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ZaidAlyafey

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Jan 17, 2013
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(AIT) Cosine Integral function

4.Integration using special functions (continued)


4.12. Cosine Integral function


[HR][/HR]

Define

$$\mathrm{ci}(x) =- \int^\infty_x \frac{\cos(t)}{t}\,dt $$

A related function is the following

$$\mathrm{Cin}(x) = \int^x_0 \frac{1-\cos(t)}{t}\,dt$$

[HR][/HR]

The derivative is

$$\frac{d}{dx}\mathrm{ci}(x) = \frac{\cos(x)}{x}$$

The integral

$$\int \mathrm{ci}(x) \,dx = x\mathrm{ci}(x)-\sin(x)$$

Prove the following

$$\mathrm{Cin}(x) = -\mathrm{ci}(x)+\log(x)+\gamma$$

Start by

$$\mathrm{Cin}(x) = \int^x_0 \frac{1-\cos(t)}{t}\,dt$$

Rewrite as

$$\mathrm{Cin}(x) =\int^\infty_0 \frac{1-\cos(t)}{t}\,dt- \int^\infty_x \frac{1-\cos(t)}{t}\,dt$$

Which simplifies to

$$\mathrm{Cin}(x) =\lim_{z \to \infty } \left[\int^z_0 \frac{1-\cos(t)}{t}\,dt- \log(z)\right] -\mathrm{ci}(x)+\log(x)$$

The limit goes to the Euler Maschorinit constant

$$\mathrm{Cin}(x) =\gamma -\mathrm{ci}(x)+\log(x)$$

Find the integral

$$\int^\infty_0 \mathrm{ci}(x) \, \cos(px) \, dx $$

Using integration by parts we get

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

Taking the limits

$$\lim_{x \to 0} \frac{\mathrm{ci}(x) \sin(px)}{p}=0$$

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

Hence we get

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

The integral

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

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

$$I = \frac{\pi}{4}+\frac{\pi}{4} = \frac{\pi}{2}$$

If $p-1<0$

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

If $p=1$ we have

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

Finally we get

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

Find the integral for, $p>1$

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

Let

$$I(p) = \int^\infty_0 \mathrm{ci}(px) \mathrm{ci}(x)\,dx$$

Differentiate with respect to $p$

$$I'(p) = \frac{1}{p}\int^\infty_0 \mathrm{cos}(px) \mathrm{ci}(x)\,dx$$

If $p>1$ from the previous example we conclude that

$$I'(p) = \frac{1}{p}\left(\frac{-\pi}{2p}\right) = -\frac{\pi}{2p^2}$$

Integrate with respect to $p$

$$I(p) = \frac{\pi}{2p} + C$$

Take the limit $p \to \infty$, so $C = 0$.
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Re: (AIT) Cosine Integral function

Prove that

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

Use the integral representation

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

Let $t = yx$

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

Switch the integrals

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

Using the Mellin transform we get

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

Prove that

$$\int^\infty_0 \mathrm{ci}(x) \log(x) \,dx = \frac{\pi}{2}$$

From the previous example we know

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

Differentiate with respect to $\alpha$

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

Take the limit $\alpha \to 1$

$$\int^\infty_0 \mathrm{ci}(x) \log(x)\,dx = 0-0+\frac{\pi}{2}\sin\left( \frac{\pi}{2}\right) = \frac{\pi}{2}$$

Prove that

$$\int^\infty_0 \mathrm{ci}(x) e^{-\alpha x}\,dx = -\frac{1}{\alpha}\log\sqrt{1+\alpha^2}$$

Use the integral representation

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

Switch the integrals

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

Use the Laplace transformation

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

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Re: (AIT) Cosine Integral function

Find the integral

$$\int^\infty_0 \mathrm{si}(qx) \mathrm{ci}(x) \, dx$$

Using the integral representation

$$\int^\infty_0 \mathrm{si}(qx) \int^\infty_1 \frac{\cos(yx)}{y}\,dy\, dx$$

Switch the integrals

$$\int^\infty_1 \frac{1}{y} \int^\infty_0 \mathrm{si}(qx) \cos(yx)\, dx\,dy$$

We also showed that

$$\frac{1}{2}\int^\infty_1 \frac{1}{y^2}\log\left( \frac{y+q}{y-q}\right)\,dy$$

We can prove that the anti-derivative

$$\left[\frac{\log(y)}{q}-\frac{1}{2q}\log(y^2-q^2)-\frac{1}{2y}\log\left( \frac{y+q}{y-q}\right)\right]^\infty_1$$

Which simplifies

$$\left[-\frac{1}{2q}\log\left(\frac{y^2-q^2}{y^2} \right)-\frac{1}{2y}\log\left( \frac{y+q}{y-q}\right)\right]^\infty_1$$

The limits

$$\lim_{y \to \infty }\frac{1}{2q}\log\left(\frac{y^2-q^2}{y^2} \right)+\frac{1}{2y}\log\left( \frac{y+q}{y-q}\right) = 0 $$

The limit $y \to 1$
$$\frac{1}{2q}\log\left(1-q^2 \right)+\frac{1}{2}\log\left( \frac{1+q}{1-q}\right)$$

Can be written as

$$\frac{1}{4q}\log\left(1-q^2 \right)^2+\frac{1}{4}\log\left( \frac{1+q}{1-q}\right)^2$$

Prove that

$$\int^\infty_0 \frac{\mathrm{ci}(\alpha x)}{x+\beta}\,dx = -\frac{1}{2}\left\{ \mathrm{si(\alpha \beta)^2+\mathrm{ci}(\alpha \beta)^2}\right\}$$

Let the following

$$I(\alpha) = \int^\infty_0 \frac{\mathrm{ci}(\alpha x)}{x+\beta}\,dx $$

Differentiate with respect to $\alpha$

$$I'(\alpha) = \frac{1}{\alpha}\int^\infty_0 \frac{\cos(\alpha x)}{x+\beta}\,dx $$

Let $x+\beta = t$

$$I'(\alpha) = \frac{1}{\alpha}\int^\infty_\beta \frac{\cos(\alpha (t-\beta)}{t}\,dt $$

Use sum to product rules

$$I'(\alpha) = \frac{1}{\alpha}\int^\infty_\beta \frac{\cos(\alpha t) \cos(\alpha \beta) + \sin(\alpha t) \sin(\alpha \beta)}{t}\,dt $$

Separate the integrals

$$I'(\alpha) = \frac{\cos(\alpha \beta)}{\alpha}\int^\infty_\beta \frac{\cos(\alpha t) }{t}\,dt+ \frac{\sin(\alpha \beta)}{\alpha}\int^\infty_\beta\frac{\sin(\alpha t) }{t}\,dt $$

