frac{\frac{\pi^2}{6} -\text{Li}_2(t) - \log(1-t) \log(t)}{t} \, dt$$

$$\displaystyle -\frac{\pi^2}{6}\log(1-x)- \int^{1}_{1-x} \frac{\text{Li}_2(t)}{t}\,dt -\int^{1}_{1-x}\frac{ \log(1-t) \log(t)}{t} \, dt$$

The first integral

• $$\displaystyle \int^{1}_{1-x} \frac{\text{Li}_2(t)}{t}\,dt =\text{Li}_3(1)-\text{Li}_3(1-x)$$

The second integral is the same as the first exercise

• $$\displaystyle \int^{1}_{1-x}\frac{ \log(1-t) \log(t)}{t} =\text{Li}_3(1) +\log(1-x)\text{Li}_2(1-x)-\text{Li}_3(1-x)$$

Collecting the results together we obtain

$$\displaystyle \int^x_0\frac{\text{Li}_2 (t)}{1-t} \, dt =-\frac{\pi^2}{6}\log(1-x)-\text{Li}_2(1-x) \log(1-x)+2\, \text{Li}_3(1-x)- 2 \zeta(3)$$

$$\displaystyle \int^x_0 \frac{\log^2(1-t)}{t}\, dt = - \log(1-x) \text{Li}_2(x) +\frac{\pi^2}{6}\log(1-x)+\text{Li}_2(1-x) \log(1-x)-2\, \text{Li}_3(1-x)+ 2 \zeta(3)$$

This section concludes the PolyLogarithm discussion , I will start the Hypergeometric function in the next thread .

Last edited:

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Integration lessons continued ....

4.Integration using special functions (continued)

4.6. Ordinary Hypergeometric function

Ordinary or Gauss hypergoemtric function is a nice generalization of power expansion or series representations of many functions . Before we start with the definition we will explain some notations .

Defined the raising factorial as follows

$$(z)_n = \begin{cases} 1 & n = 0 \\ {} \\ \frac{\Gamma(z+n)}{\Gamma(z)}=z(z+1) \cdots (z+n-1) & n > 0. \end{cases}$$

We can easily prove that $(1)_n = 1\cdot 2\cdot 3 \cdots n = n!$ and $(2)_n = 2\cdot 3\cdot 4 \cdots (n+1)=(n+1)!$

Using this definition we defnie the Gauss hypergoemtric function as follows

$${}_2F_1(a,b;c;z) = \sum_{n=0}^\infty \frac{(a)_n (b)_n}{(c)_n} \frac{z^n}{n!}.$$

4.6.1 Famous functions using Hypergemtric representations

We can represent the famous functions using the hypergeomtric function

$$\displaystyle z{}_2F_1 (1,1;2;-z)= z\sum_{n\geq 0}\frac{(1)_n (1)_n}{(2)_n}\frac{(-z)^n}{n!}=\sum_{n\geq 0}\, (-1)^n\frac{ \, n!}{(n+1)!}z^{n+1}=\sum_{n\geq 0}(-1)^{n}\frac{z^{n+1}}{n+1}=\log(1+z)$$

$$\displaystyle _2F_1(a,1;1;z)= \sum_{n\geq 0}\frac{(a)_n \, (1)_n}{(1)_n}\frac{z^n}{n!}=\sum_{n\geq 0}\frac{(a)_n }{n!}z^n=(1-z)^{-a}$$

$$\displaystyle z \, _2F_1\left(\tfrac{1}{2}, \tfrac{1}{2}; \tfrac{3}{2};z^2\right)=\sum_{n\geq 0}\frac{\left( \frac{1}{2}\right)_n\, \left( \frac{1}{2}\right)_n}{\left( \frac{3}{2}\right)_n}\frac{z^{2n+1}}{n!}= \sum_{n \geq 0}\frac{ \left( \frac{1}{2}\right)_n}{2n+1}\frac{z^{2n+1}}{n!}= \sum_{n \geq 0}\frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2^n \, n!\, (2n+1)}z^{2n+1}$$

Which can be written as

$$\displaystyle z \, _2F_1\left(\tfrac{1}{2}, \tfrac{1}{2}; \tfrac{3}{2};z^2\right)= \sum_{n \geq 0}\frac{(2n)!}{4^n \, (n!)^2\, (2n+1)}z^{2n+1}=\sum_{n \geq 0}\frac{{2n \choose n}}{4^n \, (2n+1)}z^{2n+1}= \arcsin(z)$$

To be continued ...

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Re: Integration lessons continued ....

4.Integration using special functions (continued)

4.6.1 Famous functions using Hypergemtric representations(continued)

In the previous section we discussed how to convert a hypergeometric representation into an elementary function or so called (Taylor expansion ) . Now we consider converting the Taylor expansion into the equivalent hypergeomtric representation .

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Suppose the following

$$\displaystyle _2 F_1 ( a, b; c; z ) = \sum_{k \geq 0} t_k \,\,\, , \,\,\, t_0 =1 \,\,\,\,\,\,\,\,\, (1)$$​

Now consider the ratio

$$\displaystyle \frac{t_{k+1}}{t_k} = \frac{(k+a) (k+b) }{(k+c) (k+1)} z \,\,\,\,\,\,\,\,\, (2)$$​

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Using this definition if we have a power series equivalent to (1) , we can easily find the terms $a,b,c$ and $z$.

Examples

Find the hypergeometric representation of the following

• $$\displaystyle f(z) = e^ z$$

We can easily find the Taylor expansion as

$$\displaystyle f(z) = e^ z= \sum_{k\geq 0} \frac{z^k}{k!}$$

Hence we have

$$\displaystyle \frac{t_{k+1} }{t_k} = \frac{z}{k+1}$$

Comparing to (2) we conclude

$$\displaystyle e^z ={} _2 F_1 (-,-;-;z)$$

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• $$\displaystyle f(z) = \cos(z) = \sum_{k\geq 0} \frac{(-1)^k z^{2k}}{(2k)!}$$

By the same approach

$$\displaystyle \frac{t_{k+1}}{t_k} = -\frac{1}{(2k+2)(2k+1)}z= \frac{1}{(k+1)\left(k+\frac{1}{2} \right)}\frac{-z^2}{4}$$

Hence we have by comparing to (2)

$$\displaystyle \cos(z)={}_2 F_1 \left(-,- ; \frac{1}{2}; \frac{-z^2}{4} \right)$$

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• $$\displaystyle f(z) = (1-z)^{-a}= \sum_{k\geq 0}\frac{(a)_k}{k!}z^k$$

$$\displaystyle \frac{t_{k+1}}{t_k} =\frac{(k+a)}{(k+1)}z= \frac{(k+a)(k+1)}{(k+1)(k+1)}z$$

Hence we have

$$\displaystyle (1-z)^{-a} ={} _2F_1(a,1;1;z)$$

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[HW] Find the hypergeoemtric representations of the following functions

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• $$\displaystyle \arcsin(z)$$
• $$\displaystyle \sin(z)$$

To be continued ...

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
integration lessons continued ...

4.Integration using special functions (continued)

4.6.2 Integral representation

In this section we give some transformations that will be useful to evaluate some integrals . We start this section by proving the integral representation .

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$$\displaystyle \beta(c-b,b) \, _2F_1(a,b;c;z)=\int_0^1 \frac{t^{b-1}(1-t)^{c-b-1}}{(1-tz)^a}\, dt$$​

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Proof

Start by the RHS

$$\displaystyle \int_0^1 t^{b-1}(1-t)^{c-b-1} \, (1-tz)^{-a}\, dt$$

Using the expansion of $$\displaystyle (1-tz)^{-a}$$ we have

$$\displaystyle \int_0^1 t^{b-1}(1-t)^{c-b-1} \sum_{k\geq 0}\frac{(a)_k}{k!}\, (tz)^k$$

Interchanging the integral with the series

$$\displaystyle \sum_{k\geq 0}\frac{(a)_k}{k!}\, z^k \, \int_0^1 t^{k+b-1}(1-t)^{c-b-1}\, dt$$

Recalling the beta function we have

$$\displaystyle \sum_{n\geq 0}\frac{(a)_k \Gamma(k+b) \Gamma(c-b)}{ \Gamma(k+c)}\, \frac{z^k}{k!}$$

Using the identity that

$$\displaystyle \beta(c-b,b) = \frac{\Gamma(b)\Gamma(c-b)}{\Gamma(c)}$$

$$\displaystyle \beta(c-b,b)\sum_{k\geq 0}\frac{(a)_k \Gamma(k+b) \Gamma(c)}{\Gamma(b) \Gamma(k+c)}\, \frac{z^k}{k!}= \beta(c-b, b) \int^1_0 \frac{(a)_k (b)_k}{(c)_k} \frac{z^k}{k!}$$

As required $$\displaystyle \square$$

We can consider the case $$\displaystyle z=1$$ so we have

$$\displaystyle _2F_1(a,b;c;1)= \frac{1}{\beta(c-b,b) \,}\int_0^1 \frac{t^{b-1}(1-t)^{c-b-1}}{(1-t)^a}\, dt$$

$$\displaystyle _2F_1(a,b;c;1)=\frac{1}{\beta(c-b,b) }\int_0^1 t^{b-1}(1-t)^{c-b-a-1}\, dt =\frac{\beta(b,c-b-a)}{\beta(c-b,b) }=\frac{\Gamma(c-b-a)\Gamma(c)}{\Gamma(c-a)\Gamma(c-b)}$$

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Some transformations

Start by transforming $$\displaystyle t \to 1-t$$

$$\displaystyle \beta(c-b,b) \, _2F_1(a,b;c;z)=\int_0^1 t^{c-b-1}(1-t)^{b-1}(1-(1-t)z)^{-a}\, dt =\int_0^1 t^{c-b-1}(1-t)^{b-1}(1-z+tz)^{-a}\, dt$$

Taking $$\displaystyle 1-z$$ as a common factor

$$\displaystyle \beta(c-b,b) \, _2F_1(a,b;c;z)=\frac{1}{(1-z)^a}\int_0^1 t^{c-b-1}(1-t)^{b-1}\left( 1-\frac{z}{z-1} \, t\right)^{-a}\, dt$$

Which can be written as

$$\displaystyle \beta(c-b,b) \, _2F_1(a,b;c;z)=\frac{\beta(c-b,b) {}_2F_1(a,c-b;c;z)}{(1-z)^a}$$

Evidently we have the interesting form

$$\displaystyle _2F_1(a,b;c;z)=(1-z)^{-a}{} _2F_1(a,c-b;c;z)$$

we will use this transformation later to find other special cases for $$\displaystyle z$$.

