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Barnes G and its relation to other sums and integrals


galactus User avatar
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This small tutorial is just meant to relate some interesting traits of the Barnes G and its broad implications.

I hope someone may find it useful. Assuming you are not already familiar with these tidbits.

I will add more as time permits. I have dabbled in the Barnes G off and on and have noticed how it ties into all sorts of things.

I would like to get Barnes' famous text on the topic. After all, he is the one that developed it as far as I can tell.


Let's begin with the series for \(\displaystyle x\cot(x)=\sum_{n=0}^{\infty}\frac{(-1)^{n}2^{2n}B_{2n}}{(2n)!}x^{2n}, \;\ |x|<\pi\)

Note how this series is very similar to the famous formula:

\(\displaystyle \zeta(2n)=\frac{(-1)^{n-1}2^{2n-1}\pi^{2n}B_{2n}}{(2n)!}, \;\ n\geq 1\).

Let \(\displaystyle x\to \pi x\) and we get:

\(\displaystyle \pi x\cot(\pi x)=-2\sum_{n=0}^{\infty}\zeta(2n)x^{2n} \;\ \;\ \;\ \;\ (1)\)

Now, consider the identity: \(\displaystyle 1/2\int_{a}^{b}p(x)\cot(ax/2)dx=\sum_{n=0}^{\infty}\int_{a}^{b}p(x)\sin(anx)dx \;\ \;\ \;\ \;\ (2)\), where p(x) is a polynomial.

The above identity can be mighty handy to evaluate all sorts of series.

Anyway, let \(\displaystyle a=2\pi\) in (2) and sub (1) into (2):

\(\displaystyle \frac{-1}{\pi}\int_{a}^{b}\sum_{n=0}^{\infty}\zeta(2n)x^{2n-1}dx=\sum_{n=0}^{\infty}\int_{a}^{b}p(x)\sin(2\pi nx)dx\)

Let's do an example by letting \(\displaystyle p(x)=x, \;\ a=0, \;\ b=1/2\), and change the lower sum index to n=1.

The right side results in the series \(\displaystyle \frac{1}{4\pi}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}=\frac{1}{4}\left(1-log(2)\right)\)

This gives:

\(\displaystyle \sum_{n=1}^{\infty}\frac{\zeta(2n)}{(2n+1)2^{2n+1}}=\frac{1}{4}\left(1-log(2)\right)\)


Now, consider the identity by Choi and Srivastava.

\(\displaystyle \sum_{n=1}^{\infty}\frac{\zeta(2n)}{2n+1}x^{2n+1}=\frac{1}{2}\left[\left(1-log(2\pi)\right)x+log\frac{G(1+x)}{G(1-x)}\right] \;\ \;\ \;\ \;\ (3)\)

Also, the Barnes G is given by:

\(\displaystyle G(x+1)=(2\pi)^{x/2}e^{-1/2x^{2}-\frac{\gamma}{2}x^{2}-x/2}\prod_{n=1}^{\infty}\left(1+x/n\right)^{n}e^{-x+\frac{x^{2}}{2n}}\)

Let x=1/2 in (3):

\(\displaystyle \sum_{n=1}^{\infty}\frac{\zeta(2n)}{(2n+1)2^{2n+1}}=\frac{1}{4}(1-log(2\pi))+\frac{1}{2}log\frac{G(3/2)}{G(1/2)}\)

But, by using the relation \(\displaystyle G(1+x)=\Gamma(x)G(x)\), we arrive at:

\(\displaystyle \frac{G(3/2)}{G(1/2)}=\sqrt{\pi}\)

This gives us on the right side:

\(\displaystyle \frac{1}{4}\left(1-(log(2)+log(\pi))+\frac{1}{4}log(\pi)\right)\)

\(\displaystyle =\frac{1}{4}\left(1-log(2)\right)\) as before.


We have now shown that

\(\displaystyle \sum_{n=1}^{\infty}\frac{\zeta(2n)}{2n+1}x^{2n+1}=\frac{1}{2}\left[\left(1-log(2\pi)\right)x+log\frac{G(1+x)}{G(1-x)}\right]\)

\(\displaystyle =\frac{x}{2}-\pi\sum_{n=1}^{\infty}\int_{0}^{x}t\sin(2\pi nt)dt\)

\(\displaystyle =\frac{x}{2}+\frac{x}{2}\sum_{n=1}^{\infty}\frac{\cos(2\pi nt)}{t}-\frac{1}{4\pi}\sum_{n=1}^{\infty}\frac{\sin(2\pi nt)}{n^{2}}\)

\(\displaystyle =\frac{x}{2}-\frac{x}{2}log(2\sin(\pi x))-\frac{1}{4\pi}Cl_{2}(2\pi x)\), where, as you know, \(\displaystyle Cl_{2}\) denotes the

Clausen function, \(\displaystyle Cl_{2}(x)=-\int_{0}^{x}log(2\sin(u/2))du=\sum_{n=1}^{\infty}\frac{\sin(\pi x)}{n^{2}}\).