This simplifies to

$$I'(\alpha) = -\frac{\cos(\alpha \beta)}{\alpha}\mathrm{ci}(\alpha \beta)-\frac{\sin(\alpha \beta)}{\alpha}\mathrm{si}(\alpha \beta) $$

Integrate with respect to $\alpha$


$$I(\alpha) = -\frac{1}{2}\left\{ \mathrm{si(\alpha \beta)^2+\mathrm{ci}(\alpha \beta)^2}\right\}+ C$$

If $\alpha \to \infty $ we have $C =0$.
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
(AIT) Logarithm integral function

4.Integration using special functions (continued)

4.13. Logarithm Integral function


[HR][/HR]

Define


$$\mathrm{li}(x) = \int^x_0 \frac{dt}{\log(t)}$$

[HR][/HR]

Prove that

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

Let the following

$$I(a) = \int^1_0 \int^x_0 \frac{e^{-a\log(t)}dt}{\log(t)}\,dx $$

Differentiate with respect to $a$

$$I'(a) = - \int^1_0 \int^x_0 t^{-a}\,dt\,dx$$

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

Which reduces to

$$I'(a) = \frac{1}{(a-1)(2-a)} = \frac{1}{2-a}-\frac{1}{1-a}$$

Integrate with respect to $a$

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

Take the limit $a \to \infty$ we get $C = 0$

$$\int^1_0 \int^x_0 \frac{e^{-a\log(t)}dt}{\log(t)}\,dx =\log\left( \frac{1-a}{2-a}\right)$$

Let $a \to 0$

$$\int^1_0 \mathrm{li}(x)\,dx =\log\left( \frac{1}{2}\right) = -\log(2)$$

Find the following

$$\int^1_0 x^{p-1}\mathrm{li}(x)\,dx$$

Let the following

$$I(a) = \int^1_0 x^{p-1}\int^x_0 \frac{e^{-a\log(t)}dt}{\log(t)}\,dx $$

Differentiate with respect to $a$

$$I'(a) = - \int^1_0 x^{p-1}\int^x_0 t^{-a}\,dt\,dx$$

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

Which reduces to

$$I'(a) = \frac{1}{(a-1)(p-a+1)} =\frac{1}{p} \left\{\frac{1}{p-a+1}-\frac{1}{1-a} \right\}$$

Integrate with respect to $a$

$$I(a) = \frac{1}{p}\log\left( \frac{1-a}{p-a+1}\right)+C$$

Take the limit $a \to \infty$ we get $C = 0$

$$\int^1_0 x^{p-1}\int^x_0 \frac{e^{-a\log(t)}dt}{\log(t)}\,dx =\frac{1}{p}\log\left( \frac{1-a}{p-a+1}\right)$$

Let $a \to 0$

$$\int^1_0 x^{p-1}\int^x_0 \frac{e^{-a\log(t)}dt}{\log(t)}\,dx =\frac{1}{p}\log\left( \frac{1}{p+1}\right) = -\frac{1}{p}\log\left( p+1\right)$$

Find the following


$$\int^1_0\mathrm{li}\left(\frac{1}{x}\right) \sin(a\log(x))\,dx$$

Let the following

$$I(b) = \int^1_0 \sin(a\log(x))\int^{\frac{1}{x}}_0 \frac{e^{-b\log(t)}dt}{\log(t)}\,dx $$

Differentiate with respect to $b$

$$I'(b) =-\int^1_0 \sin(a\log(x))\int^{\frac{1}{x}}_0 t^{-b}\,dt\,dx $$

$$I'(b) =\frac{1}{b-1}\int^1_0 x^{b-1}\sin(a\log(x))\,dx $$

Let $\log(x) = -t$

$$I'(b) =\frac{1}{1-b}\int^\infty_0 e^{-tb}\sin(at)\,dt $$

Now use the Laplace transform

$$I'(b) =\frac{1}{1-b}\frac{a}{a^2+b^2} $$

Integrate with respect to $b$

$$I(b) = \frac{a\log(a^2+b^2) -2a\log(b-1)+2\arctan(b/a)}{2a^2+2}+C$$

Let $b \to \infty $

$$0 = \frac{\pi}{2(a^2+1)}+C$$

Hence we have

$$\int^1_0 \sin(a\log(x))\int^{\frac{1}{x}}_0 \frac{e^{-b\log(t)}dt}{\log(t)}\,dx= \frac{a\log(a^2+b^2) -2a\log(b-1)+2\arctan(b/a)}{2a^2+2}-\frac{\pi}{2(a^2+1)}$$

Let $b \to 0$

$$\int^1_0 \sin(a\log(x))\mathrm{li}(x)\,dx= \frac{a\log(a^2)}{2a^2+2}-\frac{\pi}{2(a^2+1)} = \frac{1}{a^2+1}\left(a\log(a) -\frac{\pi}{2}\right)$$
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Re: (AIT) Logarithm integral function

Find the following integral

$$\int^1_0\frac{\mathrm{li}(x)}{x} \log^{p-1}\left(\frac{1}{x}\right)\,dx $$

Let the following

$$I(a) = \int^1_0\frac{1}{x} \left[\int^x_0 \frac{e^{-a\log(t)}}{\log(t)}dt \right]\log^{p-1}\left(\frac{1}{x}\right)\,dx $$

Differentiate with respect to $a$


$$I'(a) = -\int^1_0\frac{1}{x} \left[\int^x_0t^{-a}dt \right]\log^{p-1}\left(\frac{1}{x}\right)\,dx $$

$$I'(a) = \frac{1}{a-1}\int^1_0x^{-a}\log^{p-1}\left(\frac{1}{x}\right)\,dx $$

Let $-\log(x) = t$

$$I'(a) = \frac{-1}{1-a}\int^\infty_0 e^{-(1-a)t}t^{p-1}\,dx $$

$$I'(a) = \frac{-1}{1-a}\frac{\Gamma(p)}{(1-a)^p} = -\frac{\Gamma(p)}{(1-a)^{p+1}} $$

Integrate with respect to $a$

$$I(a) = -\frac{\Gamma(p)}{p(1-a)^p}$$

Let $a \to 0 $, Hence

$$\int^1_0\frac{\mathrm{li}(x)}{x} \log^{p-1}\left(\frac{1}{x}\right)\,dx = -\frac{\Gamma(p)}{p} $$

Prove that


$$\int^\infty_1 \mathrm{li}\left(\frac{1}{x}\right) \log^{p-1}(x)\,dx = -\frac{\pi}{\sin(\pi p)} \Gamma(p)$$

Let the following

$$I(a) = \int^\infty_1 \mathrm{li}\left(x^{-a}\right) \log^{p-1}(x)\,dx = -\frac{\pi}{\sin(\pi p)} \Gamma(p)$$