To be continued ...

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#### ZaidAlyafey

##### Well-known member
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Integration lessons (continued)

4.Integration using special functions (continued)

4.7 Error Function

The error function is an interesting function that has many applications in probability, statistics and physics.

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Definition

$$\displaystyle \text{erf}(x)=\frac{2}{\sqrt{\pi}}\int^x_0 e^{-t^2}\,dt$$

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Complementary error function

$$\displaystyle \text{erfc}(x)=1-\text{erf}(x)$$

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Imaginary error function

$$\displaystyle \text{erfi}(x) = -i \text{erf}(ix)$$

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Properties :

The error function is odd

$$\text{erf}(-x) =\frac{2}{\sqrt{\pi}}\int^{-x}_0 e^{-t^2}\,dt=-\frac{2}{\sqrt{\pi}}\int^{x}_0 e^{-t^2}\,dt=-\text{erf}(x)$$

Real part and imaginary parts

$$\Re \,\text{erf}(z)=\frac{\text{erf}(z)+\text{erf}(\bar{z})}{2}$$

$$\Im \,\text{erf}(z)=\frac{\text{erf}(z)-\text{erf}(\bar{z})}{2i}$$

Using complex variables it can be done using $\text{erf}(\bar{z})=\overline{\text{erf}(z)}\,\square$

Exercises

$$I = \int^x_0 e^{t^2}\,dt$$

The function has no elementary anti-derivative so we represent it using the error function.

So we need to find a function $f$ such that $\frac{d}{dz}f=e^{z^2}$

Consider the imaginary error function

$$\text{erfi}(x)=-i\text{erf}(ix) = -i\frac{2}{\sqrt{\pi}}\int^{ix}_0e^{-t^2}\,dt$$

By differentiating both sides we have

$$\frac{d}{dx}\text{erfi}(x) = \frac{2}{\sqrt{\pi}}e^{x^2}$$

Hence we have

$$e^{x^2}=\frac{d}{dx}\frac{\sqrt{\pi}}{2}\text{erfi}(x)$$

By integrating both sides we have

$$\int e^{x^2}\,dx=\frac{\sqrt{\pi}}{2}\text{erfi}(x)+C$$

4.7.1 Relation to other functions

Hypergeomtric function

$$\text{erf}(x) = \frac{2x}{\sqrt{\pi}} {}_1F_1\left( \frac{1}{2},\frac{3}{2},-x^2\right)$$

By expanding the hypergeometirc function

$$\frac{2x}{\sqrt{\pi}} {}_1F_1\left( \frac{1}{2},\frac{3}{2},-x^2\right)=\frac{2x}{\sqrt{\pi}}\sum_{k\geq 0}\frac{\left(\frac{1}{2} \right)_k}{\left(\frac{3}{2} \right)_k}\frac{(-x^2)^k}{k!}$$

$$\frac{2x}{\sqrt{\pi}} {}_1F_1\left( \frac{1}{2},\frac{3}{2},-x^2\right)=\frac{x}{\sqrt{\pi}} \sum_{k\geq 0}\frac{(-x^2)^k}{(\frac{1}{2}+k)k!}$$

Notice that this is actually the expanded error function

$$\frac{2}{\sqrt{\pi}}\int^x_0 e^{-t^2}\,dt =\frac{2}{\sqrt{\pi}}\int^x_0\sum_{k\geq 0}\frac{(-x^2)^k}{k!}\,dx = \frac{2}{\sqrt{\pi}}\sum_{k\geq 0}\frac{(-x)^{2k+1}}{(2k+1)k!}$$

The function can be written using the incomplete Gamma function

Incomplete Gamma function

$$\text{erf}(x)=1-\frac{\Gamma\left(\frac{1}{2},x^2 \right)}{\sqrt{\pi}}$$

Exercises

$$\int^\infty_0 \text{erfc}(x)\,dx=\frac{1}{\sqrt{\pi}}$$

Using the complementary error function

$$\int^\infty_0 (1-\text{erf}(x))\,dx$$

Integration by parts we have

$$I= x(1-\text{erf}(x))\left. \right]^\infty_0+\frac{2}{\sqrt{\pi}}\int^\infty_0xe^{-x^2}\,dx$$

Now we compute $\text{erf}(\infty)$

$$\text{erf}(\infty) =\frac{2}{\sqrt{\pi}}\int^\infty_0 e^{-t^2}\,dt = \frac{2}{\sqrt{\pi}}\times\frac{\sqrt{\pi}}{2}=1$$

Hence we have the following

$$I=\frac{2}{\sqrt{\pi}}\int^\infty_0xe^{-x^2}\,dx = \frac{1}{\sqrt{\pi}}$$

#### ZaidAlyafey

##### Well-known member
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Re: Integration lessons (continued)

4.7.2 Exercises

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Exercise 1

$$\displaystyle \int^\infty_0 \text{erfc}^2(x)\,dx = \frac{2-\sqrt{2}}{\sqrt{\pi}}$$​

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Proof

Integration by parts

$$I = x\text{erfc}^2(x)\left. \right]^\infty_0 -2\int^\infty_0 x\text{erfc}'(x)\text{erfc}(x)\,dx$$

The first integral goes to 0

$$I = -2\int^\infty_0 x\,\text{erfc}'(x)\text{erfc}(x)\,dx$$

The derivative of the complementary error function

$$\text{erfc}'(x) = (1-\text{erf}(x))'= -\frac{2}{\sqrt{\pi}}e^{-x^2}$$

$$I= \frac{4}{\sqrt{\pi}}\int^\infty_0xe^{-x^2}\text{erfc}(x)\,dx$$

Integration by parts again we have

$$I=\frac{2}{\sqrt{\pi}}e^{-x^2}\text{erfc}(x)\left. \right]^\infty_0-\frac{4}{\pi}\int^\infty_0e^{-2t^2}\,dt$$

At infinity the integral goes to 0 and at 0 we have

$$\frac{2}{\sqrt{\pi}}e^{-x^2}\text{erfc}(x)_{x=0}=\frac{2}{\sqrt{\pi}}(1-\text{erf}(0))=\frac{2}{\sqrt{\pi}}$$

$$\int^\infty_0e^{-2t^2}\,dt = \frac{\sqrt{\pi}}{2\sqrt{2}}$$

Collecting the results together we have

$$I = \frac{2}{\sqrt{\pi}}-\frac{\sqrt{\pi}}{2\sqrt{2}} \times \frac{4}{\pi}=\frac{2-\sqrt{2}}{\sqrt{\pi}}$$

Exercise 2

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$$\displaystyle \int^\infty_0 \sin(x^2) \text{erfc}(x) \,dx =\frac{\pi-2 \coth^{-1}\sqrt{2}}{4 \sqrt{2\pi}}$$​

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Proof

Using the substitution $x=\sqrt{t}$

$$\frac{1}{2}\int^\infty_0 \sin(t) t^{-\frac{1}{2}}\text{erfc}(\sqrt{t}) \,dx$$

Consider the function

$$I(a) = \frac{1}{2}\int^\infty_0 \sin(t) t^{-\frac{1}{2}}\text{erfc}(a\sqrt{t}) \,dt$$

Differentiating with respect to $a$ we have

$$I'(a) = \frac{-1}{\sqrt{\pi}}\int^\infty_0 \sin(t) e^{-a^2t}\,dt=\frac{-1}{\sqrt{\pi}}\cdot \frac{1}{a^4+1}$$

Now integrating with respect to $a$

$$I(a) = \frac{-1}{\sqrt{\pi}}\int^a_0 \frac{dx}{x^4+1}+C$$

To evaluate the constant we take $a \to \infty$

$$I(\infty) = \frac{-1}{\sqrt{\pi}}\int^\infty_0\frac{dx}{x^4+1}+C$$

The function has an anti-derivative and the value is

$$\frac{-1}{\sqrt{\pi}} \int^\infty_0\frac{dx}{x^4+1}=-\frac{\sqrt{\pi}}{2\sqrt{2}}$$

and knowing that

$$\text{erfc}(\infty)=0 \,\,\, \implies \,\,\,\, C =\frac{\sqrt{\pi}}{2\sqrt{2}}$$

Finally we get

$$I(a) = \frac{-1}{\sqrt{\pi}}\int^a_0 \frac{dx}{x^4+1}+\frac{\sqrt{\pi}}{2\sqrt{2}}$$

Plugging $a=1$ we have our integral

$$I(1) = \int^\infty_0 \sin(x^2) \text{erfc}(x) \,dx=\frac{\sqrt{\pi}}{2\sqrt{2}}-\frac{1}{\sqrt{\pi}}\int^1_0 \frac{dx}{x^4+1}$$

Also knowing that

$$\frac{1}{\sqrt{\pi}}\int^1_0 \frac{dx}{x^4+1}=\frac{\pi+2 \coth^{-1}\sqrt{2}}{4 \sqrt{2\pi}}$$

$$\int^\infty_0 \sin(x^2) \text{erfc}(x) \,dx=\frac{\pi-2 \coth^{-1}\sqrt{2}}{4 \sqrt{2\pi}}$$

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[HW]

Can you find closed forms for

$$\displaystyle \int^\infty_0 \text{erfc}^3(x)\, dx = ?$$

$$\displaystyle \int^\infty_0 \text{erfc}^4(x)\, dx = ?$$

$$\displaystyle \int^\infty_0 \text{erfc}^n(x)\, dx = ?$$

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#### ZaidAlyafey

##### Well-known member
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Re: Integration lessons (continued)

4.Integration using special functions (continued)

