Compute this integral, using a method other than the Residue Theorem, $\int_0^\infty$ $\frac{1}{1+x}$$\frac{dx}{\sqrt{x}}$
Part (a) asks to compute the integral by means of the residue at x = -1.  I have done this just now, and the answer is $\pi$.
Part (b) asks, "can you see a simpler way to do it? explain."
I tried ordinary calculus methods, making two substitutions:  once for the square root, and once for $tan(\theta)$, and get 1 in the integrand, after the trigonometric functions $\sec^2(\theta)$ cancel each other out.  But this integral diverges, obviously.
So, what is the simpler way to do it, if it's not by ordinary calculus?  Would it perhaps be another complex analysis method that's not so apparent?
Thanks,
 A: You used the correct substitutions, but you forgot to change the bounds of the integral. 
Here is how your approach should have worked out:
$\displaystyle\int_{0}^{\infty}\dfrac{1}{(1+x)\sqrt{x}}\,dx$ $= \displaystyle\int_{0}^{\infty}\dfrac{1}{(1+t^2)t}2t\,dt$ $= \displaystyle\int_{0}^{\infty}\dfrac{2}{(1+t^2)}\,dt$ 
$= \displaystyle\int_{0}^{\pi/2}\dfrac{2}{1+\tan^2\theta}\sec^2\theta\,d\theta$ $= \displaystyle\int_{0}^{\pi/2}2\,d\theta = \pi$
The 2nd substitution was $t = \tan\theta$. When $t = 0$ we have $\theta = 0$ and when $t \to \infty$, we have $\theta \to \dfrac{\pi}{2}$. Thus, the bounds for the new integral are $0 \le \theta \le \dfrac{\pi}{2}$ and not $0$ to $\infty$. So this integral does not diverge.
Also, if you remember that $\dfrac{d}{dt}\arctan t = \dfrac{1}{1+t^2}$, then you can integrate $\displaystyle\int_{0}^{\infty}\dfrac{2}{(1+t^2)}\,dt$ without the need for a substitution.
A: Let $x = y^2$. We have: $$\int_0^\infty \frac{1}{1+x}\frac{1}{\sqrt{x}}\,{\rm d}x = \int_0^\infty \frac{2}{1+y^2}\,{\rm d}y = 2 \arctan y \big|_0^\infty = 2\frac{\pi}{2} = \pi.$$
