Salvatore Boone

2022-01-03

What are the usual ways to follow in order to solve the integrals given below?
$I={\int }_{0}^{1}\mathrm{ln}\mathrm{\Gamma }\left(x\right)dx$
$J={\int }_{0}^{1}x\mathrm{\Gamma }\left(x\right)dx$

poleglit3

Expert

As an addendum of sorts to the previous answers, there is the identity
$\mathrm{log}G\left(z+1\right)=\frac{z}{2}\mathrm{log}\left(2\pi \right)-\frac{z\left(z+1\right)}{2}+z\mathrm{log}\mathrm{\Gamma }\left(z+1\right)-z\left(\mathrm{log}z-1\right)-{\int }_{0}^{z}\mathrm{log}\mathrm{\Gamma }\left(t\right)dt$
where $\mathrm{log}G\left(z\right)$ is the logarithm of the Barnes function (double gamma function) $G\left(z\right)$, the function that satisfies the functional equation $G\left(z+1\right)=\mathrm{\Gamma }\left(z\right)G\left(z\right)$. (Barnes proved this identity in his paper, where he introduced the function now named after him.) For n an integer, $G\left(n\right)$ can be expressed as
$G\left(n\right)=\prod _{k=1}^{n-2}k!$
Thus, to evaluate ${\int }_{0}^{1}\mathrm{log}\mathrm{\Gamma }\left(t\right)dt$, we have
$\mathrm{log}G\left(2\right)=\frac{1}{2}\mathrm{log}\left(2\pi \right)-1+\mathrm{log}\mathrm{\Gamma }\left(2\right)-\left(\mathrm{log}1-1\right)-{\int }_{0}^{1}\mathrm{log}\mathrm{\Gamma }\left(t\right)dt$
$0=\frac{1}{2}\mathrm{log}\left(2\pi \right)-{\int }_{0}^{1}\mathrm{log}\mathrm{\Gamma }\left(t\right)dt$
and you obtain the same solution as Andrew.
For the integral ${\int }_{0}^{1}t\mathrm{log}\mathrm{\Gamma }\left(t\right)dt$, integration by parts and taking appropriate limits yields the identity
${\int }_{0}^{1}t\mathrm{log}\mathrm{\Gamma }\left(t\right)dt=-\frac{1}{2}{\int }_{0}^{1}{t}^{2}\psi \left(t\right)dt$
Now, Victor Adamchik, in a paper on negative-order polygamma functions (the same sort of functions that appear in Argon's answer), gives the identity
${\int }_{0}^{z}{x}^{n}\psi \left(x\right)dx=$
$={\left(-1\right)}^{n}\left(\frac{{B}_{n+1}{H}_{n}}{n+1}-{\zeta }^{\prime }\left(-n\right)\right)+\sum _{k=0}^{n}{\left(-1\right)}^{k}\left(\begin{array}{c}n\\ k\end{array}\right){z}^{n-k}\left(\zeta \left(-k,z\right)-\frac{{B}_{k+1}\left(z\right){H}_{k}}{k+1}$
where ${B}_{n}$ and

lovagwb

Expert

As for the first integral, one can use the Euler's reflection formula $\mathrm{\Gamma }\left(1-z\right)\mathrm{\Gamma }\left(z\right)=\frac{\pi }{\mathrm{sin}\pi z}$
$I=\frac{1}{2}{\int }_{0}^{1}\left(\mathrm{log}\left(\mathrm{\Gamma }\left(x\right)+\mathrm{log}\mathrm{\Gamma }\left(1-x\right)\right)dx=\frac{1}{2}{\int }_{0}^{1}\mathrm{log}\frac{\pi }{\mathrm{sin}\pi x}dx=$
$=\frac{1}{2}{\int }_{0}^{1}\left(\mathrm{log}\pi -\mathrm{log}\mathrm{sin}\pi x\right)dx=\frac{1}{2}\mathrm{log}\pi -\frac{1}{2\pi }{\int }_{0}^{\pi }\mathrm{log}\mathrm{sin}xdx$
$=\frac{1}{2}\mathrm{log}\pi -\frac{1}{2\pi }\left(-\pi \mathrm{log}2\right)=\frac{1}{2}\mathrm{log}2\pi$
The last integral is well known Gauss integral.

karton

Expert

As for J, another way is to try to use the Fourier series for $\mathrm{ln}\mathrm{\Gamma }\left(x\right)$ discovered by E.E. Kummer in 1847:
$\mathrm{ln}\mathrm{\Gamma }\left(x\right)=\frac{\mathrm{ln}2\pi }{2}+\sum _{n=1}^{\mathrm{\infty }}\frac{\mathrm{cos}2\pi nx}{2n}+\sum _{n=1}^{\mathrm{\infty }}\frac{\left(\gamma +\mathrm{ln}2\pi n\right)\mathrm{sin}2\pi nx}{n\pi }\left(0
where $\gamma =0.577...$ is Euler's constant
Let's multiply this equality by x and integrate from 0 to 1.
Integrals on the right side:
${\int }_{0}^{1}xdx=\frac{1}{2}$
${\int }_{0}^{1}x\mathrm{cos}2\pi nxdx=0$
${\int }_{0}^{1}x\mathrm{sin}2\pi nxdx=-\frac{1}{2\pi n}$
Thus,
${\int }_{0}^{1}x\mathrm{ln}\mathrm{\Gamma }\left(x\right)=\frac{\mathrm{ln}2\pi }{4}-\frac{\gamma }{2{\pi }^{2}}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{{n}^{2}}-\frac{1}{2{\pi }^{2}}\sum _{n=1}^{\mathrm{\infty }}\frac{\mathrm{ln}2\pi n}{{n}^{2}}$
$=\frac{\mathrm{ln}2\pi }{4}-\frac{\gamma }{12}-\frac{1}{2{\pi }^{2}}\sum _{n=1}^{\mathrm{\infty }}\frac{\mathrm{ln}2\pi n}{{n}^{2}}$
if I am not mistaken. I don't know can this be simplified further.

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