4.8 Exponential integral function

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Definition

$$\displaystyle E(x)=\int^\infty_x\frac{e^{-t}}{t}\,dt$$​

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4.8.1 Exercises

Prove that

$$\lim_{x \to 0}\left[\log(x)+E(x) \right]=-\gamma$$

Integration by parts for $E(x)$

$$E(x)=e^{-t}\log(t) \left.\right]^\infty_x+\int^\infty_x\log(x)e^{-t}\,dt$$

$$E(x)=-e^{-x}\log(x)+\int^\infty_x\log(x)e^{-t}\,dt$$

$$\lim_{x \to 0}\left[\log(x)+E(x) \right]=\lim_{x \to 0}\left(\log(x)-e^{-x}\log(x)\right)+\int^\infty_0\log(x)e^{-t}\,dt$$

The first limit goes to 0

$$\lim_{x \to 0}\left[\log(x)+E(x) \right]=\int^\infty_0\log(x)e^{-t}\,dt=\psi(1)=-\gamma$$

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Prove that

$$\int^\infty_0 x^{p-1} E(ax)\,dx =\frac{\Gamma(p)}{pa^{p}} \,\,\, p>0$$

Integrating by parts we have

$$\int^\infty_0 x^{p-1} E(ax)\,dx=\frac{1}{p}x^pE(ax) \left. \right]^\infty_0+\frac{1}{ap} \int^\infty_0 x^{p-1}e^{-ax}\,dx$$

The first limit goes to 0

$$I=\frac{1}{ap} \int^\infty_0 x^{p-1}e^{-ax}\,dx= \frac{1}{pa^p }\int^\infty_0 x^{p-1}e^{-x}\,dx=\frac{\Gamma(p)}{p \, a^p}$$

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Prove the more general case

$$\int^\infty_0 x^{p-1}e^{ax}E(ax) \, dx = \frac{\pi}{\sin(a\pi)}\cdot \frac{\Gamma(p)}{a^p}$$

Switch to the integral representation

$$\int^\infty_0 x^{p-1}e^{ax}\int^\infty_{ax}\frac{e^{-t}}{t}\,dt \, dx$$

Use the substitution $t=ax \,y$

$$\int^\infty_0\int^\infty_{1} x^{p-1}e^{ax}\frac{e^{-ax\,y}}{y}\,dy \, dx$$

By switching the two integrals

$$\int^\infty_1\frac{1}{y}\int^\infty_{0} x^{p-1}e^{-ax(y-1)} dx\,dy$$

By the Laplace identities

$$\frac{\Gamma(p)}{a^p}\int^\infty_1 \frac{1}{y(y-1)^p} \, dy$$

Now let $y=1/x$

$$\frac{\Gamma(p)}{a^p}\int^1_0 x^{p-1}(1-x)^{-p} \, dx$$

Using the the reflection formula for the Gamma function

$$\frac{\Gamma(p)}{a^p}\int^1_0 x^{p-1}(1-x)^{-p} \, dx = \frac{\pi}{\sin(a\pi)}\cdot \frac{\Gamma(p)}{a^p}$$

#### ZaidAlyafey

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Re: Integration lessons (continued)

7.8.1 Exercises (continued)

Prove that

$$\int^\infty_0 e^{z}E^2(z) \, dz = \frac{\pi^2}{6}$$

Using the integral representation

$$E^2(z) = \int^\infty_1\int^\infty_1\frac{e^{-x z}e^{-y z}}{x y}\,dx \,d y$$

$$\int^\infty_0 e^{z}E^2(z) \, dz = \int^\infty_0 \int^\infty_1 \,\int^\infty_1\frac{e^{-z(x+y-1)}}{x y}\,dx \,d y \, dz$$

Switching the integration

$$\int^\infty_1 \frac{1}{y} \int^\infty_1 \frac{1}{x}\,\int^\infty_0e^{-z(x+y-1)}\,dz\,dx \,d y$$

$$\int^\infty_1 \frac{1}{y} \int^\infty_1\frac{1}{x(x+y-1)}\,dx \, dy$$

The inner integral is an elmetnary integral

$$\int^\infty_1\frac{1}{x(x+y-1)}\,dx = -\frac{\log(y)}{1-y}$$

$$\int^\infty_1 \frac{\log(y)}{y(y-1)} \,dy$$

Now use the substitution $y=1/x$

$$-\int^1_0 \frac{\log(x)}{(1-x)} \,dx=-\int^1_0\frac{\log(1-x)}{x}\,dx = \text{Li}_2(1)=\frac{\pi^2}{6}$$

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Prove that

$$\int^\infty_0 z^{p-1} E^2(z)\,dz=\frac{2\Gamma(p)}{p^2}\, _2F_1(p,p;p+1;-1)$$

Consider the general case

$$F(p)=\int^\infty_0 z^{p-1} E^2(z)\,dz$$

Integrating by parts

$$F(p)=\frac{2}{p} \int^\infty_0 z^{p-1}e^{-z}E(z)\,dz$$

Write the integral representation

$$\frac{2}{p} \int^\infty_0 z^{p-1}e^{-z}\int^\infty_1 \frac{e^{-zt}}{t},dtdz$$

$$\frac{2}{p} \int^\infty_1\frac{1}{t}\int^\infty_0 z^{p-1}e^{-z(1+t)} \,dz\,dt$$

Take the inner integral

$$\frac{2\Gamma(p)}{p} \int^\infty_1\frac{dt}{t(1+t)^p}$$

Use the substitution $t=1/x$

$$\frac{2\Gamma(p)}{p} \int^1_0\frac{x^{p-1}}{(1+x)^p}\,dx$$

Using the Hypergeomtirc identity

$$\beta(c-b,b) \, _2F_1(a,b;c;z)=\int_0^1 \frac{x^{b-1}(1-x)^{c-b-1}}{(1-xz)^a}\, dx$$

put $c=p+1;b=p;a=p;z=-1$

$$\beta(1,p) \, _2F_1(p,p;p+1;-1)=\int_0^1 \frac{x^{p-1}}{(1+x)^p}\, dx$$

Hence the result

$$\int^\infty_0 z^{p-1} E^2(z)\,dz=\frac{2\Gamma(p)}{p^2}\, _2F_1(p,p;p+1;-1)$$

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[HW]

Find a general formula for

$$\int^\infty _0 x^n E^2(x) \, dx = ? \,\,\,\,\,\, n \in \mathbb{N}$$

#### ZaidAlyafey

##### Well-known member
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Re: Integration lessons (continued)

4.Integration lessons (continued)

4.9.Complete elliptic integrals

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Complete elliptic of first kind

$$\displaystyle K(k)= \int^{\frac{\pi}{2}}_0\frac{d\theta}{\sqrt{1-k^2\sin^2 \theta }}=\int^1_0\frac{dx}{\sqrt{1-x^2}\sqrt{1-k^2x^2}}$$​

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Complete elliptic of second kind

$$\displaystyle E(k)= \int^{\frac{\pi}{2}}_0\sqrt{1-k^2\sin^2 \theta }\, d\theta =\int^1_0\frac{\sqrt{1-k^2x^2}}{\sqrt{1-x^2}}dx$$​

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Some special values

$$K(i) = \frac{1}{4\sqrt{2\pi}}\Gamma^2\left( \frac{1}{4}\right)$$

By definition we have

$$K(i)=\int^1_0\frac{dx}{\sqrt{1-x^2}\sqrt{1+x^2}}=\int^1_0 \frac{dx}{\sqrt{1-x^4}}$$

Let $x=\sqrt[4]{t}$ we have $dx=\frac{1}{4}t^{\frac{-3}{4}}\,dt$

$$K(i)=\frac{1}{4}\int^1_0t^{\frac{-3}{4}}(1-t)^{\frac{-1}{2}}\,dt$$

By beta function

$$K(i) = \frac{\Gamma\left( \frac{1}{4}\right)\Gamma\left( \frac{1}{2}\right)}{4\Gamma\left( \frac{3}{4}\right)}$$

By reflection formula

$$\Gamma\left( \frac{1}{4}\right)\Gamma\left( \frac{3}{4}\right)=\pi \csc\left( \frac{\pi}{4}\right)=\pi \sqrt{2}$$

$$K(i) = \frac{\Gamma^2\left( \frac{1}{4}\right)\Gamma\left( \frac{1}{2}\right)}{4\pi \sqrt{2}}=\frac{\Gamma^2\left( \frac{1}{4}\right)}{4\sqrt{2\pi}}$$

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$$E(i)=\frac{\Gamma^2\left( \frac{1}{4}\right)}{4\sqrt{2\pi}}+\frac{\Gamma^2\left(\frac{3}{4} \right)}{\sqrt{2\pi}}$$

By definition we have

$$E(i)=\int^1_0\frac{\sqrt{1+x^2}}{\sqrt{1-x^2}}dx$$

Separating the two integrals

$$E(i)=\int^1_0\frac{1+x^2}{\sqrt{1-x^4}}dx=\int^1_0\frac{1}{\sqrt{1-x^4}}dx+\int^1_0\frac{x^2}{\sqrt{1-x^4}}dx$$

The first integral is $K(i)$ for the second integral use $x=\sqrt[4]{t}$

$$\frac{1}{4}\int^1_0t^{\frac{3}{4}-1}(1-t)^{-\frac{1}{2}}\,dt=\frac{\Gamma\left( \frac{3}{4}\right)\Gamma\left( \frac{1}{2}\right)}{4\Gamma\left( \frac{5}{4}\right)}=\frac{\Gamma^2\left(\frac{3}{4} \right)}{\sqrt{2\pi}}$$

Hence we have

$$E(i) = K(i) +\frac{\Gamma^2\left(\frac{3}{4} \right)}{\sqrt{2\pi}} = \frac{\Gamma^2\left( \frac{1}{4}\right)}{4\sqrt{2\pi}}+\frac{\Gamma^2\left(\frac{3}{4} \right)}{\sqrt{2\pi}}$$

