MHB What Are Advanced Integration Techniques in Mathematics?

[alyafey22]

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

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

 

[alyafey22]

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.