4.Integration using special functions (continued)

4.3.Digamma function(continued) :

We have the following digamma property :

$\displaystyle \psi(s+1)\,=\, -\gamma \,+\, \int^{1}_{0}\frac{1-x^s}{1-x}$

Prove the following integral :

$\displaystyle \int_0^1 \frac{(1-x^a)(1-x^b)(1-x^c)}{(1-x)(-\log x)}dx= \log \left\{\frac{\Gamma(b+c+1) \Gamma(c+a+1)\Gamma(a+b+1)}{\Gamma(a+1) \Gamma(b+1) \Gamma(c+1) \Gamma(a+b+c+1)} \right\}$

Solution :

As crazy as it looks , it becomes very easy to solve if we know how to start !

First note that since there is a log in the denominator that gives as an idea to differentiate ...

$\displaystyle \text{Let : }F(c)=\int_0^1 \frac{(1-x^a)(1-x^b)(1-x^c)}{(1-x)(-\log x)}dx$

Differentiate with respect to c :

$\displaystyle F'(c)=\int_0^1 \frac{(1-x^a)(1-x^b)x^c}{(1-x)}dx$

$\displaystyle F'(c)=\int_0^1 \frac{(1-x^a-x^b+x^{a+b})x^c}{(1-x)}dx$

$\displaystyle F'(c)=\int_0^1 \frac{x^c\,-\,x^{a+c}\,-\,x^{b+c}\,+\,x^{a+b+c}}{(1-x)}dx$

$\displaystyle F'(c)=\int_0^1 \frac{(x^c\,-1)\,+\,(1-\,x^{a+c})\,+\,(1-\,x^{b+c})\,+\,(x^{a+b+c}-1)}{(1-x)}dx$

$\displaystyle F'(c)=-\int_0^1 \frac{1-x^c\,}{1-x}\,dx+\int_0^1\frac{1-x^{a+c}}{1-x}\,dx+\int_0^1\frac{1-x^{b+c}}{1-x}\,dx-\int_0^1\frac{1-x^{a+b+c}}{1-x}dx$

Now use the following result :

$\displaystyle \psi(s+1)=-\gamma+\int^{1}_0\frac{1-x^s}{1-x}\,dx$

$\displaystyle F'(c)=-\psi(c+1)-\gamma+\psi(a+c+1)+\gamma+\psi(b+c+1)+\gamma-\psi(a+b+c+1)-\gamma$

$\displaystyle F'(c)=-\psi(c+1)+\psi(a+c+1)+\psi(b+c+1)-\psi(a+b+c+1)$

Integrate with respect to c we have :

$\displaystyle F(c)=-\log\left[\Gamma(c+1)\right]+\log \left[\Gamma(a+c+1)\right] +\log\left[\Gamma(b+c+1)\right]-\log \left[\Gamma(a+b+c+1)\right] +C_1$

Which reduces to :

$\displaystyle \log\left[\frac{\Gamma(a+c+1)\Gamma(b+c+1)}{\Gamma (c+1) \Gamma (a+b+c+1)} \right] +C_1$

Now put c= 0 we have :

$\displaystyle 0=\log \left[ \frac{\Gamma(a+1)\Gamma(b+1)}{\Gamma(a+b+1)}\right]+C_1$

$\displaystyle C_1=-\log \left[ \frac{\Gamma(a+1)\Gamma(a+1)}{\Gamma(a+b+1)}\right]$

So we have the following :

$\displaystyle F(c)=\log\left[ \frac{\Gamma(a+c+1)\Gamma(b+c+1)}{\Gamma(c+1) \Gamma (a+b+c+1)}\right] -\log\left[\frac{\Gamma(a+1)\Gamma(a+1)}{\Gamma(a+b+1)}\right]$

Hence we have the result :

$\displaystyle \int_0^1 \frac{(1-x^a)(1-x^b)(1-x^c)}{(1-x)(-\log x)}dx= \log \left\{\frac{\Gamma(b+c+1) \Gamma(c+a+1)\Gamma(a+b+1)}{\Gamma(a+1) \Gamma(b+1) \Gamma(c+1) \Gamma(a+b+c+1)} \right\}$

To be Continued ...