Differentiate with respect to $a$

$$\frac{d}{da}\mathrm{li}\left(x^{-a}\right)=\frac{d}{da} \int^{x^{-a}}_0\frac{dt}{\log(t)}= \frac{x^{-a}}{a}$$

Hence we have

$$I'(a) = \frac{1}{a}\int^\infty_1 x^{-a}\log^{p-1}(x)\,dx $$

Let $\log(x) = t$

$$I'(a) = \frac{1}{a}\int^\infty_0 e^{-(a-1)t}t^{p-1}\,dt $$

Using the Laplace transform

$$I'(a) =\Gamma(p) \frac{1}{a(a-1)^p} $$

Take the integral

$$\int^\infty_1 I'(a)da = \Gamma(p)\int^\infty_1 \frac{1}{a(a-1)^p} da $$

The left hand-side

$$I(\infty)-I(1) = \Gamma(p)\int^\infty_1 \frac{1}{a(a-1)^p} da $$

Now since $I(\infty) = 0$

$$I(1) = -\Gamma(p)\int^\infty_1 \frac{1}{a(a-1)^p} da $$

Which implies that

$$\int^\infty_1 \mathrm{li}\left(\frac{1}{x}\right) \log^{p-1}(x)\,dx = -\Gamma(p)\int^\infty_1 \frac{1}{a(a-1)^p} da$$

Now let $t = a-1$

$$ \int^\infty_0 \frac{t^{-p}}{t+1} dt$$

Using the beta integral $x+y = 1$ and $x-1 = -p$ which implies that $x = 1-p,y=p$

Hence we have

$$ \int^\infty_0 \frac{t^{-p}}{t+1} dt = \beta(p,1-p) = \Gamma(p)\Gamma(1-p) = \frac{\pi}{\sin(\pi p)}$$

Finally we get

$$\int^\infty_1 \mathrm{li}\left(\frac{1}{x}\right) \log^{p-1}(x)\,dx = -\frac{\pi}{\sin(\pi p)} \Gamma(p)$$
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
(AIT) Clausen functions

4.Integration using special functions (continued)

4.14. Clausen functions

[HR][/HR]
$$\mathrm{cl}_m(\theta) = \begin{cases}
\sum_{k=1}^\infty \frac{\sin(k\theta)}{k^m} & m \text{ is even} \\
\sum_{k=1}^\infty \frac{\cos(k\theta)}{k^m} & m \text{ is odd}
\end{cases}$$
[HR][/HR]

Duplication formula

$$\mathrm{cl}_m(2 \theta) = 2^{m-1}(\mathrm{cl}_m(\theta)-(-1)^m \mathrm{cl}_m(\pi-\theta))$$

Proof

If $m$ is even then

$$\mathrm{cli}_m(\pi-\theta) = \sum_{k=1}^\infty \frac{\sin(k\pi -k\theta)}{k^m} = -\sum_{k=1}^\infty (-1)^k \frac{\sin(k\theta)}{k^m} $$

$$\sum_{k=1}^\infty (-1)^k \frac{\sin(k\theta)}{k^m} + \sum_{k=1}^\infty \frac{\sin(k\theta)}{k^m} = \frac{1}{2^{m-1}}\sum_{k=1}^\infty \frac{\sin(2k\theta)}{k^m} $$

This implies that

$$\mathrm{cl}_m(2 \theta) = 2^{m-1}(\mathrm{cl}_m(\theta)- \mathrm{cl}_m(\pi-\theta))$$

If $m$ is odd then

$$\mathrm{cli}_m(\pi-\theta) = \sum_{k=1}^\infty \frac{\cos(k\pi -k\theta)}{k^m} = \sum_{k=1}^\infty (-1)^k \frac{\cos(k\theta)}{k^m} $$

$$\sum_{k=1}^\infty \frac{\cos(k\theta)}{k^m} + \sum_{k=1}^\infty (-1)^k \frac{\cos(k\theta)}{k^m} = \frac{1}{2^{m-1}}\sum_{k=1}^\infty \frac{\sin(2k\theta)}{k^m} $$

Which implies that

$$\mathrm{cl}_m(2 \theta) = 2^{m-1}(\mathrm{cl}_m(\theta)+ \mathrm{cl}_m(\pi-\theta))$$

Collecting the results we have

$$\mathrm{cl}_m(2 \theta) = 2^{m-1}(\mathrm{cl}_m(\theta)-(-1)^m \mathrm{cl}_m(\pi-\theta))$$

Find the integral


$$\int^\pi_0 \mathrm{cl}_{m}(\theta) d\theta \,\,\,\,\,\,\,\,\,\, m \text{ is even}$$

Using the series representation

$$\int^\pi_0 \sum_{k=1}^\infty \frac{\sin(k\theta)}{k^{m}} d\theta$$

Swap the integral and the series

$$\sum_{k=1}^\infty \frac{1}{k^{m}}\int^\pi_0 \sin(k\theta) d\theta$$

The integral

$$\int^\pi_0 \cos(k\theta) d\theta =-\left[ \frac{1}{k}\sin(k\theta) \right]^\pi_0=\frac{-(-1)^k+1}{k}$$

We get the summation

$$\sum_{k=1}^\infty \frac{-(-1)^k+1}{k^{m+1}} =\zeta(m+1)+\eta(m+1)$$

Now use that

$$\eta(s) = (1-2^{1-s})\zeta(s)$$

$$\sum_{k=1}^\infty \frac{-(-1)^k+1}{k^{m+1}} =\zeta(m+1)+(1-2^{-m})\zeta(m+1) = \zeta(m+1)(2-2^{-m})$$
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Re: (AIT) Clausen functions

Find the integral for $m$ is even

$$\int^\infty_0 \mathrm{cl}_m(\theta) e^{-n\theta}\, d\theta$$

Using the series representation

$$\int^\infty_0 \sum_{k=1}^\infty \frac{\sin(k\theta)}{k^m} e^{-n\theta}\, d\theta$$

Swap the integral and the series

$$\sum_{k=1}^\infty \frac{1}{k^m}\int^\infty_0 \sin(k\theta) e^{-n\theta}\, d\theta$$

Using the Laplace transform we have

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

Add and subtract $k^2$ and divide by $n^2$

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

Distribute the numerator

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

Continue this approach to conclude that

$$\sum_{i= 1}^j(-1)^{i-1}\frac{1}{n^{2i}}\zeta(m-(2i-1))+\frac{(-1)^j}{n^{2j}}\sum_{k=1}^\infty\frac{1}{k^{m-(2j+1)}(k^2+n^2)}$$

Let $m-2j-1 = 1$ which implies that $j = m/2-1$

$$\sum_{i= 1}^{m/2-1}(-1)^{i-1}\frac{1}{n^{2i}}\zeta(m-(2i-1))+\frac{(-1)^{m/2-1}}{n^{m-2}}\sum_{k=1}^\infty\frac{1}{k(k^2+n^2)}$$

