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.

  • 1
    $\begingroup$ You're doing the math wrong, I suspect you got a wrong sign for the residue in $-i$. $\endgroup$ – Daniel Fischer Jul 26 '13 at 21:47

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:

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


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.