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DreamWeaver
Well-known member
- Sep 16, 2013
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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:
\(\displaystyle \int_0^z\log\Gamma(x)\,dx=\)
\(\displaystyle \frac{z(1-z)}{2}+\frac{z}{2}\log(2\pi)+z\,\log\Gamma(z)-\log\text{G}(z+1)\)

\(\displaystyle \frac{1}{\Gamma(x)}=xe^{\gamma x}\, \prod_{k=1}^{\infty}\Bigg\{e^{-x/k}\left(1+\frac{x}{k}\right)\Bigg\}\)
\(\displaystyle G(z+1)=(2\pi)^{z/2}\text{exp}\left(-\frac{z+z^2(1+\gamma)}{2}\right)\, \prod_{k=1}^{\infty}\left(1+\frac{z}{k}\right) \text {exp}\left(\frac{z^2}{2k}-z\right)\)
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:
\(\displaystyle \int_0^z\log\Gamma(x)\,dx=\)
\(\displaystyle \frac{z(1-z)}{2}+\frac{z}{2}\log(2\pi)+z\,\log\Gamma(z)-\log\text{G}(z+1)\)