Now let us look at the sum

$$\sum_{k=1}^\infty\frac{1}{k(k^2+n^2)} = \sum_{k=1}^\infty\frac{1}{2ink}\left\{ \frac{1}{k-in}-\frac{1}{k+in}\right\}$$

Which can be written as

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

According to the digamma function

$$\sum_{k=1}^\infty\frac{1}{k(k^2+n^2)} = \frac{1}{2n^2}\left\{ \gamma+\psi(1+in)+\psi(1-in)+\gamma\right\}$$

which simplifies to

$$\sum_{k=1}^\infty\frac{1}{k(k^2+n^2)} = \frac{\psi(1-in)+\psi(1+in)+2\gamma}{2n^2}$$

Now we we can verify $\psi(1-in) = \overline{\psi(1+in)}$ which suggests that

$$\psi(1+in) +\psi(1-in) = 2\Re\left\{\psi(1+in) \right\} $$

Hence we have the sum

$$\sum_{k=1}^\infty\frac{1}{k(k^2+n^2)} = \frac{2\Re\left\{\psi(1+in) \right\}+2\gamma}{2n^2}=\frac{\Re\left\{\psi(1+in) \right\}+\gamma)}{n^2}$$

This concludes to

$$\sum_{j= 1}^{m/2-1}(-1)^{j-1}\frac{\zeta(m-(2i-1))}{n^{2j}}+(-1)^{m/2-1}\frac{\Re\left\{\psi(1+in) \right\}+\gamma}{n^m}$$

Find the following integral

$$\int^\infty_0 \mathrm{cl}_m(\theta) \theta^{n-1}\, d\theta$$

If $m$ is even

$$\int^\infty_0 \sum_{k=1}^\infty \frac{\sin(k\theta)}{k^m}\theta^{n-1}\, d\theta$$

Using the series representation

$$\sum_{k=1}^\infty \frac{1}{k^m}\int^\infty_0 \sin(k\theta) \theta^{n-1}d\theta$$

Let $\phi = k\theta $

$$\sum_{k=1}^\infty \frac{1}{k^{m+n}}\int^\infty_0 \sin(\phi) \phi^{n-1}d\phi$$

Now using the mellin transform

$$\zeta(m+n) \sin\left( \frac{\pi}{2}n\right)\Gamma(n)$$

If $m$ is odd
$$\zeta(m+n) \cos\left( \frac{\pi}{2}n\right)\Gamma(n)$$
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Re: (AIT) Clausen functions

Clausen Integral function


$$\mathrm{cl}_2(\theta) = -\int^\theta_0 \log\left[2\sin\left(\frac{\phi}{2}\right)\right]d\phi$$

Start by the following

$$\mathrm{Li}_2(e^{i\theta}) = \sum_{k=1}^\infty \frac{e^{ik\theta}}{k^2}= \sum_{k=1}^\infty \frac{\cos(k\theta)}{k^2}+i\sum_{k=1}^\infty \frac{\sin(k\theta)}{k^2} $$

By the integral definition of the dilogarithm

$$\mathrm{Li}_2(e^{i\theta}) -\zeta(2)= -\int^{e^{i\theta}}_1 \frac{\log(1-x)}{x}\,dx$$

Let $x = e^{i\phi}$

$$\mathrm{Li}_2(e^{i\theta}) -\zeta(2)= -i\int^{\theta}_0 \log(1-e^{i\phi})d \phi$$

Let us look at the following

$$1-e^{i\phi} = 1-\cos(\phi)-i\sin(\phi) = 2\sin^2(\phi/2)-2i\sin(\phi/2)\cos(\phi/2)$$

Which simplifies to

$$1-e^{i\phi} =2\sin(\phi/2)\left[\sin(\phi/2)-i\cos(\phi/2)\right] = 2\sin(\phi/2) e^{-(i/2) (\pi-\phi)}$$

Hence our integral

$$\mathrm{Li}_2(e^{i\theta}) -\zeta(2)= -i\int^{\theta}_0 \log \left[2\sin(\phi/2) e^{-(i/2) (\pi-\phi)}\right]d \phi$$

Use the complex integral properties

$$\mathrm{Li}_2(e^{i\theta}) -\zeta(2)= -i\int^{\theta}_0 \log \left[2\sin(\phi/2)\right]d\phi +\frac{1}{4}(\pi-\theta)^2-\frac{1}{4}\pi^2$$

By equating the imaginary parts we have our result.

We can see the special value
$$\mathrm{cl}_2\left(\frac{\pi}{2}\right) = \sum_{k=1}^\infty \frac{\sin(k\pi/2)}{k^2}= \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)^2} = G$$

Where $G$ is the Catalan's constant.

Prove the following


$$\mathrm{cl}_2(2\theta) =2\mathrm{cl}_2(\theta)-\mathrm{cl}_2(\pi -\theta) $$

Use the integral representation

$$\mathrm{cl}_2(2\theta) = -\int^{2\theta}_0 \log\left[2\sin\left(\frac{t}{2}\right)\right]dt$$

Let $t= 2\phi$

$$-2\int^{\theta}_0 \log\left[2\sin\left(\phi\right)\right]d\phi$$

Use double angle identity

$$-2\int^{\theta}_0 \log\left[4\sin\left(\frac{\phi}{2}\right) \cos\left(\frac{\phi}{2}\right)\right]d\phi$$

Separate the logarithms

$$-2\int^{\theta}_0 \log\left[2\sin\left(\frac{\phi}{2}\right) \right]d\phi-2\int^{\theta}_0 \log\left[2\cos\left(\frac{\phi}{2}\right) \right]d\phi$$

We can verify that

$$\mathrm{cl}_2(\pi -\theta ) = \int^{\theta}_0 \log\left[2\cos\left(\frac{\phi}{2}\right) \right]d\phi$$

Hence

$$\mathrm{cl}_2(2\theta) =2\mathrm{cl}_2(\theta)-\mathrm{cl}_2(\pi -\theta)$$

Using that

$$\mathrm{cl}_2(3\pi) =2\mathrm{cl}_2\left(\frac{3\pi}{2}\right)-2\mathrm{cl}_2\left(-\frac{\pi}{2}\right) $$

Since $\mathrm{cl}_2(3\pi) = 0$
$$\mathrm{cl}_2\left(\frac{3\pi}{2}\right)=\mathrm{cl}_2\left(-\frac{\pi}{2}\right) = - \mathrm{cl}_2\left(\frac{\pi}{2}\right)= -G $$

Prove that

$$\int^{2\pi}_0 \mathrm{cl}_2(x)^2dx = \frac{\pi^5}{90}$$

Using the series representation

$$\sum_{k=1}^\infty \sum_{n=1}^\infty \frac{1}{(nk)^2}\int^{2\pi}_0 \sin(kx) \sin(nx) dx$$