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4.9.2.Some integrals involving elliptic integrals

$$\int^1_0 K(k) \, dk = 2G \,\,\,\,\, ;\,G=\text{Catalan's constant }$$

$$I=\int^1_0 \int^1_0\frac{1}{\sqrt{1-x^2}\sqrt{1-k^2x^2}} \,dx\, dk$$

Switching the two integrals

$$I=\int^1_0 \frac{1}{\sqrt{1-x^2}}\int^1_0\frac{1}{\sqrt{1-k^2x^2}} \,dk\, dx$$

$$I=\int^1_0 \frac{\arcsin x}{x\sqrt{1-x^2}} dx$$

Now let $\arcsin x = t$ hence we have $x=\sin t$

$$I=\int^{\frac{\pi}{2}}_0 \frac{t}{\sin \, t} dt$$

The previous integral is a representation of the constant

$$G=\frac{I}{2} \,\,\, \implies \,\,\, I=2G$$

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[HW]

Find the Values

$$E(0) \,\,\, , \,\,\,\, K \left( \frac{1}{\sqrt{2}}\right)$$

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Re: Integration lessons (continued)

4.9.3.Properties of elliptic integrals

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$$\displaystyle \tag{1}K(\sqrt{k}) = \frac{1}{\sqrt{1-k}}K\left(\sqrt{\frac{k}{k-1} }\right)\,\,;\,k\notin [1,\infty)$$​

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Starting by the integral representation

$$K(k)=\int^1_0 \frac{dx}{\sqrt{1-x^2}\sqrt{1-k^2x^2}}$$

By using that $x=\sqrt{1-y^2}$

$$\int^1_0\frac{y\,dy}{\sqrt{1-y^2}\sqrt{1-(1-y^2)}\sqrt{1-k^2(1-y^2)}}$$

$$\int^1_0\frac{ dy}{\sqrt{1-y^2}\sqrt{1-k^2+k^2y^2}}=\int^1_0\frac{ dy}{\sqrt{1-k^2}\sqrt{1-y^2}\sqrt{1-\frac{k^2}{k^2-1}y^2}}$$

$$K(k)=\frac{1}{\sqrt{1-k^2}}K\left( \sqrt{\frac{k^2}{k^2-1}}\right)$$

which is equivalent to the property by taking the root

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$$\displaystyle \tag{2} E(\sqrt{k}) = \sqrt{1-k}\,E\left(\sqrt{\frac{k}{k-1}} \right)\,\,;\,k\notin [1,\infty)$$​

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Staring by the integral representation

$$E(\sqrt{k})=\int^1_0 \frac{\sqrt{1-kx^2}}{\sqrt{1-x^2}}\,dx$$

By using that $x=\sqrt{1-y^2}$

$$E(\sqrt{k})=\sqrt{1-k}\int^1_0 \frac{\sqrt{1-\frac{k}{k-1}y}}{\sqrt{1-y^2}}\,dx=\sqrt{1-k}\,E\left(\sqrt{\frac{k}{k-1} }\right)$$

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Using the above formulas we can have some values

For the value $k=-1$

$$K(i) = \frac{1}{\sqrt{2}}K\left(\frac{1}{\sqrt{2}} \right)$$

$$K\left(\frac{1}{\sqrt{2}} \right) = \sqrt{2}K(i)= \frac{1}{4\sqrt{\pi}}\Gamma^2\left( \frac{1}{4}\right)$$

Similarly we have

$$E\left ( \frac{1}{\sqrt{2}}\right)= \frac{\Gamma^2\left( \frac{1}{4}\right)}{8\sqrt{\pi}}+\frac{\Gamma^2\left(\frac{3}{4} \right)}{2\sqrt{\pi}}$$

Hence using $k\to -k$

$$K(\sqrt{k}\,i) = \frac{1}{\sqrt{1+k}}K\left(\frac{\sqrt{k}}{\sqrt{1+k}}\right)\,\,$$

$$E(\sqrt{k}\,i) = \sqrt{1+k}\,E\left(\frac{\sqrt{k}}{\sqrt{1+k}}\right)\,\,$$

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#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Re: Integration lessons (continued)

4.9.4.Elliptic integrals as a Hypergeometric function

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Definition

$$\displaystyle K(k)=\frac{\pi}{2}{}_2F_1\left(\frac{1}{2},\frac{1}{2},1,k^2 \right)$$​

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Using the integral representation of the hypergeometric function

$$\beta(c-b,b) \, _2F_1(a,b,c,z)=\int_0^1 \frac{t^{b-1}(1-t)^{c-b-1}}{(1-tz)^a}\, dt$$

Now use the substitution $t=x^2$ and $z=k^2$

$$\beta(c-b,b) \, _2F_1(a,b,c,k^2)=2\int_0^1 \frac{x^{2b-1}(1-x^2)^{c-b-1}}{(1-k^2x^2)^a}\, dx$$

Put $a=\frac{1}{2}\,;$ $b=\frac{1}{2}$ and $c=1$

$$\int_0^1 \frac{dx}{\sqrt{1-x^2}\sqrt{1-k^2x^2}}\, =\frac{1}{2}\beta(1/2,1/2) \, _2F_1\left(\frac{1}{2},\frac{1}{2},1,k^2 \right)$$

By the beta function we have

$$\frac{1}{2}\beta(1/2,1/2)=\frac{1}{2}\Gamma^2\left(\frac{1}{2} \right)=\frac{\pi}{2}$$

Hence the result

$$K(k) =\frac{\pi}{2}\, _2F_1\left(\frac{1}{2},\frac{1}{2},1,k^2 \right)$$

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By the same approach we have

$$\displaystyle E(k) =\frac{\pi}{2}\, _2F_1\left(\frac{1}{2},-\frac{1}{2},1,k^2 \right)$$​

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A relation using the Quadratic transformation

$$_2F_1\left( a,b,2b,\frac{4z}{(1+z)^2} \right)=(1+z)^{2a}\,_2F_1\left(a,a-b+\frac{1}{2},b+\frac{1}{2},z^2 \right).$$

Hence we can deduce by putting $a=b=1/2$

$$K\left( \frac{2\sqrt{k}}{1+k} \right)=(1+k)K(k)$$

Or we have

$$K(k)=\frac{1}{k+1}K\left( \frac{2\sqrt{k}}{1+k} \right)$$

Hence we have for $k=\frac{1}{\sqrt{2}}$

$$K\left(2 \sqrt{-4+3 \sqrt{2}} \right)=\frac{1+\sqrt{2}}{\sqrt{2}}K\left(\frac{1}{\sqrt{2}} \right)= \frac{1+\sqrt{2}}{4\sqrt{2\pi}}\Gamma^2\left( \frac{1}{4}\right)$$

For the elliptic integral of second kind using the hypergeomtric representation with $a=\frac{-1}{2}$ and $b=\frac{1}{2}$

$$_2F_1\left( -1/2,1/2,1,\frac{4z}{(1+z)^2} \right)=(1+z)^{-1}{}_2F_1\left(-1/2,-1/2,1,z^2 \right)$$

The later hypergeometric series can be written in terms of elliptic integrals using some general contiguity relations

$${}_2F_1\left(-1/2,-1/2,1,z^2 \right)=\frac{2}{\pi}\left( 2 E(k)+(k^2-1)K(k)\right)$$

So we have

$$2 E(k)+(k^2-1)K(k)=(k+1)E\left(\frac{2\sqrt{k}}{1+k} \right)$$

For $k=\frac{1}{\sqrt{2}}$

$$E\left(2 \sqrt{-4+3 \sqrt{2}} \right)=\frac{\sqrt{2}}{1+\sqrt{2}}\left[\frac{\Gamma^2\left( \frac{1}{4}\right)}{8\sqrt{\pi}}+\frac{\Gamma^2\left(\frac{3}{4} \right)}{\sqrt{\pi}} \right]$$

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Values for complex arguments

Start by the following

$$K\left(\frac{2\sqrt{k}}{1+k} \right)=\int^1_0 \frac{1}{\sqrt{1-x^2}\sqrt{1-\frac{4k}{(1+k)^2}x^2}}\,dx$$

By some simplifications we have

$$E\left(\frac{2\sqrt{k}}{1+k} \right)=(1+k)\int^1_0 \frac{1}{\sqrt{1-x^2}\sqrt{(1+k)^2-4k \, x^2}}\,dx$$

Use $x=\sqrt{1-y^2}$

$$\int^1_0 \frac{1+k}{\sqrt{1-y^2}\sqrt{(1+k)^2-4k\,(1-y^2)}}\,dy=\frac{1+k}{1-k}\int^1_0 \frac{1}{\sqrt{1-y^2}\sqrt{1+\frac{4k}{(1-k)^2}y^2}}\,dy$$

Hence we have

$$K\left(\frac{2\sqrt{k}}{1+k} \right)=\frac{1+k}{1-k}K\left(\frac{2\sqrt{-k}}{1-k} \right)$$

Similarly we have

$$E\left(\frac{2\sqrt{k}}{1+k} \right)=\frac{1-k}{1+k}E\left(\frac{2\sqrt{-k}}{1-k} \right)$$

Using these formulas and the results we got earlier we have for $x=1/\sqrt{2}$
$$K\left(2 \sqrt{-4-3 \sqrt{2}} \right)=\frac{\pi\,\sqrt{\pi}}{4} \cdot \frac{2-\sqrt{2}}{\Gamma^2\left( \frac{3}{4}\right)}$$

$$E\left(2 \sqrt{-4-3 \sqrt{2}} \right)=\frac{(2+\sqrt{2})\left(\pi^2+4 \Gamma^4\left( \frac{3}{4}\right)\right)}{4\sqrt{\pi}\Gamma^2\left( \frac{3}{4}\right)}$$

Last edited:

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Re: Integration lessons (continued)

4.9.5.Differentiation of elliptic integrals

Note : We should remove the variable $k$ and denote elliptic integrals $E$ and $K$ once there is no confusion. It is assumed that the variable is $k$ when we use these symbols.