This brings us to:

\(\displaystyle log\frac{G(1+x)}{G(1-x)}=x-xlog(2\sin(\pi x))-x+xlog(2\pi)+\int_{0}^{x}log(2\sin(\pi t))dt\)

The x and xlog(2) can be cancelled and we arrive at:

\(\displaystyle log\frac{G(1+x)}{G(1-x)}=-xlog\left(\frac{\sin(\pi x)}{2\pi}\right)+\int_{0}^{x}log(\sin(\pi t))dt\)

For example, letting x=1/2, we can find a value of the above log-sin integral:

\(\displaystyle \int_{0}^{1/2}log(\sin(\pi x))dx=1/2log(\pi)+1/2log\left(\frac{\sin(\pi/2)}{2\pi}\right)=-\frac{1}{2}log(2)\).




To be continued. Next we will look at Alexeiewsky's theorem. A theorem relating integrals of log-gamma to the Barnes G.

Random Variable Integration Guru
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Posts: 381
I thought I'd add a derivation of (3).

The only thing you need is the infinite product representation.


\(\displaystyle \log G(1+z) = \frac{z}{2} \log(2 \pi) - \frac{z^{2}}{2} - \frac{\gamma z^{2}}{2} - \frac{z}{2} + \sum_{n=1}^{\infty} \Big[ n \log \Big(1 + \frac{z}{n} \Big) -z + \frac{z^{2}}{2n} \Big]\)

\(\displaystyle = \frac{z}{2} \log(2 \pi) - \frac{z^{2}}{2} - \frac{\gamma z^{2}}{2} - \frac{z}{2} + \sum_{n=1}^{\infty} \Big[ n \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k} \Big( \frac{z}{n} \Big)^{k} -z + \frac{z^{2}}{2n} \Big]\) for \(\displaystyle |z| < 1\)

\(\displaystyle = \frac{z}{2} \log(2 \pi) - \frac{z^{2}}{2} - \frac{\gamma z^{2}}{2} - \frac{z}{2} + \sum_{n=1}^{\infty} \sum_{k=3}^{\infty} \frac{(-1)^{k-1}}{k} \frac{z^{k}}{n^{k-1}}\)

\(\displaystyle = \frac{z}{2} \log(2 \pi) - \frac{z^{2}}{2} - \frac{\gamma z^{2}}{2} - \frac{z}{2} + \sum_{k=3}^{\infty} (-1)^{k-1} \frac{z^{k}}{k} \sum_{n=1}^{\infty} \frac{1}{n^{k-1}}\)

\(\displaystyle = \frac{z}{2} \log(2 \pi) - \frac{z^{2}}{2} - \frac{\gamma z^{2}}{2} - \frac{z}{2} + \sum_{k=3}^{\infty} (-1)^{k-1} \zeta(k-1) \frac{z^{k}}{k}\)

\(\displaystyle = \frac{z}{2} \log(2 \pi) - \frac{z^{2}}{2} - \frac{\gamma z^{2}}{2} - \frac{z}{2} + \sum_{k=2}^{\infty} (-1)^{k} \zeta(k) \frac{z^{k+1}}{k+1}\)


Then

\(\displaystyle \log G(1-z) = - \frac{z}{2} \log(2 \pi) - \frac{z^{2}}{2} - \frac{\gamma z^{2}}{2} + \frac{z}{2} - \sum_{k=2}^{\infty} \zeta(k) \frac{z^{k+1}}{k+1}\)


And

\(\displaystyle \log \frac{G(1+z)}{G(1-z)} = \log G(1+z) - \log G(1-z) = z \log(2 \pi) -z + \sum_{k=2}^{\infty} (-1)^{k} \zeta(k) \frac{z^{k+1}}{k+1} + \sum_{k=2}^{\infty} \zeta(k) \frac{z^{k+1}}{k+1}\)

\(\displaystyle = z \log(2 \pi) -z + 2 \sum_{k=1}^{\infty} \zeta(2k) \frac{z^{2k+1}}{2k+1}\)


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