Consider the integral

$$\int_0^{2\pi} \sin(kx)\sin(nx)\,dx = \frac{1}{2}\int_0^{2\pi} \cos((k-n)x)-\cos((k+n)x)\,dx $$

Consider two cases

If $n = k $ then

$$\frac{1}{2}\int_0^{2\pi} 1-\cos(2nx)\,dx = \pi $$

If $n \neq k$

$$\frac{1}{2}\int_0^{2\pi} \cos((k-n)x)-\cos((k+n)x)\,dx =\frac{1}{2}\left[\frac{\sin((k-n)x)}{k-n}-\frac{\sin((k+n)x)}{k+n} \right]^{2\pi}_0 =0 $$

Hence we have

$$\int_0^{2\pi} \sin(kx)\sin(nx)\,dx=\begin{cases} 0 & n \neq k\\ \pi & n=k\end{cases}$$

We can write the series as

$$\sum_{n=1}^\infty \sum_{k=1}^\infty \frac{1}{(nk)^2} = \sum_{n\neq k}^\infty \frac{1}{(nk)^2}+ \sum_{n=1}^\infty \frac{1}{n^4} $$

Now since the integral $n\neq k $ goes to zero the result reduces to

$$\pi \sum_{n=1}^\infty \frac{1}{n^4} =\pi \zeta(4) = \frac{\pi^5}{90} $$
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
(AIT) Barnes G function

4.Integration using special functions (continued)


4.15. Barnes G function


[HR][/HR]
Define the following

$$ G(z+1)=(2\pi)^{z/2}\exp\left(-\frac{z+z^2(1+\gamma)}{2}\right)\prod_{n=1}^{\infty}\left\{\left(1+\frac{z}{n} \right)^n\exp\left(\frac{z^2}{2n}-z\right) \right\}$$

[HR][/HR]

Difference formula

$$G(z+1) = \Gamma(z) G(z)$$

From the series representation we have

$$ \frac{G(z+1)}{G(z)} = \sqrt{2 \pi} \exp \left(-z - \gamma z + \frac{\gamma}{2} \right) \prod_{k=1}^{\infty} \left(\frac{k+z}{k+z-1} \right)^{k} \exp \left( \frac{2z-1-2k}{2k} \right). $$

This can be written as

$$\frac{G(z+1)}{G(z)} = z\sqrt{2 \pi} \exp \left(-z + \frac{\gamma}{2} \right) \prod_{k=1}^{\infty} \left(\frac{k+z}{k+z-1} \right)^{k} \exp \left( -\frac{1+2k}{2k} \right)\left(1+\frac{z}{k}\right)\\ \frac{e^{- \gamma z }}{z} \prod_{k=1}^{\infty} \left(1+\frac{z}{k} \right)^{-1} e^{\left( \frac{z}{k} \right)}$$

Which simplifies to

$$\frac{G(z+1)}{G(z)} = z\Gamma(z)\sqrt{2 \pi} \exp \left(-z + \frac{\gamma}{2} \right) \prod_{k=1}^{\infty} \left(\frac{k+z}{k+z-1} \right)^{k} \exp \left( -\frac{1+2k}{2k} \right)\left(1+\frac{z}{k}\right)$$

It suffices to prove that

$$ z\sqrt{2 \pi} \exp \left(-z + \frac{\gamma}{2} \right) \prod_{k=1}^{\infty} \left(\frac{k+z}{k+z-1} \right)^{k} \exp \left( -\frac{1+2k}{2k} \right)\left(1+\frac{z}{k}\right) = 1$$

or

$$ \prod_{k=1}^{\infty} \left(\frac{k+z}{k+z-1} \right)^{k} \exp \left( -\frac{1+2k}{2k} \right)\left(1+\frac{z}{k}\right) = \frac{\exp \left(z -\frac{\gamma}{2} \right)}{z\sqrt{2 \pi}}$$


Start by

$$\lim_{N \to \infty }\prod_{k=1}^{N} \left(\frac{k+z}{k+z-1} \right)^{k} \exp \left( -\frac{1+2k}{2k} \right)\left(1+\frac{z}{k}\right) $$

Notice

\begin{align}
\prod_{k=1}^{N} \left(\frac{k+z}{k+z-1} \right)^{k} \left(1+\frac{z}{k}\right)&= \frac{\prod_{k=1}^{N} (k+z)^k \prod_{k=1}^{N}\left(1+\frac{z}{k}\right)}{\prod_{k=1}^{N}(k+z-1)^k} \\[1em]
&= \frac{\prod_{k=1}^{N} (k+z)^k \prod_{k=1}^{N} (k+z)}{zN!\prod_{k=1}^{N-1}(k+z)^{k+1}} \\[1em]
&= \frac{(N+z)^{N+1}\prod_{k=1}^{N-1} (k+z)^k\prod_{k=1}^{N-1} (k+z)}{zN!\prod_{k=1}^{N-1}(k+z)^{k+1}}\\[1em]
&= \frac{(N+z)^{N+1}\prod_{k=1}^{N-1} (k+z)^{k+1}}{zN!\prod_{k=1}^{N-1}(k+z)^{k+1}}\\[1em]
&= \frac{(N+z)^{N+1}}{zN!}
\end{align}

The second product

$$\prod_{k=1}^{N} \exp \left( -\frac{1+2k}{2k} \right) =\exp \left( -\sum_{k=1}^N\frac{1+2k}{2k} \right) = e^{-\frac{1}{2}H_N-N} $$

Hence we have the following
$$e^{-\frac{1}{2}H_N-N}\times \frac{(N+z)^{N+1}}{z N! } $$

According to Stirling formula we have

$$e^{-\frac{1}{2}H_N-N}\frac{(N+z)^{N+1}}{z N! } \sim e^{-\frac{1}{2}H_N-N}\frac{(N+z)^{N+1}}{z(N/e)^N}\times \frac{1}{\sqrt{2\pi N}} $$

By some simplifications we have

$$\frac{e^{-\frac{1}{2}(H_N-\log N)}}{z}\left(1+\frac{z}{N}\right) \times \left(1+ \frac{z}{N}\right)^N\times \frac{1}{\sqrt{2\pi}} \sim \frac{\exp\left(-\frac{\gamma}{2} +z\right)}{z\sqrt{2\pi}}$$

Where we used that

$$\lim_{n \to \infty }H_n -\log(n)= \gamma$$

Reflection formula


$$\log \left\{\frac{G(1-z)}{G(1+z)} \right\} = z \log\left(\frac{\sin(\pi z)}{\pi}\right) + \frac{\mathrm{cl}_2(2\pi z)}{2\pi}$$