Differentiation:

Interestingly the derivative of elliptic integrals can be written in terms of elliptic integrals

Derivative of complete elliptic integral of second kind

$$\frac{d}{dk}E=\int^1_0 \frac{\frac{\partial\,}{\partial\,k}\sqrt{1-k^2 x^2}}{\sqrt{1-x^2}}\,dx$$

$$\frac{d}{dk}E=\int^1_0 \frac{-k\,x^2}{\sqrt{1-x^2}\sqrt{1-k^2\,x^2}}\,dx$$

Adding and subtracting 1 results in

$$\frac{1}{k}\int^1_0 \frac{\sqrt{1-k^2x^2}}{\sqrt{1-x^2}}\,dx-\frac{1}{k}\int^1_0 \frac{dx}{\sqrt{1-x^2}\sqrt{1-k^2\,x^2}}$$

Upon realizing the relation to elliptic integrals we conclude

$$\displaystyle \frac{d}{dk}E=\frac{E-K}{k}$$​

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For the complete elliptic integral of first kind we need more work

Start by the following

$$\frac{d}{dk}K=\int^1_0 \frac{1}{\sqrt{1-x^2}}\,\frac{\partial\,}{\partial\,k}\left[\frac{1}{\sqrt{1-k^2 x^2}}\right]dx$$

$$\frac{d}{dk}K=\int^1_0 \frac{kx^2}{\sqrt{1-x^2}\sqrt{1-k^2x^2}(1-k^2x^2)}\,dx$$

Adding and subtracting 1 we have

$$\frac{-1}{k}\int^1_0 \frac{1-kx^2-1}{\sqrt{1-x^2}\sqrt{1-k^2x^2}(1-k^2x^2)}\,dx=\frac{1}{k}\int^1_0 \frac{dx}{\sqrt{1-x^2}\sqrt{1-k^2x^2}(1-k^2x^2)}-\frac{K}{k}$$

Let us focus on the first integral

$$\int^1_0 \frac{1}{\sqrt{1-x^2}(1-\, k^2 \, x^2)^{\frac{3}{2}}}\,dx$$

Let $x=\sqrt{t}$ and we have $dx= \frac{1}{2\sqrt{t}}\,dt$

$$\frac{1}{2}\int^1_0 \frac{t^{-\frac{1}{2}}}{\sqrt{1-t}(1-\, k^2 \, t)^{\frac{3}{2}}}\,dx$$

Using the hypergeometric integral representation

$$\frac{1}{2}\int^1_0 \frac{t^{-\frac{1}{2}}}{\sqrt{1-t}(1-\, k^2 \, t)^{\frac{3}{2}}}=\frac{\pi}{2}{}_2F_1\left(\frac{3}{2},\frac{1}{2},1,k^2 \right)$$

Using the linear transformation

$$_2F_1\left(a,b,c,z \right)=(1-z)^{c-a-b}\,{}_2F_1\left(c-a,c-b,c,z \right)$$

We get by putting $k'=\sqrt{1-k^2}$

$$\frac{\pi}{2}{}_2F_1\left(\frac{3}{2},\frac{1}{2},1,k^2 \right)=\frac{1}{1-k^2}\frac{\pi}{2}{}_2F_1\left(-\frac{1}{2},\frac{1}{2},1,k^2 \right)=\frac{E}{k'^2}$$

So finally we get

$$\displaystyle \frac{d}{dk}K=\frac{1}{k}\left( \frac{E}{ k'^2}-K \right)$$​

Note : $k'$ is called the complementary modulus and it should be of interest for us in the next set of lectures.

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Integration lessons continued ...

4.Integration using special functions (continued)

4.10. Euler sums

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General definition:

$$\displaystyle S_{p^{\,r},q} = \sum_{k\geq 1} \frac{(H_k^{(p)})^r}{k^q}$$​

Where we define the general harmonic number

$$\displaystyle H_k^{(p)} = \sum_{n=1}^k \frac{1}{n^p} \,\,\,;\,\, H^{(1)}_k \equiv H_k = \sum_{n=1}^k \frac{1}{n}$$​

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Euler sums were greatly studied by Euler, hence the name.

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Generating function:

$$\displaystyle \sum_{k\geq 1} H_k^{(p)}x^k = \frac{\mathrm{Li}_p(x)}{1-x}$$​

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Proof

$$\sum_{k\geq 1} H_k^{(p)}x^k = \sum_{k\geq 1} \sum_{n=1}^k \frac{1}{n^p}x^k$$

By interchanging the two series we have

$$\sum_{n\geq 1} \sum_{k\geq n} \frac{x^k}{n^p} =\sum_{n\geq 1}\frac{1}{n^p} \sum_{k\geq n}x^k$$

The inner sum is a geometric series

$$\frac{1}{1-x} \sum_{n\geq 1} \frac{x^n}{n^p} =\frac{\mathrm{Li}_p(x)}{1-x}$$

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We can use this to generate some more functions by integrating. Hence assume $p=1$

$$\sum_{k\geq 1} H_k x^k = -\frac{\log(1-x)}{1-x}$$

Divide by $x$ and integrate to get

$$\sum_{k\geq 1} \frac{H_k}{k} x^k =\mathrm{Li}_2(x)+\frac{1}{2}\log^2(1-x)$$

Now divide by $x$ and integrate again

$$\sum_{k\geq 1} \frac{H_k}{k^2} x^k =\mathrm{Li}_3(x)+\frac{1}{2}\int^x_0 \frac{\log^2(1-t)}{t}\,dt$$

Now let us look at the integral

$$\int^x_0 \frac{\log^2(1-t)}{t}\, dt$$

Integrating by parts we get the following

$$\int^x_0 \frac{\log^2(1-t)}{t}\, dt = - \log(1-x) \text{Li}_2(x) -\int^x_0\frac{\text{Li}_2 (t)}{1-t} \, dt$$

Now we are left with the following integral

$$\int^x_0\frac{\text{Li}_2 (t)}{1-t} \, dt = \int^{1}_{1-x} \frac{\text{Li}_2 (1-t)}{t} \, dt$$

$$\int^{1}_{1-x} \frac{\frac{\pi^2}{6} -\text{Li}_2(t) - \log(1-t) \log(t)}{t} \, dt$$

$$-\frac{\pi^2}{6}\log(1-x)- \int^{1}_{1-x} \frac{\text{Li}_2(t)}{t}\,dt -\int^{1}_{1-x}\frac{ \log(1-t) \log(t)}{t} \, dt$$

The first integral

• $$\int^{1}_{1-x} \frac{\text{Li}_2(t)}{t}\,dt =\text{Li}_3(1)-\text{Li}_3(1-x)$$

The second integral by parts we obtain

• $$\int^{1}_{1-x}\frac{ \log(1-t) \log(t)}{t} =\text{Li}_3(1) +\log(1-x)\text{Li}_2(1-x)-\text{Li}_3(1-x)$$

Collecting the results together we obtain

$$\int^x_0\frac{\text{Li}_2 (t)}{1-t} \, dt =-\frac{\pi^2}{6}\log(1-x)-\text{Li}_2(1-x) \log(1-x)+2\, \text{Li}_3(1-x)- 2 \zeta(3)$$

Hence we solved the integral

$$\int^x_0 \frac{\log^2(1-t)}{t}\, dt = - \log(1-x) \text{Li}_2(x) +\frac{\pi^2}{6}\log(1-x)+\text{Li}_2(1-x) \log(1-x)-2\, \text{Li}_3(1-x)+ 2 \zeta(3)$$

So we have got our Harmonic sum

$$\sum_{k\geq 1} \frac{H_k}{k^2} \, x^{k} = \text{Li}_3(x)+\frac{1}{2} \left( - \log(1-x) \text{Li}_2(x) +\frac{\pi^2}{6}\log(1-x)+\text{Li}_2(1-x) \log(1-x)-2\, \text{Li}_3(1-x)+ 2 \zeta(3) \right)$$

$$\sum_{k\geq 1} \frac{H_k}{k^2} \, x^{k} = \text{Li}_3(x)-\, \text{Li}_3(1-x)+\, \log(1-x) \text{Li}_2(1-x) +\frac{1}{2}\log(x) \log^2(1-x)+\zeta(3)$$

The expression can be further simplified but I will leave it for the reader.

To be continued ...

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
4.Integration using special functions (continued)

4.10. Euler sums (continued)

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General definition:

$$\displaystyle \sum_{n=1}^\infty \frac{H_n}{n^q}= \left(1+\frac{q}{2} \right)\zeta(q+1)-\frac{1}{2}\sum_{k=1}^{q-2}\zeta(k+1)\zeta(q-k)\,\,;q\geq 2$$​

This can be proved using complex analysis or using basic techniques of series manipulations as in this answer.

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Where by definition we have for $q=2$

$$\sum_{n=1}^\infty \frac{H_n}{n^2}= 2\zeta(3)$$

Let us look at a way of solving that sum using integration

First we need that

$$H_n = \int^1_0 \frac{1-x^n}{1-x}\,dx$$

proof

$$\int^1_0 \frac{1-x^n}{1-x}\,dx =\sum_{k\geq 0} \int^1_0 x^k-x^{n+k}\,dx$$

$$\sum_{k\geq 0} \frac{1}{k+1}-\frac{1}{n+k+1} = H_n$$

Plugging the result in the sum we get

$$\int^1_0 \frac{1}{1-x}\sum_{n = 1}^\infty\frac{1-x^n }{n^2}\,dx = \int^1_0 \frac{\zeta(2)-\mathrm{Li}_2(x)}{1-x}dx$$

Now use the duplication formula for dilogarithm

$$\zeta(2)-\mathrm{Li}_2(x) = \mathrm{Li}_2(1-x)+\log(x)\log(1-x)$$

Hence we have

$$\int^1_0 \frac{\mathrm{Li}_2(1-x)+\log(x)\log(1-x)}{1-x}dx = \int^1_0 \frac{\mathrm{Li}_2(x)+\log(1-x)\log(x)}{x}dx$$

First integral

$$\int^1_0 \frac{\mathrm{Li}_2(x)}{x} = \mathrm{Li}_3(1) =\zeta(3)$$

Second integral

$$\int^1_0 \frac{\log(1-x)\log(x)}{x}dx =\int^1_0 \frac{\mathrm{Li}_2(x)}{x} = \zeta(3)$$

