# Evaluate $\int_{0}^{\infty} \frac{x^{1/2}}{1 + x^2}\,\mathrm dx$ using Residue Theorem.

Evaluate $\int_{0}^{\infty} \frac{x^{1/2}}{1 + x^2}\,\mathrm dx$ using the Residue Theorem.

I have been given a formula to compute integrals of this type:

$I = \int_{0}^{\infty} R(x) x^{\alpha}\,\mathrm dx = \frac{2\pi i}{1 - e^{2\pi i \alpha}} \sum_{a\in\mathbb{C}}\operatorname{Res}(R(z)z^{\alpha}, a)$ for $0 < \alpha < 1$,

however when I use this formula to compute the above integral I am finding that the answer is $\frac{i\pi}{\sqrt{2}}$. I know that the answer should not be complex so either I am doing the math wrong or this formula is incorrect.

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You're doing the math wrong, I suspect you got a wrong sign for the residue in $-i$. –  Daniel Fischer Jul 26 '13 at 21:47

On $\mathbb{C} \setminus \mathbb{R}_{\geqslant 0}$, we choose the branch of $\sqrt{z}$ with $\sqrt{i} = e^{i\pi/4} = \frac{1+i}{\sqrt{2}}$. Then the residues of $\frac{\sqrt{z}}{z^2+1}$ are

\begin{align} \operatorname{Res}_i \frac{\sqrt{z}}{z^2+1} &= \frac{\sqrt{i}}{2i} = \frac{1+i}{2i\sqrt{2}}\\ \operatorname{Res}_{-i} \frac{\sqrt{z}}{z^2+1} &= \frac{\sqrt{-i}}{-2i} = \frac{-1+i}{-2i\sqrt{2}} = \frac{1-i}{2i\sqrt{2}}, \end{align}

the sum of the residues is then

$$\frac{1+i}{2i\sqrt{2}} + \frac{1-i}{2i\sqrt{2}} = \frac{1}{i\sqrt{2}}$$

and the formula yields $$\frac{2\pi i}{1 - (-1)} \frac{1}{i\sqrt{2}} = \frac{\pi}{\sqrt{2}} \in \mathbb{R}.$$

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You can avoid dealing with square roots by substituting $x=u^2$ to get

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

In this case, we may use a simple semicircular contour in the upper half-plane, with two poles at $z=e^{i \pi/4}$ and $z=e^{i 3 \pi/4}$. Theiintegral is simply $i 2 \pi$ times the sum of the residues at these poles:

$$i 2 \pi \left [\frac{e^{i \pi/2}}{4 e^{i 3 \pi/4}} + \frac{e^{i 3 \pi/2}}{4 e^{i 9 \pi/4}} \right ] = \pi \cos{\frac{\pi}{4}} = \frac{\pi}{\sqrt{2}}$$

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