Loggamma integral...

[DreamWeaver]

Assuming the canonical product forms for the Gamma Function \(\displaystyle \Gamma(z)\) and Double Gamma Function (Barnes' G-Function) \(\displaystyle \text{G}(z)\):


\(\displaystyle \frac{1}{\Gamma(x)}=xe^{\gamma x}\, \prod_{k=1}^{\infty}\Bigg\{e^{-x/k}\left(1+\frac{x}{k}\right)\Bigg\}\)


\(\displaystyle G(z+1)=(2\pi)^{z/2}\text{exp}\left(-\frac{z+z^2(1+\gamma)}{2}\right)\, \prod_{k=1}^{\infty}\left(1+\frac{z}{k}\right) \text {exp}\left(\frac{z^2}{2k}-z\right)\)


Where the Euler-Mascheroni constant is defined by the limit:


\(\displaystyle \gamma=\lim_{n\to\infty}\, \left(1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n}-\log n \right)\)


And the Barnes' G-Function has the property:


\(\displaystyle \text{G}(z+1)=\text{G}(z)\,\Gamma(z)\)


Prove the following parametric evaluation:



\(\displaystyle \int_0^z\log\Gamma(x)\,dx=\)


\(\displaystyle \frac{z(1-z)}{2}+\frac{z}{2}\log(2\pi)+z\,\log\Gamma(z)-\log\text{G}(z+1)\)

 

[Random Variable]

Well, for $0 < x <1 $, $\displaystyle \ln \Gamma(x)$ has the Fourier expansion

$$ \ln \Gamma(x) = \frac{\ln 2 \pi}{2} + \sum_{n=1}^{\infty} \frac{1}{2n} \cos(2 \pi n x) + \sum_{n=1}^{\infty} \frac{\gamma + \ln(2 \pi n) }{n \pi} \sin(2 \pi n x)$$


So

$$\int_{0}^{z} \ln \Gamma(x) \ dx = \frac{z}{2} \ln (2 \pi) + \frac{1}{4 \pi} \sum_{n=1}^{\infty} \frac{\sin(2 \pi n z)}{n^{2}} + \frac{\gamma}{2 \pi^{2}} \sum_{n=1}^{\infty} \frac{\cos(2 \pi n z)-1}{n^{2}} $$

$$ + \frac{1}{2 \pi^{2}} \sum_{n=1}^{\infty} \frac{\ln(2 \pi n)}{n^{2}} \Big(\cos(2 \pi nz) -1 \Big)$$




I don't know what to do from there, though.

 

[DreamWeaver]

Here's a little hint, if you like...

Spoiler

Take the logarithm of canonical product for the Barnes G function. Keep a note of that result, integrate the logarithm of the Weierstrauss product form for the Gamma function from 0 to z, and then compare the two...

 

[Random Variable]

Using the product representation which I still can't derive,

$\displaystyle \ln G(z+1) = \frac{z}{2} \ln(2 \pi) - \frac{1}{2} z(z+1) - \frac{1}{2}\gamma z^{2} + \sum_{k=1}^{\infty} \Bigg( -z+\frac{z^{2}}{2k} + k \ln \Big( 1+ \frac{z}{k} \Big) \Bigg)$


And from the Weierstrass product representation of the Gamma function,

$ \displaystyle \ln \Gamma(z+1) = - \gamma z + \sum_{k=1}^{\infty} \Bigg( \frac{z}{k} - \ln \Big(1+ \frac{z}{k} \Big) \Bigg) $

EDIT: $ \displaystyle \int_{0}^{z} \ln \Gamma(x+1) \ dx = -\frac{\gamma z^{2}}{2} + \sum_{k=1}^{\infty} \Bigg( \frac{z^{2}}{2k} - (z+1)\log \Big( 1+ \frac{z}{k} \Big) - z \Bigg)$


So $\displaystyle z \ln \Gamma(z+1) - \int_{0}^{z} \ln \Gamma(z+1) = - \frac{1}{2} \gamma z^{2} + \sum_{k=1}^{\infty} \Bigg(\frac{z^{2}}{2k} + k \ln \Big( 1+ \frac{z}{k} \Big) -x \Bigg)$


Rearranging we have

$ \displaystyle \int_{0}^{z} \ln \Gamma (x+1) \ dx = \int_{0}^{z} \ln \Gamma(x+1) \ dx + \int_{0}^{z} \ln z = \int_{0}^{z} \ln \Gamma(x+1) \ dx + z \ln z -z $

$ \displaystyle = \frac{z}{2} \ln(2 \pi)- \frac{1}{2} z(z+1) + z \ln \Gamma(z+1) - G (z+1)$

$ \displaystyle = \frac{z}{2} \ln(2 \pi)- \frac{1}{2} z(z+1) + z \ln z + z \ln \Gamma(z) - G(z+1)$


$ \displaystyle \implies \int_{0}^{z} \ln \Gamma(x+1) \ dx = \frac{z}{2} \ln(2 \pi) - \frac{z(z-1)}{2} + z \ln \Gamma(z) - G(z+1) $

 

[DreamWeaver]

I'm feeling quite stupid.