4.Integration using special functions (continued)

4.3.Digamma function(continued) :

We continue with some more exercises :

Find the following integral :

$\displaystyle \int^{\infty}_0\left(e^{-bx}-\frac{1}{1+ax}\right)\,\frac{dx}{x}$

Solution :

Let us first use the substitution t = ax so we get the following :

$\displaystyle \int^{\infty}_0\left(e^{-\frac{bt}{a}}-\frac{1}{1+t}\right)\,\frac{dt}{t}$

Now we must realize the result we get earlier in this series that :

$\displaystyle \int^{\infty}_0 \frac{e^{-x}-(1+x)^{-a}}{x}\,dx = \psi(a)$

But there we don't have $\displaystyle e^{-x}$ so let us add and subtract it :

$\displaystyle \int^{\infty}_0\left(e^{-t}-e^{-t}+e^{-\frac{bt}{a}}-\frac{1}{1+t}\right)\,\frac{dt}{t}$

$\displaystyle \int^{\infty}_0 \left(e^{-t}-\frac{1}{1+t}\right)\,\frac{dt}{t}+\int^{\infty}_0 \frac{e^{-\frac{bt}{a}}-e^{-t}}{t}\, dt$

$\displaystyle \int^{\infty}_0 \left(e^{-t}-\frac{1}{1+t}\right)\frac{dt}{t}=-\gamma$

We already also proved several posts ago that :

$\displaystyle \int^{\infty}_0 \frac{e^{-\frac{bt}{a}}- e^{-t}}{t}\, dt= -\log\left(\frac{b}{a}\right) = \log\left(\frac{a}{b}\right)$

Hence the result :

$\displaystyle \int^{\infty}_0 \left(e^{-bx}-\frac{1}{1+ax} \right)\, \frac{dx}{x} = \log\left(\frac{a}{b} \right)-\gamma$

Prove the following integral :

$\displaystyle \int^{\infty}_0 \, e^{-ax} \left(\frac{1}{x}-\text{coth}(x) \right) \, dx = \psi\left(\frac{a}{2}\right) - \ln\left(\frac{a}{2}\right) + \frac{1}{a}$

Solution :

We know the following hyperoblic identity : (hopefully )

$\displaystyle \text{coth}(x)= \frac{1+e^{-2x}}{1-e^{-2x}}$

$\displaystyle \int^{\infty}_0 \, e^{-ax} \left(\frac{1}{x}-\frac{1+e^{-2x}}{1-e^{-2x}} \right) \, dx$

Now let 2x =t so we have :

$\displaystyle \int^{\infty}_0 \, e^{-\left(\frac{at}{2}\right)} \left(\frac{1}{t}-\frac{1+e^{-t}}{2(1-e^{-t})} \right) \, dt$

$\displaystyle \int^{\infty}_0 \, \frac{e^{-\left(\frac{at}{2}\right)}}{t}-\frac{e^{-\left(\frac{at}{2}\right)}+e^{\left(-\frac{at}{2}-t\right)}}{2(1-e^{-t})} \, dt$

I will add and subtract some terms :

$\displaystyle \int^{\infty}_0 \, \frac{e^{-t}+e^{-\left(\frac{at}{2}\right)}-e^{-t}}{t}\,-\,\frac{e^{-\left(\frac{at}{2}\right)}+e^{-\left(\frac{at}{2}\right)}-e^{-\frac{at}{2}}+e^{-\left(\frac{at}{2}\right)-t}}{2(1-e^{-t})} \, dt$

$\displaystyle \int^{\infty}_0 \, \frac{e^{-t}}{t}-\frac{e^{-\left(\frac{at}{2}\right)}}{1-e^{-t}}\,dt +\int^{\infty}_0 \frac{e^{-\left(\frac{at}{2}\right)}-e^{\left(-\frac{at}{2}-t\right)}} {2(1-e^{-t})}\, dt +\int^{\infty}_0 \frac{e^{-\left(\frac{at}{2}\right)}-e^{-t}}{t} \, dt$

First : We use the identity :

$\displaystyle \psi \left(x\right)=\int^{\infty}_0 \, \frac{e^{-t}}{t}-\frac{e^{-\left(x t\right)}}{1-e^{-t}}\, dt$

$\displaystyle \int^{\infty}_0 \, \frac{e^{-t}}{t}-\frac{e^{-\left(\frac{at}{2}\right)}}{1-e^{-t}}\, dt=\psi \left(\frac{a}{2}\right)$

The second integral can easily be proven :

$\displaystyle \int^{\infty}_0\frac{e^{-\left(\frac{at}{2}\right)}-e^{-\left(\frac{at}{2}-t\right)}}{2(1-e^{-t})}=\int^{\infty}_0 \frac{e^{-\left(\frac{at}{2}\right)}}{2}\,dt=\frac{1}{a}$

We also proved earlier that :

$\displaystyle \int^{\infty}_0\frac{e^{-t}\,-\,e^{-st}}{t} \, dt = \ln \left( s \right)$