Finally

$$\sum_{n=1}^\infty \frac{H_n}{n^2} = \zeta(3)+\zeta(3) = 2\zeta(3)$$

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Examples

$$\int^1_0\frac{\log^2(1-x)\log(x)}{x} = -\frac{\pi^4}{180}$$

Using the generating function

$$\sum_{k\geq 1}H_k x^{k-1} = -\frac{\log(1-x)}{x(1-x)}$$

By integrating both sides

$$\sum_{k\geq 1}\frac{H_k}{k} x^{k} =\mathrm{Li}_2(x) +\frac{1}{2}\log^2(1-x)$$

Or

$$\log^2(1-x) =2\sum_{k\geq 1}\frac{H_k}{k} x^{k} -2\mathrm{Li}_2(x)$$

plugging this in our integral we have

$$2\int^1_0\left(\sum_{k\geq 1}\frac{H_k}{k} x^{k} -\mathrm{Li}_2(x)\right)\frac{\log(x)}{x}\,dx$$

First integral

$$-2\int^1_0\mathrm{Li}_2(x)\frac{\log(x)}{x}\,dx =2\int^1_0\frac{\mathrm{Li}_3(x)}{x}\,dx = 2\zeta(4)$$

Second integral

$$2\sum_{k\geq 1}\frac{H_k}{k}\int^1_0x^{k-1}\log(x)\,dx$$

Using integration by parts twice and the formula presented first

$$-2\sum_{k\geq 1}\frac{H_k}{k^3} =-5\zeta(4)+\zeta^2(2)$$

Finally we get

$$\int^1_0\frac{\log^2(1-x)\log(x)}{x} =-5\zeta(4)+\zeta^2(2)+2\zeta(4) =\zeta^2(2)-3\zeta(4)$$

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
4.Integration using special functions (continued)

4.10. Euler sums (continued)

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Show that

$$\displaystyle \int_{0}^{\infty} \, e^{-at} \sin(bt) \frac{\ln t}{t}\, dt =- \left( \frac{ \log(a^2+b^2) }{2} +\gamma \right) \arctan \left( \frac{b}{a} \right)$$​

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Proof

We can start by the following integral

$$I(s) = \int_{0}^{\infty} t^{s-1} \, e^{-at} \, \sin(bt) dt$$

By using the the expansion of the sine function

$$I(s)=\int_{0}^{\infty}t^{s-1} \, e^{-at}\sum_{n\geq 0} \frac{(-1)^n (bt)^ {2n+1}}{\Gamma(2n+2)}$$

By swapping the summation and integration

$$I(s)= \sum_{n\geq 0} \frac{(-1)^n (b)^{2n+1}}{\Gamma(2n+2)} \int_{0}^ {\infty}t^{s+2n} \, e^{-at} dt= \frac{1}{a^s} \sum_{n\geq 0} \frac{(-1)^n (b)^{2n+1} \Gamma (s+2n+1)}{\Gamma(2n+2) a^{2n+1}}$$

By differentiating and plugging $s=0$ we have
$$I'(0) = \sum_{n\geq 0} \frac{(-1)^n \psi_0(2n+1)}{2n+1} \left( \frac{b} {a} \right)^{2n+1} -\log(a) \sum_{n\geq 0} \frac{(-1)^n }{2n+1} \left( \frac {b}{a} \right)^{2n+1}$$

Now use that $\psi(n+1) +\gamma = H_n$

$$I'(0) = \sum_{n\geq 0} \frac{(-1)^n H_{2n}-\gamma}{2n+1} \left( \frac{b} {a} \right)^{2n+1} -\log(a) \arctan \left( \frac{b}{a} \right)$$
$$I'(0) = \sum_{n\geq 0} \frac{(-1)^n H_{2n}}{2n+1} \left( \frac{b}{a} \right)^{2n+1}-(\gamma +\log(a))\arctan \left( \frac{b}{a} \right)$$

Now we look at the harmonic sum

\begin{align} \sum_{k\geq 0}(-1)^k H_{2k} x^{2k}&= \sum_{k\geq 0}(-1)^k x^ {2k} \int^1_0 \frac{1-t^{2k}}{1-t} \, dt\\ &= \int^1_0 \frac{1}{1-t} \sum_{k\geq 0}(-1)^k x^{2k} \left(1-t^{2k}\right) \, dt\\ &= \int^1_0 \frac{1}{1-t} \sum_{k\geq 0}(-1)^k \left(x^{2k}-(xt)^{2k}\right) \, dt\\ &= \int^1_0 \frac{1}{1-t}\left(\frac{1}{1+x^2}-\frac{1}{1+t^2x^2}\right) \, dt\\ &=\frac{1}{1+x^2} \int^1_0 \frac{1+t^2x^2-1-x^2}{(1-t)(1+t^2x^2)} \, dt\\ &=\frac{-x^2}{1+x^2} \int^1_0 \frac{(1-t^2)}{(1-t)(1+t^2x^2)} \, dt\\ &=\frac{-x^2}{1+x^2} \int^1_0 \frac{1+t}{(1+t^2x^2)} \, dt\\ &=\frac{-x^2}{1+x^2} \left( \int^1_0 \frac{1}{1+t^2x^2}+\frac{t}{1+t^2x^2} \, dt \right)\\ &= \frac{-1}{2(1+x^2)} \left(2x \arctan (x) + \log(1+x^2) \right) \end{align}

Using this we conclude by integrating

$$\sum_{k\geq 0}\frac{(-1)^k H_{2k}}{2k+1} x^{2k}=-\frac{1}{2} \log(1+x^2) \arctan(x)$$

Hence the following

$$\sum_{k\geq 0}\frac{(-1)^k H_{2k}}{2k+1} \left(\frac{b}{a} \right)^{2k+1}=- \frac{1}{2} \log \left( \frac{a^2+b^2}{a^2} \right) \arctan \left(\frac{b}{a} \right)$$

Substituting that in our integral

\begin{align} \int_{0}^{\infty} \, e^{-at} \sin(bt) \frac{\ln t}{t}\, dt &= -\left( \frac{1}{2} \log \left( \frac{a^2+b^2}{a^2} \right) + \gamma +\log(a) \right) \arctan \left( \frac{b}{a} \right)\\ &=- \left( \frac{ \log(a^2+b^2) }{2} +\gamma \right) \arctan \left( \frac{b} {a} \right) \end{align}

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Prove that

\displaystyle \begin{align} \int^1_0 \frac{\mathrm{Li}_p(x)\,\, \mathrm{Li}_q(x)\, }{x}\, dx&= \sum_{n=1}^{p-1}(-1)^{n-1}\zeta(p-n+1)\zeta(q+n) -\frac{1}{2}\sum_{n=1}^{{p+q}-2}(-1)^{p-1}\zeta(n+1)\zeta({p+q}-n)\\ &+(-1)^{p-1}\left(1+\frac{{p+q}}{2} \right)\zeta({p+q}+1)\end{align}​

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proof

We can see that

$$\int^1_0 \frac{\mathrm{Li}_p(x)\,\, \mathrm{Li}_q(x)\, }{x}\, dx = \sum_{k\geq 1}\sum_{n\geq 1}\frac{1}{k^{q}n^{p}(n+k)}$$

Let us first look at the following

$$\mathscr{C}(\alpha , k) =\sum_{n\geq 1}\frac{1}{n^{\alpha}(n+k)}\,\,\, ; \,\,\,\,\mathscr{C}(1, k)=\frac{H_k}{k}$$

We can find a general formula to find the sum

\begin{align}
\mathscr{C}(\alpha , k) &=\sum_{n\geq 1}\frac{1}{k\, n^{\alpha-1}}\left( \frac{1}{n}-\frac{1}{n+k}\right)\\ &= \frac{1}{k}\zeta(\alpha)-\frac{1}{k}\mathscr{C}(\alpha-1 , k)\\ &= \frac{1}{k}\zeta(\alpha)-\frac{1}{k^2}\zeta(\alpha-1)+\frac{1}{k^2}\mathscr{C}(\alpha-2 , k)\\ &= \sum_{n=1}^{\alpha-1}(-1)^{n-1}\frac{\zeta(\alpha-n+1)}{k^n}+(-1)^{\alpha-1}\frac{H_k}{k^\alpha}
\end{align}

Hence we have the general formula

$$\mathscr{C}(\alpha , k) = \sum_{n=1}^{\alpha-1}(-1)^{n-1}\frac{\zeta(\alpha-n+1)}{k^n}+(-1)^{\alpha-1}\frac{H_k}{k^\alpha}$$

Dividing by $k^{\beta}$ and summing w.r.t to $k$

$$\sum_{k\geq 1}\frac{\mathscr{C}(\alpha , k)}{k^{\beta}} = \sum_{n=1}^{\alpha-1}(-1)^{n-1}\zeta(\alpha-n+1)\zeta(\beta+n)+(-1)^{\alpha-1}\sum_{k\geq 1}\frac{H_k}{k^{\alpha+\beta}}$$

Now we use that

$$\sum_{n=1}^\infty \frac{H_n}{n^q}= \left(1+\frac{q}{2} \right)\zeta(q+1)-\frac{1}{2}\sum_{k=1}^{q-2}\zeta(k+1)\zeta(q-k)$$

Hence we have

$$\sum_{n=1}^\infty \frac{H_n}{n^{\alpha+\beta}}= \left(1+\frac{{\alpha+\beta}}{2} \right)\zeta({\alpha+\beta}+1)-\frac{1}{2}\sum_{k=1}^{{\alpha+\beta}-2}\zeta(k+1)\zeta({\alpha+\beta}-k)$$

And the generalization is the following formula

\begin{align}
\sum_{k\geq 1}\frac{\mathscr{C}(\alpha , k)}{k^{\beta}} &= \sum_{n=1}^{\alpha-1}(-1)^{n-1}\zeta(\alpha-n+1)\zeta(\beta+n) -\frac{1}{2}\sum_{n=1}^{{\alpha+\beta}-2}(-1)^{\alpha-1}\zeta(n+1)\zeta({\alpha+\beta}-n)\\ &+(-1)^{\alpha-1}\left(1+\frac{{\alpha+\beta}}{2} \right)\zeta({\alpha+\beta}+1)\end{align}

We conclude by putting that

$$\displaystyle \sum_{k\geq 1}\sum_{n\geq 1}\frac{1}{k^{q}n^{p}(n+k)} = \sum_{k\geq 1}\frac{\mathscr{C}(p, k)}{k^{q}}$$

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
4.Integration using special functions (continued)

4.10. Euler sums (continued)

We can relate the generalized harmonic number to the polygamma function

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$$\displaystyle H^{(p)}_k = \zeta(p) +(-1)^{p-1}\frac{\psi_{p-1}(k+1)}{ (p-1)!}$$​

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proof

$$H^{(p)}_k = \sum^k_{n=1}\frac{1}{n^p} =\zeta(p)-\sum^\infty_{n=k+1}\frac{1}{n^p}$$

Now let $n=i+k+1$

$$H^{(p)}_k = \sum^k_{n=1}\frac{1}{n^p} =\zeta(p)-\sum_{i\geq 0}\frac{1}{(i+k+1)^p}$$

We know that

$$(-1)^{p}\frac{\psi_{p-1}(k+1)}{ (p-1)!} =\sum_{i\geq 0} \frac{1}{(i+k+1)^{p}}\,\,p\geq 1$$

Hence we have

$$H^{(p)}_k = \zeta(p) +(-1)^{p-1}\frac{\psi_{p-1}(k+1)}{ (p-1)!}$$

We can use that to obtain a nice integral representation.

