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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.

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$$\frac13\Gamma\left(\frac13\right)\Gamma\left(\frac23\right) = \frac{\pi}3\csc\frac{\pi}3$$ from the usual reflection formula. – J. M. Apr 13 '13 at 17:14
How did you get $\Gamma\left(\frac{1}{3}\right)$ ? – aziiri Apr 13 '13 at 17:35
$\Gamma(n+1)=n\,\Gamma(n)$ – J. M. Apr 13 '13 at 17:36
Do you want to evaluate this integral using the $\beta$ function? – Mhenni Benghorbal Apr 13 '13 at 17:50
up vote 5 down vote accepted

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

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

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Using the change of variables $t=e^{3x}$ and $\frac{1}{1+t}=u$ in a row,we have

$$ \int\limits_{ - \infty }^\infty {\frac{{e^{2x} }}{{(e^{3x} + 1)^2 }}} dx = \frac{1}{3}\int\limits_{ 0 }^\infty {\frac{{t^{-1/3} }}{{( t + 1)^2 }}} dx = \frac{1}{3}\int\limits_{0 }^1 {{{u^{1/3} }}{{( 1-u)^{-1/3} }}} dx=\frac{1}{3}\beta(4/3,2/3) $$

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I think OP asked how to find $\text{B}\left(\frac{4}{3},\frac{2}{3}\right)$... – Cortizol Apr 13 '13 at 18:05
Anyway, thank you, I did exactly what you've done. – aziiri Apr 13 '13 at 18:42
@aziiri: You are welcome. – Mhenni Benghorbal Apr 13 '13 at 22:27
@Cortizol: Thanks for the comment. Anyways, Tavares did the job. – Mhenni Benghorbal Apr 13 '13 at 22:29
What's the down vote for? – Mhenni Benghorbal Apr 14 '13 at 23:00

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