TITLE: About the Beta function : $\text{B}\left(\frac{4}{3},\frac{2}{3}\right)$.
QUESTION [5 upvotes]: Find the value of : $\text{B}\left(\frac{4}{3},\frac{2}{3}\right)$, where $\text{B}(x,y)$ is the Beta function.

Why do I need this ? Because I want to calculate : $$
\int\limits_{ - \infty }^\infty  {\frac{{e^{2x} }}{{(e^{3x}  + 1)^2 }}} dx.$$
The exercise says : Calculate it with Beta function, I did calculate it using other means, but trying to use Beta I failed finding the exact value which I mentioned in the first line.
Please do not try to find it by calculating the integral in a different way (I've done it before), I'm just wondering if there are any quick method to do it.

REPLY [5 votes]: If you want to find $\mathrm{B}(\frac{4}{3},\frac{2}{3})$ you can use:

the relationship between $\mathrm{B}(x,y)$ and the gamma function $\Gamma (x)$
$$\begin{equation*}
\mathrm{B}(x,y)=\frac{\Gamma (x)\Gamma (y)}{\Gamma (x+y)},
\end{equation*}$$
the gamma function functional equation 
$$\begin{equation*}
\Gamma (x+1)=x\Gamma (x),\qquad \Gamma (1)=1,
\end{equation*}$$
Euler's reflection formula
$$\begin{equation*}
\Gamma (1-x)\Gamma (x)=\frac{\pi }{\sin (\pi x)},
\end{equation*}$$

to get successively
$$\begin{eqnarray*}
\mathrm{B}(\frac{4}{3},\frac{2}{3}) &=&\frac{\Gamma (\frac{4}{3})\Gamma (
\frac{2}{3})}{\Gamma (\frac{4}{3}+\frac{2}{3})}, \\
&=&\frac{\frac{1}{3}\Gamma (\frac{1}{3})\Gamma (\frac{2}{3})}{\Gamma (2)}, \\
&=&\frac{1}{3}\frac{\pi }{\sin (\frac{2}{3}\pi )}, \\
&=&\frac{2}{9}\pi \sqrt{3}.
\end{eqnarray*}$$