$\displaystyle \int^{\infty}_0\frac{e^{-\left(\frac{at}{2}\right)}-e^{-t}}{t} \, dt = -\ln \left(\frac{a}{2}\right)$

Hence we have the full integral :

$\displaystyle \int^{\infty}_0 \, e^{-ax} \left(\frac{1}{x}-\text{coth}(x) \right) \, dx = \psi\left(\frac{a}{2}\right) - \ln\left(\frac{a}{2}\right) + \frac{1}{a}$

[HW] Prove : $\displaystyle \psi \left(x\right)=\int^{\infty}_0 \, \frac{e^{-t}}{t}-\frac{e^{-\left(x t\right)}}{1-e^{-t}}\, dt$

Zeta function ,,, is on the way

4.Integration using special functions (continued)

4.4.Riemann Zeta function :

You surely had heard about this interesting function, not only has it resulted in many interesting founding but also the yet to be solved Riemann Hypothesis makes it one of the most celebrated functions in history of mathematics. Its interesting aspect is the relation to prime numbers.

In this tutorial we will not go so deep in the proofs and analyticity of it, we will try to describe its properties and mainly focus on the integral representation and its relation to other functions.

Let us first start by defining the zeta function as the following :

$\displaystyle \zeta(s) \,=\, \sum_{n=1}^{\infty}\,\frac{1}{n^s}\,\,\, \,\, \text{Re}(s)>1$

There are more general representation of the zeta function but we will stick to the one we defined above. For the first glance it seems that zeta has no relations to other functions but it turned out that it has a strong relation to other functions such as gamma and digamma functions.

4.4.1 zeta and gamma representation :

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

This relation turns out to be so much interesting since we can evaluate the right-hand integral using known results for both zeta and gamma.

This most famous value for zeta function is when $s=2$ which represents the infinite sum of reciprocals of squares :

$\displaystyle \zeta(2) =\sum^{\infty}_{n=1} \frac{1}{n^2}\,= \frac{\pi^2}{6}$

Actually we can derive all the values of $\displaystyle \zeta(2k) \,\,\,\, k\in \mathbb{Z}$ but unfortunately there is no known way to find the zeta for odd integers $\displaystyle \zeta(2k+1)\,$ .

Now let us use this result to find some integrals :

$\displaystyle \int^{\infty}_0 \frac{t}{e^t-1} \,dt$

It follows directly from the gamma-zeta relation that taking $s=2$ we have :

$\displaystyle \Gamma(2) \, \zeta(2) \, = \int^{\infty}_{0}\, \frac{t}{e^t-1}\, dt$

$\displaystyle \int^{\infty}_{0}\, \frac{t}{e^t-1}\, dt=\zeta(2)=\frac{\pi^2}{6}$

Solve the following integral :

$\displaystyle \int_0^{\frac{\pi}{2}} \,\frac{\log \left(\sec(x) \right)}{\tan(x)} \,dx$

use the substitution : $\displaystyle \sec(x)=e^t$

$\displaystyle \int_0^{\infty} \, \frac{t}{e^{2t}-1}\,dt\, =\, \frac{1}{4} \cdot \int_0^{\infty}\,\frac{t}{e^t-1}\,dt=\frac{\pi^2}{24}$

Another similar relation between zeta and gamma is the following identity :

$\displaystyle \Gamma(s)\, \zeta(s) \, (1-2^{1-s})\, = \int^{\infty}_0\, \frac{t^{s-1}}{e^t+1}\,$

It follows from that equation that :

$\displaystyle \int^{\infty}_0\, \frac{t}{e^t+1}\,dt = \frac{\pi^2}{12}$

To be continued ...

4.4.2.Zeta function and Bernoulli numbers

We will see in this section some values of the zeta function using an equation due to Euler , but let us first define the Bernoulli numbers .

It is usually defined as $\displaystyle B_{k}$ and the easiest way to derive them is find the coefficients of the power series of $\displaystyle \frac{x}{e^x-1}$

Definition

$\displaystyle \frac{x}{e^x-1} = \sum_{k \geq 0} \frac{B_k}{k!}x^k$

Now let us derive some values for the Bernoulli numbers , rewrite the power series as

$\displaystyle x = (e^x-1) \, \sum_{k\geq 0} \frac{B_k}{k!}x^k$

$\displaystyle x = \left( x+\frac{1}{2!}x^2 + \frac{1}{3!} x^3+ \frac{1}{4!} x^4+\cdots \right) \cdot \left( B_0 +B_1 \, x+\frac{B_2}{2!}\,x^2 +\frac{B_3}{3!} x^3+ \cdots \right)$