Start by the series expansion

$$\frac{G(1-z)}{G(1+z)} = \frac{(2\pi)^{-z/2}\exp\left(\frac{z-z^2(1+\gamma)}{2}\right)\prod_{n=1}^{\infty}\left\{\left(1-\frac{z}{n} \right)^n\exp\left(\frac{z^2}{2n}+z\right) \right\}}{(2\pi)^{z/2}\exp\left(-\frac{z+z^2(1+\gamma)}{2}\right)\prod_{n=1}^{\infty}\left\{\left(1+\frac{z}{n} \right)^n\exp\left(\frac{z^2}{2n}-z\right) \right\}}$$

This simplifies to

$$\frac{G(1-z)}{G(1+z)} = (2\pi)^{-z}e^z\prod_{n=1}^{\infty}\frac{(n-z)^n}{(n+z)^n} e^{2z}$$

Take the log of both sides

$$\log \left\{\frac{G(1-z)}{G(1+z)} \right\} = -z \log(2\pi)+z + \log\left \{ \prod_{n=1}^{\infty}\frac{(n-z)^n}{(n+z)^n} e^{2z} \right\}$$
Let the following

$$f(z) = \log\left\{\prod_{n=1}^{\infty}\frac{(n-z)^n}{(n+z)^n} e^{2z} \right\} = \sum_{n=1}^\infty n\log(n-z)-n\log(n+z)+2z$$

Differentiate with respect to $z$

$$f'(z) =\sum_{n=1}^{\infty}\frac{-n}{n-z}-\frac{n}{n+z}+2 = \sum_{n=1}^{\infty}\frac{-n(n+z)-n(n-z)+2(n^2-z^2)}{n^2-z^2} $$

Hence we have

$$\sum_{n=1}^{\infty}\frac{-2z^2}{n^2-z^2}$$

Now we can use the following

$$z\pi \cot \pi z = 1+\sum_{n=1}^{\infty}\frac{-2z^2}{n^2-z^2}
$$

Hence we conclude that

$$f'(z) = z\pi \cot \pi z-1$$

Integrate with respect to $z$

$$f(z) = \int^z_0 x\pi \cot(\pi x) \, dx -z$$

Hence we have

$$\log \left\{\frac{G(1-z)}{G(1+z)} \right\} = -z \log(2\pi)+\int^z_0 z\pi \cot(\pi x) \, dx$$

Now use integration by parts for the integral
\begin{align}
\int^z_0 x\pi \cot(\pi x) \, dx
&=z\log(\sin\pi z)-\int^z_0 \log(\sin\pi x) dx\\
&=z\log(2\sin\pi z)-\int^z_0 \log(2\sin\pi x) dx\\
&=z\log(2\sin\pi z)-\frac{1}{2\pi }\int^{2\pi z}_0 \log\left(2\sin\frac{x}{2}\right) dx\\
&=z\log(2\sin\pi z)+\frac{\mathrm{cl}_2(2\pi z)}{2\pi}\\
\end{align}

That implies

$$\log \left\{\frac{G(1-z)}{G(1+z)} \right\} = z\log(2\sin\pi z)-z \log(2\pi)+ \frac{\mathrm{cl}_2(2\pi z)}{2\pi}$$

which implies our result.
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Re: (AIT) Barnes G function

Values at positive integers

$$G(n) = \prod^{n-1}_{k=1} \Gamma(k)$$

It can be proved by induction. For $G(1) = 1$, suppose

$$G(n) = \prod^{n-1}_{k=1} \Gamma(k)$$

We want to show

$$G(n+1) = \prod^{n}_{k=1} \Gamma(k)$$

By the difference formula

$$G(n+1) = \Gamma(n)G(n) = \Gamma(n) \prod^{n-1}_{k=1} \Gamma(k) =\prod^{n}_{k=1} \Gamma(k) $$

Loggamma integral

$$\int^z_0 \log \Gamma(x) dx =\frac{z}{2}\log(2\pi)+\frac{z(z-1)}{2}+z\log \Gamma(z)-\log G(z+1)$$

Take the log to the series representation

$$ \log G(z+1)=\frac{z}{2}\log(2\pi)-\frac{z+z^2(1+\gamma)}{2}+\sum_{n=1}^{\infty}n \log\left(1+\frac{z}{n} \right)+\frac{z^2}{2n}-z$$

Let the following

$$f(z) = \sum_{n=1}^{\infty}n \log\left(1+\frac{z}{n} \right)+\frac{z^2}{2n}-z$$

Differentiate with respect to $z$

$$f(z) = \sum_{n=1}^{\infty}\frac{n}{z+n}+\frac{z}{n}-1 =\sum_{n=1}^{\infty}\frac{z^2}{n(n+z)} $$

Now use the following

$$\psi(z) = -\gamma -\frac{1}{z}+\sum_{n=1}^\infty \frac{z}{n(n+z)}$$

which implies that

$$\sum_{n=1}^\infty \frac{z^2}{n(n+z)} = z\psi(z)+\gamma z+1$$

Hence we have

$$f(z) = z\psi(z)+\gamma z+1$$

Integrate with respect to $z$

$$f(z) = \int^z_0 x\psi(x) dx +\frac{\gamma z^2}{2}+z$$

which implies that

$$f(z) = z\log \Gamma(z) -\int^z_0 \log \Gamma(x) dx +\frac{\gamma z^2}{2}+z$$

Hence we have

$$ \log G(z+1)=\frac{z}{2}\log(2\pi)-\frac{z+z^2(1+\gamma)}{2}+z\log \Gamma(z) -\int^z_0 \log \Gamma(x) dx +\frac{\gamma z^2}{2}+z$$

By some rearrangements we have

$$\int^z_0 \log \Gamma(x) dx =\frac{z}{2}\log(2\pi)+\frac{z(z-1)}{2}+z\log \Gamma(z)-\log G(z+1)$$
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Re: (AIT) Barnes G function

Relation to Hyperfactorial function


Prove for $n$ is a positive integer

$$G(n+1) = \frac{(N!)^n}{H(n)}$$

Where $H(n)$ is the hyperfactorial function

$$H(n) = \prod^n_{k=1}k^k$$

proof

We can prove it by induction for $n=0$ we have, $G(1)=1$,

suppose that

$$G(n) = \frac{\Gamma(n)^{n-1}}{H(n-1)}$$

we want to find

$$G(n+1) = \Gamma(n)G(n) = \Gamma(n) \frac{\Gamma(n)^{n-1}}{H(n-1)}$$

Notice that

$$H(n-1) = \prod^{n-1}k^k = \frac{\prod^{n}k^k}{n^n} = \frac{H(n)}{n^n }$$

We deduce that

$$G(n+1) = \Gamma(n)G(n) = \frac{\Gamma(n)^{n}\times n^n}{H(n)} = \frac{(N!)^n}{H(n)}$$

