Residues

**Chandru1** · May 27th 2010, 03:31 AM

Evaluate $\int\limits_{0}^{\infty} \frac{x^{1/3}}{1+x^{2}}\ dx$

**chiph588@** · May 27th 2010, 06:58 AM

Originally Posted by Chandru1

Evaluate $\int\limits_{0}^{\infty} \frac{x^{1/3}}{1+x^{2}}\ dx$

Hint:

Let $a=\frac13$ and $x=e^t \implies dx = e^tdt$

Thus $\int_0^\infty \frac{x^a}{x^2+1}\ dx = \int_{-\infty}^\infty \frac{e^{(a+1)t}}{e^{2t}+1}\ dt$

**simplependulum** · May 27th 2010, 07:14 AM

Let $x= t^3$

The integral

$I = 3 \int_0^{\infty} \frac{ t^3 }{1+ t^6} ~dt$

Sub $t = 1 /s$

$I = 3 \int_0^{\infty} \frac{s}{1+s^6}~ds$

Sub $s^2 = u$

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

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

$I = \frac{3}{2} \left[ \int_0^{\infty} \frac{1+u}{(1+u)(1-u+u^2)}~du - \int_0^{\infty} \frac{u}{1+u^3}~du \right]$

But by substituting $u= 1/y$ in the integral $\int_0^{\infty} \frac{du}{1+u^3}$ , we have

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

Thus we have

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

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

$I = \frac{3}{4} \cdot \frac{4\pi}{3\sqrt{3}}$

$I = \frac{\pi}{\sqrt{3}}$

**chiph588@** · May 27th 2010, 02:37 PM

This is a much better way than my method! It requires no complex analysis.

My way does generalize this problem though:

Given $-1<\Re(a)<1$ , $\int_0^\infty \frac{x^a}{x^2+1}\ dx = \frac12\pi\sec\left(\frac{\pi a}{2}\right)$

p.s. May I ask how you came up with your solution?

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