Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
2  
$$\frac13\Gamma\left(\frac13\right)\Gamma\left(\frac23\right) = \frac{\pi}3\csc\frac{\pi}3$$ 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
add comment

2 Answers

up vote 4 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

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

share|improve this answer
add comment

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

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

share|improve this answer
2  
I think OP asked how to find $\text{B}\left(\frac{4}{3},\frac{2}{3}\right)$... –  Cortizol Apr 13 '13 at 18:05
1  
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
show 5 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.