A related constant


We define the Glaisher-Kinkelin constant as

$$A = \lim_{n \to \infty}\frac{H(n)}{n^{n^2/2+n/2+1/12}e^{-n^2/4}}$$

Prove that

$$\lim_{n \to \infty}\frac{G(n+1)}{(2\pi)^{n/2}n^{n^2/2-1/12} e^{-3n^2/4}} = \frac{e^{1/12}}{A}$$

proof

From the previous result we have

$$\lim_{n \to \infty}\frac{(N!)^n}{H(n)(2\pi)^{n/2}n^{n^2/2-1/12} e^{-3n^2/4}}$$

Now use the Stirling approximation

$$(N!)^n \sim (2\pi)^{n/2}n^{n^2+n/2}e^{-n^2+1/12}$$

Hence we deduce that

$$\lim_{n \to \infty}\frac{(2\pi)^{n/2}n^{n^2+n/2}e^{-n^2+1/12}}{H(n)}\times\frac{1}{(2\pi)^{n/2}n^{n^2/2-1/12} e^{-3n^2/4}}$$

By simplifications we have

$$e^{1/12}\lim_{n \to \infty}\frac{n^{n^2/2+n/2+1/12}e^{-n^2/4}}{H(n)} = \frac{e^{1/12}}{A}$$

Exercise

$$\zeta'(2) = \frac{\pi^2}{6}\left(\log(2\pi)+\gamma-12\log A \right)$$

We already proved that

$$\log \left[\frac{G(n+1)}{(2\pi)^{n/2}n^{n^2/2-1/12} e^{-3n^2/4}} \right] \sim \frac{1}{12}-\log A$$

Let the following

$$f(n) = \log \left[\frac{G(n+1)}{(2\pi)^{n/2}n^{n^2/2-1/12} e^{-3n^2/4}} \right]$$

Use the series representation of the Barnes functions

$$f(n) = \log \left[ \frac{(2\pi)^{n/2}\exp\left(-\frac{n+n^2(1+\gamma)}{2}\right)\prod_{k=1}^{\infty}\left\{\left(1+\frac{n}{k} \right)^k\exp\left(\frac{n^2}{2k}-n\right) \right\}}{(2\pi)^{n/2}n^{n^2/2-1/12} e^{-3n^2/4}} \right]$$

Which reduces to

$$f(n) = -\frac{n+n^2(1+\gamma)}{2}+\sum^\infty_{k=1}\left\{k \log \left(1+\frac{n}{k} \right)+\frac{n^2}{2k}-n \right\}\\ -\left( \frac{n^2}{2}-\frac{1}{12}\right)\log(n)+\frac{3n^2}{4}$$

Differentiate with respect to $n$

$$f'(n) = -\frac{1}{2}-n-\gamma n+n\psi(n)+\gamma n+1-n\log(n)-\frac{n}{2}+\frac{1}{12n}+\frac{3n}{2}$$

Note that we already showed that

$$\frac{d}{dn}\sum^\infty_{k=1}\left\{k \log \left(1+\frac{n}{k} \right)+\frac{n^2}{2k}-n\right\}= n\psi(n)+\gamma n+1$$

By simplifications we have

$$f'(n) = n\psi(n)-n\log(n)+\frac{1}{12n}+\frac{1}{2}$$

Now use that

$$\psi(n) = \log(n)-\frac{1}{2n}-2\int^\infty_0 \frac{z dz}{(n^2+z^2)(e^{2\pi z}-1)}dz$$

Hence we deduce that

$$f'(n) =-2\int^\infty_0 \frac{nz dz}{(n^2+z^2)(e^{2\pi z}-1)}dz+\frac{1}{12n}$$

Integrate with respect to $n$

$$f(n) =-\int^\infty_0 \frac{z\log(n^2+z^2)}{(e^{2\pi z}-1)}dz+\frac{1}{12}\log(n)+C$$

Take the limit $n \to 0$

$$C = \lim_{n \to 0}f(n)-\frac{1}{12}\log(n)+\int^\infty_0 \frac{z\log(z^2)}{(e^{2\pi z}-1)}dz $$

Hence we have the limit

$$\lim_{n \to 0}\frac{G(n+1)}{(2\pi)^{n/2}n^{n^2/2-1/12} e^{-3n^2/4}}-\frac{1}{12}\log(n) = \lim_{n \to 0}\log \frac{G(n+1)}{(2\pi)^{n/2}n^{n^2/2} e^{-3n^2/4}} = 0$$

Hence we see that

$$C = 2\int^\infty_0 \frac{z\log(z)}{(e^{2\pi z}-1)}dz $$

Finally we have

$$f(n) =-\int^\infty_0 \frac{z\log(n^2+z^2)}{(e^{2\pi z}-1)}dz+\frac{1}{12}\log(n)+ 2\int^\infty_0 \frac{z\log(z)}{(e^{2\pi z}-1)}dz$$

$$f(n) =-\int^\infty_0 \frac{z\log\left(1+\frac{z^2}{n^2} \right)}{(e^{2\pi z}-1)}dz-\log(n^2)\int^\infty_0 \frac{z}{(e^{2\pi z}-1)}dz+\frac{1}{12}\log(n)\\+ 2\int^\infty_0 \frac{z\log(z)}{(e^{2\pi z}-1)}dz$$

Also we have

$$\int^\infty_0 \frac{z}{(e^{2\pi z}-1)}dz = \frac{1}{24}$$

That simplifies to

$$f(n)=-\int^\infty_0 \frac{z\log\left(1+\frac{z^2}{n^2} \right)}{(e^{2\pi z}-1)}dz+2\int^\infty_0 \frac{z\log(z)}{(e^{2\pi z}-1)}dz$$

Take the limit $n \to \infty $

$$2\int^\infty_0 \frac{z\log(z)}{(e^{2\pi z}-1)}dz = \frac{1}{12}-\log A$$

Now use that

$$
\begin{align}
2\int^\infty_0 \frac{z\log(z)}{(e^{2\pi z}-1)}dz
&= 2\int^\infty_0 \frac{z\log(z)}{e^{2\pi z}}\times \frac{1}{1-e^{-2\pi z}}dz\\
&= 2\sum_{n=0}^\infty \int^\infty_0 e^{-2\pi z(n+1)}z\log(z)\,dz\\
&= \sum_{n=1}^\infty \frac{\psi(2) − \log(2\pi)+\log(n)}{2\pi^2 n^2}\\
&=\frac{(\psi(2)-\log(2\pi))\zeta(2)+\zeta'(2)}{2\pi^2}\\
\end{align}
$$

