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Loggamma integral...

DreamWeaver

Well-known member
Sep 16, 2013
337
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:



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


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




(Hug) (Hug) (Hug)
 

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
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

Well-known member
Sep 16, 2013
337
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...
 

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
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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DreamWeaver

Well-known member
Sep 16, 2013
337
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


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


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

Well-known member
MHB Math Helper
Jan 31, 2012
253
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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DreamWeaver

Well-known member
Sep 16, 2013
337
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...
 

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
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

Well-known member
Sep 16, 2013
337
DreamWeaver


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

Oooops!!! :eek: :eek: :eek:

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! :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)
 

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
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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ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Very nice RV !
 

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253

DreamWeaver

Well-known member
Sep 16, 2013
337
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.