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$$\displaystyle \sum_{k\geq 1}\frac{H^{(p)}_k}{k^q} = \zeta(p)\zeta(q) +(-1)^{p}\frac{1}{ (p-1)!}\int^1_0\frac{\mathrm{Li}_q(x)\log(x)^{p-1}}{1-x}\,dx$$​

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Note that

$$\psi_0(a+1)= \int^1_0\frac{1-x^a}{1-x}\,dx$$

By differentiating with respect to $a$ , $p$ times we have

$$\psi_p(a+1) = \frac{\partial}{\partial a^p}\int^1_0\frac{1-x^a}{1-x}\,dx$$

$$\psi_p(a+1) = -\int^1_0\frac{x^a\log(x)^{p}}{1-x}\,dx$$

Let $a=k$

$$\psi_{p-1}(k+1) = -\int^1_0\frac{x^k\log(x)^{p-1}}{1-x}\,dx$$

Substituting that in our formula

$$H^{(p)}_k = \zeta(p) +(-1)^{p}\frac{1}{ (p-1)!}\int^1_0\frac{x^k\log(x)^{p-1}}{1-x}\,dx$$

Now divide by $k^q$ and sum with respect to $k$

$$\sum_{k\geq 1}\frac{H^{(p)}_k}{k^q} = \zeta(p)\zeta(q) +(-1)^{p}\frac{1}{ (p-1)!}\int^1_0\frac{\mathrm{Li}_q(x)\log(x)^{p-1}}{1-x}\,dx$$

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Symmetric formula

$$\displaystyle \sum_{k\geq 1} \frac{H^{(p)}_k}{k^q}+\sum_{k\geq 1} \frac{H^{(q)}_k}{k^p} =\zeta(p)\zeta(q)+\zeta(p+q)$$​

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proof

Take the left side and swap the finite and infinite sums

$$\sum_{i\geq 1} \,\sum_{k\geq i}\frac{1}{i^p} \frac{1}{k^q}=\sum_{i\geq 1} \,\sum_{k\geq 1}\frac{1}{i^p} \frac{1}{k^q}-\sum_{i\geq 1}\frac{1}{i^p} \,\sum_{1\leq k \leq i-1} \frac{1}{k^q}$$

The second sum can be written as

$$\sum_{i\geq 1}\frac{1}{i^p} \,\sum_{1\leq k \leq i-1} \frac{1}{k^q} = \sum_{i\geq 1}\frac{1}{i^p} \,\left(\sum_{1\leq k \leq i}\frac{1}{k^q}-\frac{1}{i^p}\right)$$

By separating and changing the index we get

$$\sum_{k\geq 1}\frac{H^{(q)}_k}{k^p}-\zeta(p+q)$$

Hence we have

$$\sum_{k\geq 1} \frac{H^{(p)}_k}{k^q} =\zeta(p)\zeta(q)-\sum_{k\geq 1}\frac{H^{(q)}_k}{k^p}+\zeta(p+q)$$

$$\sum_{k\geq 1} \frac{H^{(p)}_k}{k^q}+\sum_{k\geq 1}\frac{H^{(q)}_k}{k^p} =\zeta(p)\zeta(q)+\zeta(p+q)$$

for the special case $p=q=n$

$$\sum_{k\geq 1} \frac{H^{(n)}_k}{k^n} =\frac{\zeta^2(n)+\zeta(2n)}{2}$$

Last edited:

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Examples

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$$\displaystyle \sum_{k\geq 1}\frac{H_k^{(3)}}{k^2} = \frac{11\zeta(5)}{2}-2\zeta(2)\zeta(3)$$​

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Using the symmetry formula

$$\sum_{k\geq 1}\frac{H_k^{(3)}}{k^2} = \zeta(2)\zeta(3)+\zeta(5)-\sum_{k\geq 1}\frac{H_k^{(2)}}{k^3}$$

Using the integral formula on the second sum

$$\sum_{k\geq 1}\frac{H_k^{(2)}}{k^3} =\zeta(2)\zeta(3)+ \int^1_0 \frac{\mathrm{Li}_3(x)\log(x)}{1-x}\,dx$$

Using integration by parts on the integral

$$\int^1_0 \frac{\mathrm{Li}_3(x)\log(x)}{1-x}\,dx =- \int^1_0\frac{\mathrm{Li}_2(x)\mathrm{Li}_2(1-x)}{x}\,dx$$

Let us think of solving

$$\int^1_0\frac{\mathrm{Li}_2(x)\mathrm{Li}_2(1-x)}{x}\,dx$$

Using the duplication formula

$$\mathrm{Li}_2(1-x) = \zeta(2)-\mathrm{Li}_2(x)-\log(x)\log(1-x)$$

$$\int^1_0\frac{\mathrm{Li}_2(x)(\zeta(2)-\mathrm{Li}_2(x)-\log(x)\log(1-x))}{x}\,dx$$

The first integral

$$\zeta(2)\int^1_0 \frac{\mathrm{Li}_2(x)}{x}\,dx = \zeta(2)\zeta(3)$$

The third integral

$$\int^1_0\frac{\mathrm{Li}_2(x)\log(x)\log(1-x)}{x}\,dx = \frac{1}{2}\int^1_0\frac{\mathrm{Li}^2_2(x)}{x}\,dx$$

Finally we get

$$\int^1_0 \frac{\mathrm{Li}_3(x)\log(x)}{1-x}\,dx =\frac{3}{2}\int^1_0\frac{\mathrm{Li}^2_2(x)}{x}\,dx-\zeta(2)\zeta(3)$$

So

$$\sum_{k\geq 1}\frac{H_k^{(2)}}{k^3} = \frac{3}{2}\int^1_0\frac{\mathrm{Li}^2_2(x)}{x}\,dx$$

Hence we finally get that

$$\sum_{k\geq 1}\frac{H_k^{(3)}}{k^2}=\zeta(2)\zeta(3)+\zeta(5)-\frac{3}{2}\int^1_0\frac{\mathrm{Li}^2_2(x)}{x}\,dx$$

Let us solve the integral

$$\int^1_0 \frac{\mathrm{Li}^2_2(x)}{x}\,dx$$

By series expansion

$$\sum_{k,n\geq 1}\frac{1}{(nk)^2}\int^1_0x^{n+k-1}\,dx =\sum_{k,n\geq 1}\frac{1}{(nk)^2(n+k)}$$

By some manipulations we get

$$\sum_{k\geq 1}\frac{1}{k^3}\sum_{n\geq 1}\frac{k}{n^2(n+k)}= \sum_{k\geq 1}\frac{1}{k^3}\sum_{n\geq 1}\frac{1}{n^2}-\sum_{k\geq 1}\frac{1}{k^3}\sum_{n\geq 1}\frac{1}{n(n+k)}$$

This can be simplified to conclude that

$$\int^1_0 \frac{\mathrm{Li}^2_2(x)}{x}\,dx = \zeta(2)\zeta(3)-\sum_{k\geq 1}\frac{H_k}{k^4}$$

Now using that

$$\sum_{k\geq 1}\frac{H_k}{k^4} = 3\zeta(5)-\zeta(2)\zeta(3)$$

Hence

$$\int^1_0 \frac{\mathrm{Li}^2_2(x)}{x}\,dx = 2\zeta(2)\zeta(3)-3\zeta(5)$$

Finally we get

$$\sum_{k\geq 1}\frac{H_k^{(3)}}{k^2}=\zeta(2)\zeta(3)+\zeta(5)-\frac{3}{2}\left( 2\zeta(2)\zeta(3)-3\zeta(5)\right)=\frac{11\zeta(5)}{2}-2\zeta(2)\zeta(3)$$

Last edited:

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Advanced Integration Techniques (back with new lessons)

4.Integration using special functions (continued)

4.11. Sine Integral function

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We define the following

$$\mathrm{Si}(z) = \int^z_0 \frac{\sin(x) }{x}\, dx$$

A closely related function is the following

$$\mathrm{si}(z) = -\int^\infty_z \frac{\sin(x) }{x}\, dx$$

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These function are related through the equation

$$\mathrm{Si}(z) = \mathrm{si}(z)+\frac{\pi}{2}$$

A closely related function is the $\mathrm{sinc}$ function

$$\mathrm{sinc}=\begin{cases} 1 & x=0 \\ \frac{\sin(x)}{x} & x \neq 0 \end{cases}$$

Using that we conclude that

$$\frac{d}{dx} \mathrm{Si}(x) = \mathrm{sinc}(x)$$

For the integration we conclude that

$$\int \mathrm{Si}(x)\,dx = \cos(x)+ x \,\mathrm{Si(x)}$$

Examples

Prove that

$$\int^\infty_0 \sin(x) \mathrm{si}(x) \, dx = -\frac{\pi}{4}$$

Using integration by parts we get

$$-\int^\infty_0 \frac{\sin(x)\cos(x)}{x} dx = -\frac{1}{2} \int^\infty_0 \frac{\sin(2x)}{x}\,dx$$