Hence we conclude that

$$\zeta'(2) =(\log(2\pi)-\psi(2))\zeta(2) + 2\pi^2\left(\frac{1}{12}-\log A \right) = \zeta(2) (\log(2\pi)+\gamma-12 \log A)$$
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Re: (AIT) Barnes G function

Relation to hurwitz zeta function


Prove that

$$\log G(z+1) - z \log \Gamma(z) = \zeta'(-1) - \zeta'(-1,z) $$

Start by the following

$$\zeta(s,z) = \frac{z^{-s}}{2}+\frac{z^{1-s}}{s-1}+2\int^\infty_0 \frac{\sin(s\arctan(x/z))}{(z^2+x^2)^{s/2}(e^{2\pi x}-1)}\,dx$$

Hence we have

$$\zeta'(-1,z) = -\frac{z\log(z)}{2}+\frac{z^2\log(z)}{2}-\frac{z^2}{4}+\int^\infty_0 \frac{x\log(x^2+z^2)+2z\arctan(x/z)}{(e^{2\pi x}-1)}\,dx$$

Now use that

$$\psi(z) = \log(z)-\frac{1}{2z}-2\int^\infty_0 \frac{x }{(z^2+x^2)(e^{2\pi x}-1)}dx$$

Which implies that

$$\int^\infty_0 \frac{2zx }{(z^2+x^2)(e^{2\pi x}-1)}dx=z\log(z)-\frac{1}{2}-z\psi(z)$$

By taking the integral

$$\int^\infty_0 \frac{x\log(x^2+z^2) -x\log(x^2)}{(e^{2\pi x}-1)}dx=\int^z_0x\log(x)\,dx-\int^z_0x\psi(x)\,dx-\frac{z}{2}$$

Which simplifies to

$$\int^\infty_0 \frac{x\log(x^2+z^2) }{(e^{2\pi x}-1)}dx=\zeta'(-1)-\frac{z^2}{4}+\frac{1}{2} z^2 \log(z)-z\log\Gamma(z)+\int^z_0\log\Gamma(x)\,dx-\frac{z}{2}$$

Also we have

$$2\int^\infty_0 \frac{x}{(x^2+z^2)(e^{2\pi x}-1)}dx=\log(z)-\frac{1}{2z}-\psi(z)$$

By integration we have

$$2\int^\infty_0 \frac{\arctan(x/z)}{(e^{2\pi x}-1)}dx=z+\frac{\log(z)}{2}-z \log(z)+\log\Gamma(z)+C$$

Let $z \to 1$ to evaluate the constant

$$2\int^\infty_0 \frac{\arctan(x/z)}{(e^{2\pi x}-1)}dx=z+\frac{\log(z)}{2}-z \log(z)+\log\Gamma(z)-\frac{1}{2}\log(2\pi)$$

Multiply by $z$

$$2\int^\infty_0 \frac{z\arctan(x/z)}{(e^{2\pi x}-1)}dx=z^2+\frac{z\log(z)}{2}-z^2 \log(z)+z\log\Gamma(z)-\frac{z}{2}\log(2\pi)$$

Substitute both integrals our formula

$$\zeta'(-1,z) =-\frac{z\log(z)}{2}+\frac{z^2\log(z)}{2}-\frac{z^2}{4}+z^2+\frac{z\log(z)}{2}-z^2 \log(z)+z\log\Gamma(z)\\-\frac{z}{2}\log(2\pi)+\zeta'(-1)-\frac{z^2}{4}+\frac{1}{2} z^2 \log(z)-z\log\Gamma(z)+\int^z_0\log\Gamma(x)\,dx-\frac{z}{2}$$

Which reduces to

$$\int_{0}^{z} \log \Gamma(x) \, \mathrm dx = \frac{z}{2} \log(2 \pi) + \frac{z(1-z)}{2} - \zeta^{'}(-1) + \zeta^{'}(-1,z)$$

We also showed that

$$\int_{0}^{z} \log \Gamma(x) \, \mathrm dx = \frac{z}{2} \log (2 \pi) + \frac{z(1-z)}{2} + z \log \Gamma(z) - \log G(z+1)$$

By equating the equations we get our result.

Prove that


$$\zeta'(-1) = \frac{1}{12}-\log A$$

Start by

$$\zeta(s) = \lim_{m \to \infty} \left( \sum_{k=1}^{m} k^{-s} - \frac{m^{1-s}}{1-s} - \frac{m^{-s}}{2} + \frac{sm^{-s-1}}{12} \right) \ , \ \text{Re}(s) >-3.$$

Differentiate with respect to $s$

$$\zeta'(s) = \lim_{m \to \infty} \left(- \sum_{k=1}^{m} k^{-s}\log(k) - \frac{m^{1-s}}{(1-s)^2} +\frac{m^{1-s}}{1-s}\log(m)+\frac{m^{-s}}{2}\log(m)\\ + \frac{m^{-s-1}}{12}-\frac{m^{-s-1}}{12}\log(m)\right) $$

Now let $s \to -1$

$$\zeta'(-1) = \lim_{m \to \infty} \left(- \sum_{k=1}^{m} k\log(k) - \frac{m^{2}}{4} +\frac{m^{2}}{2}\log(m)+\frac{m}{2}\log(m)\\ + \frac{1}{12}-\frac{1}{12}\log(m)\right) $$

Take the exponential of both sides

$$e^{\zeta'(-1)}= e^{1/12}\lim_{m \to \infty} \frac{m^{m^2/2+m/2-1/12} e^{-m^2/4}}{ e^{\sum_{k=1}^{m} k\log(k)}} = e^{1/12}\lim_{m \to \infty} \frac{m^{m^2/2+m/2-1/12} e^{-m^2/4}}{ H(m+1)} = \frac{ e^{1/12}}{A} $$

We conclude that

$$\zeta'(-1) = \frac{1}{12}-\log A$$

Prove that

$$G\left(\frac{1}{2}\right) = 2^{1/24} \pi^{-1/4}e^{1/8}A^{-3/2}$$

We know that

$$\log G(z)+ \log \Gamma(z)- z \log \Gamma(z) = \zeta'(-1) - \zeta'(-1,z) $$

Note that

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

Which implies that

$$\zeta' \left(-1, \frac{1}{2} \right) = \frac{\log(2)}{2}\zeta(-1)-\frac{1}{2}\zeta'(-1)$$

Hence we have

$$\log G\left(\frac{1}{2} \right)+ \frac{1}{2}\log \Gamma\left(\frac{1}{2} \right)= \frac{3}{2}\zeta'(-1)-\frac{\log(2)}{2}\zeta(-1) $$

Using that we have

$$G\left(\frac{1}{2}\right) = 2^{1/24}\pi^{-1/4}e^{\frac{3}{2}\zeta'(-1)}$$

Note that

$$\zeta(-1) = -\frac{1}{12}$$

This can be proved by the functional equation of the zeta function.
 
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