About the Beta function : $\text{B}\left(\frac{4}{3},\frac{2}{3}\right)$.

Question

Find the value of : $\text{B}\left(\frac{4}{3},\frac{2}{3}\right)$, where $\text{B}(x,y)$ is the Beta function.

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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.

Answer 1

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

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

to get successively

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