Salvatore Boone

Answered

2022-01-03

What are the usual ways to follow in order to solve the integrals given below?

$I={\int}_{0}^{1}\mathrm{ln}\mathrm{\Gamma}\left(x\right)dx$

$J={\int}_{0}^{1}x\mathrm{\Gamma}\left(x\right)dx$

Answer & Explanation

poleglit3

Expert

2022-01-04Added 32 answers

As an addendum of sorts to the previous answers, there is the identity

$\mathrm{log}G(z+1)=\frac{z}{2}\mathrm{log}\left(2\pi \right)-\frac{z(z+1)}{2}+z\mathrm{log}\mathrm{\Gamma}(z+1)-z(\mathrm{log}z-1)-{\int}_{0}^{z}\mathrm{log}\mathrm{\Gamma}\left(t\right)dt$

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 $G(z+1)=\mathrm{\Gamma}\left(z\right)G\left(z\right)$ . (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

$\mathrm{log}G\left(2\right)=\frac{1}{2}\mathrm{log}\left(2\pi \right)-1+\mathrm{log}\mathrm{\Gamma}\left(2\right)-(\mathrm{log}1-1)-{\int}_{0}^{1}\mathrm{log}\mathrm{\Gamma}\left(t\right)dt$

$0=\frac{1}{2}\mathrm{log}\left(2\pi \right)-{\int}_{0}^{1}\mathrm{log}\mathrm{\Gamma}\left(t\right)dt$

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

${\int}_{0}^{z}{x}^{n}\psi \left(x\right)dx=$

$={(-1)}^{n}(\frac{{B}_{n+1}{H}_{n}}{n+1}-{\zeta}^{\prime}(-n))+\sum _{k=0}^{n}{(-1)}^{k}\left(\begin{array}{c}n\\ k\end{array}\right){z}^{n-k}(\zeta (-k,z)-\frac{{B}_{k+1}\left(z\right){H}_{k}}{k+1}$

where$B}_{n$ and

where

Thus, to evaluate

and you obtain the same solution as Andrew.

For the integral

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

where

lovagwb

Expert

2022-01-05Added 50 answers

As for the first integral, one can use the Euler's reflection formula $\mathrm{\Gamma}(1-z)\mathrm{\Gamma}\left(z\right)=\frac{\pi}{\mathrm{sin}\pi z}$

$I=\frac{1}{2}{\int}_{0}^{1}(\mathrm{log}(\mathrm{\Gamma}\left(x\right)+\mathrm{log}\mathrm{\Gamma}(1-x))dx=\frac{1}{2}{\int}_{0}^{1}\mathrm{log}\frac{\pi}{\mathrm{sin}\pi x}dx=$

$=\frac{1}{2}{\int}_{0}^{1}(\mathrm{log}\pi -\mathrm{log}\mathrm{sin}\pi x)dx=\frac{1}{2}\mathrm{log}\pi -\frac{1}{2\pi}{\int}_{0}^{\pi}\mathrm{log}\mathrm{sin}xdx$

$=\frac{1}{2}\mathrm{log}\pi -\frac{1}{2\pi}(-\pi \mathrm{log}2)=\frac{1}{2}\mathrm{log}2\pi$

The last integral is well known Gauss integral.

The last integral is well known Gauss integral.

karton

Expert

2022-01-09Added 439 answers

As for J, another way is to try to use the Fourier series for

where

Let's multiply this equality by x and integrate from 0 to 1.

Integrals on the right side:

Thus,

if I am not mistaken. I don't know can this be simplified further.

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