MHB What is the Relationship Between Loggamma Integral and Barnes' G-Function?

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The discussion explores the relationship between the Loggamma integral and Barnes' G-Function, focusing on their canonical product forms. Participants derive a parametric evaluation of the integral of the logarithm of the Gamma function, linking it to the properties of Barnes' G-Function. The conversation includes various mathematical expansions and representations, such as Fourier series and infinite products, to establish connections between these functions. Additionally, there are hints at alternative derivations and functional equations related to the G-Function. The thread concludes with a suggestion for further exploration of Fourier expansions related to the Barnes G-Function.
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Assuming the canonical product forms for the Gamma Function $$\Gamma(z)$$ and Double Gamma Function (Barnes' G-Function) $$\text{G}(z)$$:$$\frac{1}{\Gamma(x)}=xe^{\gamma x}\, \prod_{k=1}^{\infty}\Bigg\{e^{-x/k}\left(1+\frac{x}{k}\right)\Bigg\}$$$$G(z+1)=(2\pi)^{z/2}\text{exp}\left(-\frac{z+z^2(1+\gamma)}{2}\right)\, \prod_{k=1}^{\infty}\left(1+\frac{z}{k}\right) \text {exp}\left(\frac{z^2}{2k}-z\right)$$Where the Euler-Mascheroni constant is defined by the limit:$$\gamma=\lim_{n\to\infty}\, \left(1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n}-\log n \right)$$And the Barnes' G-Function has the property:$$\text{G}(z+1)=\text{G}(z)\,\Gamma(z)$$Prove the following parametric evaluation:
$$\int_0^z\log\Gamma(x)\,dx=$$$$\frac{z(1-z)}{2}+\frac{z}{2}\log(2\pi)+z\,\log\Gamma(z)-\log\text{G}(z+1)$$

(Hug) (Hug) (Hug)
 
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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.
 
Here's a little hint, if you like... (Bandit)
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...
 
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) $
 
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Random Variable said:
I'm feeling quite stupid.
With the sort of problems you solve? (Wait) I hope you're not selling that, cos I'm not buying it... Meh!

It's just late, is all... (Bandit)Here's a slightly different angle: looking at the closed form given, express the series difference$$z\log\Gamma(z)-\log G(z+1)$$using both canonical products...

Can you spot the similarity between that and $$\int_0^z\log\Gamma(x)\,dx=\int_0^z\log\left[ xe^{\gamma x} \prod_{k=1}^{\infty} \Bigg\{ e^{-x/k}\left(1+\frac{x}{k}\right)\Bigg\}\right]\,dx$$...? ;)If you write$$z\log\Gamma(z)-\log G(z+1)-\int_0^z\log\left[ xe^{\gamma x} \prod_{k=1}^{\infty} \Bigg\{ e^{-x/k}\left(1+\frac{x}{k}\right)\Bigg\}\right]\,dx$$what do you get...?
 
You know what I did? I differentiated $ \displaystyle \ln \Big(1+ \frac{z}{k} \Big)$ instead of integrating. (Doh)

EDIT: I fixed my post.
 
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Random Variable said:
You know what I did? I differentiated $ \displaystyle \ln \Big(1+ \frac{z}{k} \Big)$ instead of integrating. (Doh)

EDIT: I fixed my post.

Easily done! Been there, bought - and regretted buying - the t-shirt...
 
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)$
 
Random Variable said:
DreamWeaverAnd so you're aware, you're missing the exponent $k$.
Oooops! :o :o :o

Sorry about that...
Random Variable said:
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! :D It's been a very long day at work, mind, so I'll give this one a go tomorrow. If I get started now... (Wait)
 
  • #10
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 $0 < \text{Re}(z) < 1$.
 
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  • #11
Very nice RV !
 
  • #13
Very nice indeed, RV...! (Muscle)

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.
 

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