# wonderful integral

• Jul 2nd 2010, 05:32 AM
dapore
wonderful integral
• Jul 2nd 2010, 05:45 AM
Prove It
Quote:
To start...

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

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

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

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

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

which can be solved using partial fractions.
• Jul 2nd 2010, 10:20 PM
simplependulum
Quote:
Sub . $\displaystyle e^{3x} + 1 = \frac{1}{t}$

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

We have

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

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

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

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

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

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

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

Integration by parts ,

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

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

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

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

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

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

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

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

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