Let $2x = t$

$$-\frac{1}{2}\int^\infty_0 \frac{\sin(t)}{t} dx = -\frac{\pi}{4}$$

Prove that

$$\int^\infty_0 x^{\alpha -1}\,\mathrm{si}(x) \, dx =- \frac{\Gamma(\alpha)}{\alpha}\sin\left(\frac{\pi \alpha}{2}\right)$$

Using the integral representation

$$-\int^\infty_0 x^{\alpha -1}\,\int^\infty_x \frac{\sin(t)}{t}\,dt \, dx$$

Let $xy = t$

$$-\int^\infty_0 x^{\alpha -1}\,\int^\infty_1 \frac{\sin(xy)}{y}\,dy \, dx$$

Switching the integrals we get

$$-\int^\infty_1 \frac{1}{y} \int^\infty_0 x^{\alpha -1}\, \sin(xy) \, dx \,dy$$

Now let $xy = t$

$$-\int^\infty_1 \frac{1}{y^{\alpha+1}} \int^\infty_0 t^{\alpha -1}\, \sin(t) \, dx \,dy$$

The Mellin transform of the sine function is defined as

$$\mathcal{M}_s(\sin (x)) = \int^\infty_0 x^{s-1}\sin(x)\,dx = \Gamma (s) \sin\left( \frac{\pi s}{2} \right)$$

Hence we conclude that

$$-\Gamma (\alpha) \sin\left( \frac{\pi \alpha}{2} \right)\int^\infty_1 \frac{1}{y^{\alpha+1}} = -\frac{\Gamma (\alpha)}{\alpha} \sin\left( \frac{\pi \alpha}{2} \right)$$

Prove that

$$\int^\infty_0 e^{-\alpha \, x}\,\mathrm{si}(x) \, dx =- \frac{\arctan (\alpha)}{\alpha}$$

Using the integral representation

$$-\int^\infty_0 e^{-\alpha \, x}\,\int^\infty_x \frac{\sin(t)}{t}\,dt \, dx$$

Let $xy = t$

$$-\int^\infty_0 e^{-\alpha \, x}\,\int^\infty_1 \frac{\sin(xy)}{y}\,dy \, dx$$

Switching the integrals

$$-\int^\infty_1 \frac{1}{y}\int^\infty_0 e^{-\alpha \, x}\, \sin(xy)\,dx \, dy$$

The inner integral is the laplace transform of the sine function

$$\mathcal{L}_s(\sin(at)) = \frac{a}{s^2+a^2}$$

Hence we conclude that

$$-\int^\infty_1 \frac{1}{y^2+\alpha^2} \,dy = -\frac{\arctan(\alpha)}{\alpha}$$

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Re: Advanced Integration Techniques (back with new lessons)

4.Integration using special functions (continued)

4.11. Sine Integral function (continued)

Prove that

$$\int^\infty_0 \mathrm{si}(x) \log(x) \,dx =\gamma+1$$

We know that

$$\int^\infty_0 x^{\alpha -1}\,\mathrm{si}(x) \, dx =- \frac{\Gamma(\alpha)}{\alpha}\sin\left(\frac{\pi \alpha}{2}\right)$$

Differentiate with respect to $\alpha$

$$\int^\infty_0 x^{\alpha -1}\,\mathrm{si}(x) \,\log(x) dx = \frac{\Gamma(\alpha)}{\alpha^2}\sin\left(\frac{\pi \alpha}{2}\right)-\frac{\Gamma(\alpha)\psi(\alpha)}{\alpha}\sin\left(\frac{\pi \alpha}{2}\right) +\frac{\pi}{2}\cos\left(\frac{\pi \alpha}{2}\right)$$

Let $\alpha \to 1$

$$\int^\infty_0 \mathrm{si}(x) \,\log(x) dx = 1- \psi(1) = 1-(-\gamma) = 1+\gamma$$

Find the integral

$$\int^\infty_0 \mathrm{si}(x) \, \sin(px) \, dx$$

Using integration by parts we get

$$-\left[\frac{\mathrm{si}(x) \cos(px)}{p} \right]^\infty_0+\frac{1}{p}\int^\infty_0 \frac{\sin(x)}{x} \, \cos(px) \, dx$$

Taking the limits

$$\lim_{x \to 0} \frac{\mathrm{si}(x) \cos(px)}{p} = \frac{\mathrm{si}(0)}{p} = -\frac{\pi}{2p}$$

$$\lim_{x \to \infty} \frac{\mathrm{si}(x) \cos(px)}{p} = 0$$

Hence we get

$$-\frac{\pi}{2p}+\frac{1}{p}\int^\infty_0 \frac{\sin(x)}{x} \, \cos(px) \, dx$$

The integral

$$\int^\infty_0 \frac{\sin(x)}{x} \, \cos(px) \, dx = \frac{1}{2}\int^\infty_0 \frac{\sin((p+1)x) -\sin((p-1)x)}{x}dx$$

Separate the integrals

$$I = \frac{1}{2}\int^\infty_0 \frac{\sin((p+1)x)}{x}dx -\frac{1}{2}\int^\infty_0 \frac{\sin((p-1)x)}{x}dx$$

If $p-1>0$ we get

$$I = \frac{\pi}{4}-\frac{\pi}{4} = 0$$

If $p-1<0$

$$I = \frac{1}{2}\int^\infty_0 \frac{\sin((p+1)x)}{x}dx +\frac{1}{2}\int^\infty_0 \frac{\sin((p-1)x)}{x}dx = \frac{\pi}{2}$$

If $p=1$ we have

$$I = \frac{1}{2}\int^\infty_0 \frac{\sin(2x)}{x}dx +0 = \frac{\pi}{4}$$

Finally we get

$$\int^\infty_0 \mathrm{si}(x)\sin(px) dx= \begin{cases} -\frac{\pi}{2p} & p > 1 \\ -\frac{\pi}{4p} & p < 1 \\ 0 & p = 1 \end{cases}$$

Last edited:

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Re: Advanced Integration Techniques (back with new lessons)

4.Integration using special functions (continued)

4.11. Sine Integral function (continued)

Prove that for $0<a<2$

$$\int^\infty_0 \mathrm{si}^2(x) \, \cos(ax) \, dx = \frac{\pi}{2a}\log(a+1)$$

Using integration by parts we get

$$\left[\frac{\mathrm{si}^2(x) \sin(ax)}{a} \right]^\infty_0-\frac{2}{a}\int^\infty_0 \frac{\mathrm{si}(x)\sin(x)}{x} \, \sin(ax) \, dx$$

Taking the limits

$$\lim_{x \to 0} \frac{\mathrm{si}^2(x) \sin(ax)}{a} = 0$$

$$\lim_{x \to \infty} \frac{\mathrm{si}^2(x) \sin(ax)}{a} = 0$$

Let the integral

$$I(a) = \int^\infty_0 \frac{\mathrm{si}(x)\sin(x)}{x} \, \sin(ax) \, dx$$

Differentiate with respect to $a$

$$I'(a) = \int^\infty_0 \mathrm{si}(x)\sin(x) \, \cos(ax) \, dx$$

Now use the product to sum trigonometric rules

$$I'(a) =\frac{1}{2} \int^\infty_0 \mathrm{si}(x)(\sin((a+1)x)-\sin((a-1)x)) \, dx$$

From the previous exercise we have

$$\int^\infty_0 \mathrm{si}(x)\sin((a+1)x) dx = \frac{-\pi}{4(a+1)}\,\,\,;\, a>0$$

$$\int^\infty_0 \mathrm{si}(x)\sin((a+1)x) dx = 0\,\,\,;\, a<2$$

Hence we conclude that for $0<a<2$

$$I'(a) =-\frac{\pi}{4(a+1)}$$

Integrate with respect to $a$

$$I(a) =-\frac{\pi}{4}\log(a+1)+C$$

Let $a \to 0$

$$I(0) =0+C \,\,\, \to \,\,\, C = 0$$

Hence we have

$$\int^\infty_0 \frac{\mathrm{si}(x)\sin(x)}{x} \, \sin(ax) \, dx = -\frac{\pi}{4}\log(a+1)$$

Which implies that

$$\int^\infty_0 \mathrm{si}^2(x) \, \cos(ax) \, dx = \frac{-2}{a}\left( -\frac{\pi}{4}\log(a+1)\right) = \frac{\pi}{2a}\log(a+1)$$

Find the integral, for $a \neq 1$

$$\int^\infty_0 \mathrm{si}(x) \cos(ax)\,dx$$

Use integration by parts to obtain

$$\frac{1}{a}\int^\infty_0\frac{\sin(x) \sin(ax)}{x}\,dx$$

Let the integral

$$I(t) = \int^\infty_0e^{-tx}\frac{\sin(x) \sin(ax)}{x}\,dx$$

Differentiate with respect to $t$

$$I'(t) = -\int^\infty_0e^{-tx}\sin(x) \sin(ax)\,dx$$

Use product to sum rules

$$I'(t) = \frac{1}{2}\int^\infty_0e^{-tx}(\cos((a+1)x)-\cos((a-1)x))\,dx$$

Now we can use the Laplace transform

$$I'(t) = \frac{1}{2}\left(\frac{t}{t^2+(a+1)^2}-\frac{t}{t^2+(a-1)^2} \right)$$

Integrate with respect to $t$

$$I(t) = -\frac{1}{4} \log\left(\frac{t^2+(a+1)^2}{t^2+(a-1)^2} \right) +C$$

After verifying the constant goes to 0, we have

$$\int^\infty_0e^{-tx}\frac{\sin(x) \sin(ax)}{x}\,dx = -\frac{1}{4} \log\left(\frac{t^2+(a+1)^2}{t^2+(a-1)^2} \right)$$

Let $t \to 0$

$$\int^\infty_0\frac{\sin(x) \sin(ax)}{x}\,dx = -\frac{1}{4} \log\left(\frac{a+1}{a-1} \right)^2$$

We conclude that

$$\int^\infty_0 \mathrm{si}(x) \cos(ax)\,dx = -\frac{1}{4a} \log\left(\frac{a+1}{a-1} \right)^2$$

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