$\displaystyle x = B_0 x + \left(B_1+\frac{B_0}{2!} \right)x^2 + \left(\frac{B_0}{3!}+ \frac{B_2}{2!}+\frac{B_1}{2!}\right)x^3 +\left( \frac{B_0}{4!}+\frac{B_1}{3!}+\frac{B_2}{2!\, 2!}+\frac{B_3}{3!}\right) x^4+ \cdots$

By comparing the terms we get the following values

$\displaystyle {B_0 = 1 \, , \, B_1 =-\frac{1}{2} \, + \, B_2 = \frac{1}{6} \, , \, B_3=0 \, , \, B_4 = -\frac{1}{30} , \cdots }$

Actually continuing with this we deduce

• $\displaystyle B_{2k+1}=0 \,\, \, \, \, \,\,\, \, \, \, \, \forall \,\, \, \, \, \, k\in \mathbb{Z}^+$
• Every Bernoulli number depends on all the numbers before it so it can be defined recursively .
• $\displaystyle B_{2k} \, , \, k\geq 1$ are alternating .

Now according to Euler we have the following interesting result :

$\displaystyle \zeta(2k) \, = \, (-1)^{k-1} B_{2k} \frac{2^{2k-1}}{(2k)!}{\pi}^{2k}$

PROOF

Let us start by the following definition due to Euler

$\displaystyle \frac{\sin(z)}{z} = \prod_{n\geq1} \left(1-\frac{z^2}{n^2 \, \pi^2} \right)$

Take the logarithm to both sides

$\displaystyle \ln(\sin(z)) - \ln(z) = \sum_{n\geq1} \ln \left(1-\frac{z^2}{n^2 \, \pi^2} \right)$

By differentiation with respect to z

$\displaystyle \cot(z) -\frac{1}{z} = \sum_{n\geq1} \frac{-2 \frac{z}{n^2 \, \pi^2 }}{1-\frac{z^2}{n^2 \, \pi^2}}$

By simple algaberic manipulation we have

$\displaystyle z\cot(z) = 1+2\sum_{n\geq1} \frac{z^2}{z^2-n^2 \, \pi^2}$

$\displaystyle z\cot(z) = 1-2\sum_{n\geq1} \frac{z^2}{ n^2 \, \pi^2} \left(\frac{1}{1-\frac{z^2}{\pi^2 \, n^2}} \right)$

Now using the power series expansion

$\displaystyle \frac{1}{1-\frac{z^2}{\pi^2 \, n^2}}= \sum_{k\geq 0} \frac{1}{n^{2k}\, \pi^{2k}} z^{2k}$ converges for $\displaystyle |z|< \pi \, n$

$\displaystyle \frac{z^2}{ n^2 \, \pi^2}\left(\frac{1}{1-\frac{z^2}{\pi^2 \, n^2}} \right) = \sum_{k\geq 1}\frac{1}{n^{2k}\, \pi^{2k}} z^{2k}$ notice the change in the index !

So now the sums becomes

$\displaystyle z\cot(z) = 1-2\sum_{n\geq1} \sum_{k \geq 1} \frac{1}{n^{2k}} \, \frac{z^{2k}}{\pi^{2k}}$

Now if we invert the order of summation we have

$\displaystyle z\cot(z) = 1-2\sum_{k\geq1} \sum_{n \geq 1} \frac{1}{n^{2k}} \, \frac{z^{2k}}{\pi^{2k}}$ (Wow do you notice !)

$\displaystyle z\cot(z) = 1-2\sum_{k\geq1} \frac{\zeta(2k) }{\pi^{2k}}z^{2k} \text{ }\cdots(1)$

Euler didn't stop here , he used power series estimation for $\displaystyle z\cot(z)$ using the Bernoulli numbers

staring by the equation $\displaystyle \frac{x}{e^x-1} = \sum_{k\geq 0} \frac{B_k}{k!}x^k$

by putting $\displaystyle x=2iz$ we have

$\displaystyle \frac{2iz}{e^{2iz}-1} = \sum_{k\geq0} \frac{B_k}{k!}{(2iz)}^k$

Which can be reduced directly to the following by noticing that $\displaystyle B_{2k+1}=0, \,\,\, k>0$

$\displaystyle z \cot(z) = 1- \sum_{k\geq 1}(-1)^{k-1} B_{2k} \frac{2^{2k}}{(2k)!}z^{2k} \text{ } \cdots(2)$

The result is immediate by comparing (1) and (2) $\displaystyle \square$

[HW] find $\displaystyle \zeta(4) \, ,\, \zeta(6) \, , \, B_5 \, , \, B_6$

To be continued ...

