1. ## Residues

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

2. Originally Posted by Chandru1
Evaluate $\displaystyle \int\limits_{0}^{\infty} \frac{x^{1/3}}{1+x^{2}}\ dx$
Hint:

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

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

3. Let $\displaystyle x= t^3$

The integral

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

Sub $\displaystyle t = 1 /s$

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

Sub $\displaystyle s^2 = u$

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

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

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

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

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

Thus we have

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

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

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

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

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

My way does generalize this problem though:

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