wonderful integral

**Prove It** · July 2nd 2010, 05:45 AM

Originally Posted by dapore

$\120dpi prove \: \: that \: \: \int_{-\infty }^{\infty }\left ( \frac{e^{x}}{e^{3x}+1} \right )^{2}dx=\frac{2\pi }{9\sqrt{3}}$

To start...

$\int{\left(\frac{e^x}{e^{3x} + 1}\right)^2\,dx} = \int{\frac{e^x}{(e^{3x} + 1)^2}\,e^x\,dx}$ .

Now make the substitution $u = e^x$ so that $du = e^x\,dx$ and the integral becomes

$\int{\frac{u}{(u^3 + 1)^2}\,du}$

$= \int{\frac{u}{[(u + 1)(u^2 - u + 1)]^2}\,du}$

$= \int{\frac{u}{(u + 1)^2(u^2 - u + 1)^2}\,du}$

which can be solved using partial fractions.

**simplependulum** · July 2nd 2010, 10:20 PM

Originally Posted by dapore

$\120dpi prove \: \: that \: \: \int_{-\infty }^{\infty }\left ( \frac{e^{x}}{e^{3x}+1} \right )^{2}dx=\frac{2\pi }{9\sqrt{3}}$

Sub . $e^{3x} + 1 = \frac{1}{t}$

$3e^{3x} ~dx = \frac{dt}{t^2}$

We have

$I = \frac{1}{3} \int_0^1 \left( \frac{ \sqrt[3]{t} }{ \sqrt[3]{1-t}}\right) ~dt$

$= \frac{1}{3} \frac{ \Gamma(\frac{4}{3}) \Gamma(\frac{2}{3})}{\Gamma(2)}$

$= \frac{1}{9} \Gamma(\frac{1}{3})\Gamma(\frac{2}{3})$

$= \frac{1}{9} \frac{\pi}{\sin(\frac{\pi}{3}) } = \frac{2\pi}{9\sqrt{3}}$

ps : or straightforward computation for $I = \int_0^{\infty} \frac{u}{(u^3+1)^2} ~du$

Use $u \mapsto \frac{1}{u}$

$I = \int_0^{\infty} \frac{u^3}{(u^3+1)^2}~du$

Integration by parts ,

$I = -\frac{1}{3} \left[ \frac{u}{u^3+1} \right]_0^{\infty} + \frac{1}{3} \int_0^{\infty} \frac{du}{u^3+1}$

$= \frac{1}{3} \int_0^{\infty} \frac{du}{u^3+1}$

Consider $J = \int_0^{\infty} \frac{du}{u^3+1}$

Use again $u \mapsto \frac{1}{u}$ and we have

$J = \int_0^{\infty} \frac{u}{u^3+1}~du$ so

$J + J = \int_0^{\infty} \frac{u+1 }{u^3+1}~du$

$= \int_0^{\infty} \frac{du}{u^2 - u + 1 } = \frac{4\pi}{3\sqrt{3}}$

$J = \frac{2\pi}{3\sqrt{3 }}$

Finally , $I = \frac{1}{3} J = \frac{2\pi}{9\sqrt{3 }}$

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