4.4.3. Hurwitz zeta and polygamma functions

Hurwitz zeta is a generalization of the zeta function by adding a parameter . This intimate relation between the two functions arises from multiple differentiations of the the digamma function .

Let us first start by defining the Hurwitz zeta function

Definition

$\displaystyle \zeta(a,z) \, = \, \sum_{n\geq 0} \frac{1}{(n+z)^a}$

Note : according to this definition we have $\displaystyle \zeta(a,1) = \zeta(a)$

Let us define the polygamma function as the function produced by differentiating the digamma function and it is often denoted by $\displaystyle \psi_{n}(z) \,\,\,\, \forall \,\, n\geq 0$ . We define the digamma function by setting $n=0$ so it's denoted by $\psi_0(z)$ .

Other values can be found by the following recurrence relation $\displaystyle \psi'_{n}(z) = \psi_{n+1}(z)$ , so we have $\displaystyle \psi_{1}(z) = \psi'_{0}(z)$

We have the following relation between Hurwitz zeta and the polygamma function

DEFINITION

$\displaystyle \psi_{n}(z) \, = \, (-1)^{n+1}n!\,\zeta(n+1,z) \,\,\, \, \forall \,\, n\geq 1$

PROOF

We have already proved the following relation

$\displaystyle \psi_{0}(z) = -\gamma -\frac{1}{z}+ \sum_{n\geq 1}\frac{z}{n(n+z)}$

This can be written as the following

$\displaystyle \psi_{0}(z) = -\gamma + \sum_{k\geq 0}\frac{1}{k+1}-\frac{1}{k+z}$

By differentiating with respect to $z$

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

$\displaystyle \psi_{2}(z) = -2\sum_{k\geq 0}\frac{1}{(k+z)^3}$

$\displaystyle \psi_{3}(z) = 2 \cdot 3 \, \sum_{k\geq 0}\frac{1}{(k+z)^4}$

$\displaystyle \psi_{4}(z) = -2 \cdot 3 \cdot 4 \,\sum_{k\geq 0}\frac{1}{(k+z)^5}$

$\displaystyle \text{ }\cdot$

$\displaystyle \text{ }\cdot$

$\displaystyle \text{ }\cdot$

$\displaystyle \psi_{n}(z) = (-1)^{n+1}n!\,\sum_{k\geq 0}\frac{1}{(k+z)^{n+1}}$

We realize the RHS is just the Hurwitz zeta function

$\displaystyle \psi_{n}(z) = (-1)^{n+1}n!\,\zeta(n+1,z)$

As required to prove $\displaystyle \square$.

By setting $z=1$ we have an equation in terms of the ordinary zeta function

$\displaystyle \psi_{n}(1) = (-1)^{n+1}n!\,\zeta(n+1)$

Now since we already proved in the preceding section that

$\displaystyle \zeta(2k) \, = \, (-1)^{k-1} B_{2k} \frac{2^{2k-1}}{(2k)!}{\pi}^{2k}$

we can easily verify the following

$\displaystyle \psi_{2k-1}(1)= (2k-1)!\, (-1)^{k-1} B_{2k} \frac{2^{2k-1}}{(2k)!}{\pi}^{2k}$

$\displaystyle \psi_{2k-1}(1)= (-1)^{k-1} B_{2k} \frac{2^{2k-2}}{k}{\pi}^{2k} \,\,\, \, , \,\, k\geq 1$

This can be used to evaluate some values for the polygamma function

$\displaystyle \psi_{1}(1)= \frac{\pi^2}{6} \,\, , \,\, \psi_{3}(1) = \frac{\pi^4}{15}$

Other values can be evaluated in terms of the zeta function

$\displaystyle \psi_{2}(1)= -2 \zeta(3) \,\, , \,\, \psi_{4}(1) = -24 \zeta(5)$

To Be Continued ...

4.4.4. integrals involve zeta computations

Prove that

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

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 ...

4.4.5 Dirichlet eta function( an alternating form )

Dirichlet eta function is the alternating form of the zeta function

DEFINITION :

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

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

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

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

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

We shall look at the polylogarithms in the next thread

4.Integration using special functions (continued)

4.5. Polylogarithm

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

DEFINITION

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

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

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

There is an interesting recursive representation of this function

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

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

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

Square formula

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

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 ...

4.5.1.Dilogarithms

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

DEFINITION

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

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

Some Functional equations

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

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 .

Another function equation

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

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 ...

4.5.1.Dilogarithms (continued)

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

Yet another functional equation

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

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

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

Well, that doesn't end here , prove

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

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

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

Prove

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

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 $\displaystyle \square$

[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 ...