With the sort of problems you solve??? I hope you're not selling that, cos I'm not buying it... Meh!

It's just late, is all...


Here's a slightly different angle: looking at the closed form given, express the series difference


\(\displaystyle z\log\Gamma(z)-\log G(z+1)\)


using both canonical products...

Can you spot the similarity between that and


\(\displaystyle \int_0^z\log\Gamma(x)\,dx=\int_0^z\log\left[ xe^{\gamma x} \prod_{k=1}^{\infty} \Bigg\{ e^{-x/k}\left(1+\frac{x}{k}\right)\Bigg\}\right]\,dx\)


...?


If you write


\(\displaystyle z\log\Gamma(z)-\log G(z+1)-\int_0^z\log\left[ xe^{\gamma x} \prod_{k=1}^{\infty} \Bigg\{ e^{-x/k}\left(1+\frac{x}{k}\right)\Bigg\}\right]\,dx\)


what do you get...?

 

[Random Variable]

You know what I did? I differentiated $ \displaystyle \ln \Big(1+ \frac{z}{k} \Big)$ instead of integrating.

EDIT: I fixed my post.

 

[DreamWeaver]

You know what I did? I differentiated $ \displaystyle \ln \Big(1+ \frac{z}{k} \Big)$ instead of integrating.

EDIT: I fixed my post.

Easily done! Been there, bought - and regretted buying - the t-shirt...

 

[Random Variable]

DreamWeaver

Could you show either in this thread or in the thread on the other forum that the infinite product representation of $G(x)$ satisfies the functional equation?

And so you're aware, you're missing the exponent $k$.


$ \displaystyle G(z+1)=(2\pi)^{z/2}\text{exp}\left(-\frac{z+z^2(1+\gamma)}{2}\right)\, \prod_{k=1}^{\infty}\left(1+\frac{z}{k}\right)^{ {\color{red}{k}} } \text {exp}\left(\frac{z^2}{2k}-z\right)$

 

[DreamWeaver]

DreamWeaver


And so you're aware, you're missing the exponent $k$.

Oooops!!!

Sorry about that...

DreamWeaver

Could you show either in this thread or in the thread on the other forum that the infinite

product representation of $G(x)$ satisfies the functional equation?

Good idea! It's been a very long day at work, mind, so I'll give this one a go tomorrow. If I get started now...

 

[Random Variable]

Here's an alternative derivation.


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

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

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

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

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

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


Now I'm going to use the generating function $ \displaystyle\sum_{k=2}^{\infty} \zeta(k) z^{k-1} = - \gamma - \psi(1-z)$.

A derivation can be found here.


Then

$\displaystyle \sum_{k=2}^{\infty} (-1)^{k} \zeta(k) z^{k} = \gamma z + z \psi(z+1) $

And upon integrating both sides from $0$ to $z$,

$ \displaystyle \sum_{k=2}^{\infty} \frac{(-1)^{k}}{k+1} \zeta(k) z^{k+1} = \frac{\gamma z^{2}}{2} + z \log \Gamma(z+1) - \int_{0}^{z} \log \Gamma(x+1) \ dx $


So

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

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

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


And by analytic continuation, the equation should be valid for $\text{Re}(z) > 0 $, not just $0 < \text{Re}(z) < 1$.

 

[ZaidAlyafey]

Very nice RV !

 

[Random Variable]

Thanks.

If anyone was interested, about a month ago I asked on Stack Exchange how to show that the infinite product representation satisfies the functional equation. I got a very helpful response. It turns out that it's not hard to show at all.

The infinite product representation of the Barnes G function - Mathematics Stack Exchange

 

[DreamWeaver]

Very nice indeed, RV...!!

On the subject of the Barnes G-function, have you - in your travails - come across a Fourier expansion? I'm sure there must be one... Would be mighty useful, methinks.