Welcome to mathimatikoi.org forum; Enjoy your visit here.

## Integral involving $\Gamma$ function

Real Analysis
Grigorios Kostakos
Founder
Articles: 0
Posts: 460
Joined: Mon Nov 09, 2015 1:36 am
Location: Ioannina, Greece

### Integral involving $\Gamma$ function

Evaluate $\displaystyle\int_{0}^{1}\log({\Gamma(x+1)})\,dx\,,$ where $\Gamma(x)$ is the gamma function.
Grigorios Kostakos
ZardoZ
Articles: 0
Posts: 13
Joined: Wed Nov 11, 2015 12:47 pm

### Re: Integral involving $\Gamma$ function

We will use the identity $\displaystyle \Gamma(z+1)=z\cdot \Gamma(z)$, so

$\int_{0}^{1}\log \left(\Gamma(x+1)\right)\;\mathbb{d}x = \int_{0}^{1}\ln\left(x\cdot \Gamma(x)\right)\;\mathbb{d}x=\int_{0}^{1}\log(x)\;\mathbb{d}x+\int_{0}^{1}\log\left(\Gamma(x)\right)\;\mathbb{d}x=-1+\underbrace{\int_{0}^{1}\log\left(\Gamma(x)\right)\;\mathbb{d}x}_{\rm I}\;\;\left(\color{red}\dagger\right)$ and we will have to compute the integral denoted by I. Using the formula $\displaystyle \Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)}$, we will make the "trick" presented in the computations below
$\begin{eqnarray*} \textrm{I} &=& \int_{0}^{1}\log\left(\Gamma(x)\right)\;\mathbb{d}x\\ &=& \frac{1}{2}\left(\int_{0}^{1}\log\left(\Gamma(x)\right)\;\mathbb{d}x+\int_{0}^{1}\log\left(\Gamma(1-x)\right)\;\mathbb{d}x\right)\\&=&\frac{1}{2}\int_{0}^{1}\log\left(\Gamma(x)\Gamma(1-x)\right)\;\mathbb{d}x\\&=& \frac{1}{2}\int_{0}^{1}\log\left(\frac{\pi}{\sin(\pi x)}\right)\;\mathbb{d}x\\ &=& \frac{1}{2}\int_{0}^{1}\log(\pi)-\log(\sin(\pi x ))\;\mathbb{d}x\\&=& \frac{1}{2}\log(\pi)-\frac{1}{2}\int_{0}^{1}\log(\sin(\pi x ))\;\mathbb{d}x\\&=&\frac{1}{2}\log(\pi)-\frac{1}{2\pi } \underbrace{\int_{0}^{\pi}\log(\sin( x ))\;\mathbb{d}x}_{\textrm{J}}\;\;\left(\color{red}\ddagger\right)\end{eqnarray*}$
and we are left with the computation of the integral J.
$\begin{eqnarray*} \textrm{J} &=& \int_{0}^{\pi}\log(\sin( x ))\;\mathbb{d}x\\ &=&2 \int_{0}^{\frac{\pi}{2}}\log(\sin(x))\;\mathbb{d}x\;\;\color{yellow}{(*)}\\ &=&2 \int_{0}^{\frac{\pi}{2}}\log\left(\sin\left(\frac{\pi}{2}-x\right)\right)\;\mathbb{d}x\\ &=&2 \int_{0}^{\frac{\pi}{2}}\log\left(\cos(x)\right)\;\mathbb{d}x\;\;\color{yellow}{(**)}\end{eqnarray*}$
so $\displaystyle 2\textrm{J}=2\int_{0}^{\frac{\pi}{2}}\log(\sin(x))+\log(\cos(x))\;\mathbb{d}x\Rightarrow \textrm{J}=\int_{0}^{\frac{\pi}{2}}\log\left(\sin(x)\cos(x)\right)\;\mathbb{d}x=\int_{0}^{\frac{\pi}{2}}\log\left(\frac{\sin(2x)}{2}\right)\;\mathbb{d}x$.
$\begin{eqnarray*} \textrm{J} &=& \int_{0}^{\frac{\pi}{2}}\log\left(\frac{\sin(2x)}{2}\right)\;\mathbb{d}x\\ &=& \int_{0}^{\frac{\pi}{2}}\log(\sin(2x))\;\;\mathbb{d}x-2\int_{0}^{\frac{\pi}{2}}\log(2)\;\mathbb{d}x\\&=&\frac{1}{2}\underbrace{\int_{0}^{\pi}\log(\sin(x))\;\mathbb{d}x}_{\textrm{J}}-\frac{\pi}{2}\log(2)\end{eqnarray*}$
and $\textrm{J}=\frac{1}{2}\textrm{J}-\frac{\pi}{2}\log(2)\Rightarrow \textrm{J}=-\pi\log(2)$. And finally going back to $\displaystyle \left(\ddagger\right)$, we get $\displaystyle \textrm{I}=\frac{1}{2}\log(\pi)-\frac{1}{2\pi }\left(-\pi \log(2)\right)=\frac{1}{2}\log(2\pi)$ and plugging this in the equation $\displaystyle \left(\dagger\right)$ we get
$\boxed{\displaystyle\int_{0}^{1}\log\left(\Gamma(x+1)\right)\;\mathbb{d}x=\frac{1}{2}\log(2\pi)-1}$
Grigorios Kostakos
Founder
Articles: 0
Posts: 460
Joined: Mon Nov 09, 2015 1:36 am
Location: Ioannina, Greece

### Re: Integral involving $\Gamma$ function

An alternative calculation based on some "quite heavy" equalities:
\begin{align*}
\displaystyle\int_{0}^{1}\log({\Gamma(x+1)})\,dx&=\int_{0}^{1}-\gamma x +\sum_{n=2}^\infty \frac{\zeta(n)}{n} \, (-x)^{n}\,dx\\
& =-\frac{\gamma}{2}+\sum_{n=2}^\infty \frac{\zeta(n)}{n}\int_{0}^{1}(-x)^{n}\,dx\\
& =-\frac{\gamma}{2}+\sum_{n=2}^\infty \frac{\zeta(n)}{n}\,\frac{(-1)^n}{n+1}\\
& =-\frac{\gamma}{2}+\sum_{n=2}^\infty \frac{(-1)^n\,\zeta(n)}{n\,(n+1)}\\
& =-\frac{\gamma}{2}+\frac{1}{2}\,(\log2+\log\pi+\gamma-2)\\
& =-\frac{1}{2}\,\log(2\pi)-1\,,
\end{align*} where $\zeta$ is Riemann zeta function and $\gamma$ is Euler–Mascheroni constant.
Grigorios Kostakos
akotronis
Former Team Member
Articles: 0
Posts: 36
Joined: Mon Nov 09, 2015 12:29 am

### Re: Integral involving $\Gamma$ function

Is there a "direct" way computing of $\displaystyle\sum_{n=2}^\infty \frac{(-1)^n\,\zeta(n)}{n\,(n+1)}$ ?
Grigorios Kostakos
Founder
Articles: 0
Posts: 460
Joined: Mon Nov 09, 2015 1:36 am
Location: Ioannina, Greece

### Re: Integral involving $\Gamma$ function

akotronis wrote:Is there a "direct" way computing of $\displaystyle\sum_{n=2}^\infty \frac{(-1)^n\,\zeta(n)}{n\,(n+1)}$ ?
See Series involving Riemann zeta function.
Grigorios Kostakos