# Math Help - Integral

1. ## Integral

Hi I have another integral question that I wanted to ask.

Question: Prove that $\displaystyle\int_{-\infty}^\infty\frac{e^{2x}}{(e^{3x}+1)^2}dx=\frac{ 2\pi}{9\sqrt{3}}$

2. Originally Posted by nonsingular
Hi I have another integral question that I wanted to ask.

Question: Prove that $\displaystyle\int_{-\infty}^\infty\frac{e^{2x}}{(e^{3x}+1)^2}dx=\frac{ 2\pi}{9\sqrt{3}}$

$\displaystyle\int_{-\infty}^0\frac{e^{2x}}{(e^{3x}+1)^2}dx
+\int_{0}^\infty\frac{e^{2x}}{(e^{3x}+1)^2}dx$

the first one:
$\displaystyle\int_{-\infty}^0\frac{e^{2x}}{(e^{3x}+1)^2}dx =$
$\displaystyle\int_{-\infty}^0(\frac{e^x}{(e^{3x}+1)})^2dx =$
$\displaystyle\int_{0+}^1\frac{u^2}{(u^3+1)^2}dx =$

where $u=e^x$, and bounds change as well, and $dx= e^{-x}du =u^{-1} du$
$\displaystyle\int_{0+}^1\frac{u^2}{(u^3+1)^2}u^{-1}du =$
$\displaystyle\int_{0+}^1\frac{u}{(u^3+1)^2}du =$
For the indefinite, Wolfram gives:

integral x/(x^3+1)^2 dx = 1/18 (log(x^2-x+1)+(6 x^2)/(x^3+1)-2 log(x+1)+2 sqrt(3) tan^(-1)((2 x-1)/sqrt(3)))+constant

Pretty messy, but you can use it for the remaining half.

Hope this helps.

3. Hi

complex analysis

The integral $\int_{0}^{\infty} \frac{x}{(x^3+1)^2} ~dx$

Substitute $x = t^{\frac{1}{3}}$

The integral becomes $\frac{1}{3} \int_{0}^{\infty} \frac{t^{- \frac{1}{3}}}{(t+1)^2} ~ dt$

Then use this formula :

$\int_{0}^{\infty} x^{p-1} Q(x) ~dx = \frac{2\pi i}{1 - e^{2p \pi i}} \sum_{all~poles} Res[(z)^{p-1} Q(z)]$

Now sub. p into $\frac{2}{3}$

$= \frac{1}{3} \frac{2\pi i}{e^{\frac{2 \pi i}{3}}(e^{-\frac{2 \pi i}{3}} - e^{\frac{2 \pi i}{3}} )} ( (z)^{- \frac{1}{3}} )'_{z= -1}$

$= \frac{1}{3} \frac{2\pi i}{e^{\frac{2 \pi i}{3}}(-2i sin{ \frac{2 \pi}{3}} )} ( -\frac{1}{3} ) (-1)^{-\frac{4}{3} }$

$= \frac{ 2 \pi i}{ -6 i \sin{ \frac{ 2\pi }{3}} (-3) }$

$= \frac{2 \pi}{9 \sqrt3 }$

4. Originally Posted by nonsingular
Hi I have another integral question that I wanted to ask.

Question: Prove that $\displaystyle \int_{-\infty}^\infty\frac{e^{2x}}{(e^{3x}+1)^2}dx=\frac{ 2\pi}{9\sqrt{3}}$

let $e^{3x}=\tan^2t.$ then $\int_{-\infty}^\infty\frac{e^{2x}}{(e^{3x}+1)^2}dx=\frac{ 2}{3}\int_0^{\frac{\pi}{2}}(\sin t)^{\frac{1}{3}} (\cos t)^{\frac{5}{3}} \ dt=\frac{1}{3}B \left(\frac{2}{3},\frac{4}{3} \right)=\frac{1}{3} \Gamma \left(\frac{2}{3} \right) \Gamma \left(\frac{4}{3} \right),$ and the result follows because $\Gamma \left(\frac{4}{3} \right)=\frac{1}{3}\Gamma \left(\frac{1}{3} \right)$ and by Euler's reflection formula:

$\Gamma \left(\frac{2}{3} \right) \Gamma \left(\frac{1}{3} \right)=\frac{\pi}{\sin(\frac{\pi}{3})}=\frac{2\pi }{\sqrt{3}}.$