I'm working through Boas Chapter 14, and I came across an interesting issue in problem 33.
I get a value for the integral $I = \int_{0}^{\infty} \frac{x^{1/2}}{1+x^2} dx = \frac{i\pi\sqrt{2}}{2}$. While the answer in the book (and elsewhere online including this website) has no $i$ component, and everything works just fine.
This question has been asked before: Evaluate $\int_{0}^{\infty} \frac{x^{1/2}}{1 + x^2}\,\mathrm dx$ using Residue Theorem. but the answer on that question from 2013 has an error in the calculation of the Residue at $-i$.
Specifically, the answer claims
$$ \text{Res}_{-i} \frac{\sqrt{z}}{z^2+1} = \frac{\sqrt{-i}}{-2i} = \frac{-1+i}{-2i\sqrt{2}} = \frac{1-i}{2i\sqrt{2}} $$ However $$ \sqrt{-i} = (e^{-i\pi/2})^{1/2} = e^{-i\pi/4} = \cos{\frac{\pi}{4}} - i\sin{\frac{\pi}{4}} = \frac{1}{\sqrt{2}} -\frac{i}{\sqrt{2}} $$ Thus $$ \frac{\sqrt{-i}}{-2i} = \frac{1-i}{-2i\sqrt{2}} = \frac{-1+i}{2i\sqrt{2}} $$
With this correction, the sum of the residues is $\frac{\sqrt{2}}{2}$, and therefore the value of the contour integral $i\pi\sqrt{2}$.
If you carry out the rest of the contour integration, you pick up the $1/2$ term, and end up with $I = \frac{i\pi\sqrt{2}}{2}$.
Am I missing something here? I saw elsewhere online people mistakenly calculating the Residue at $-i$, is there another way to lose this imaginary component of the answer?
EDIT:
My friend pointed out my error - incase anyone else makes the same mistake as me.
The polar angle must be measured in the positive direction. Meaning
$$ -i = e^{3\pi/2} \qquad \text{and not}\qquad -i=e^{-\pi/2} $$
Then, when you apply the square root, you end up with
$$ -i = e^{3\pi/4} = \frac{-1 + i}{\sqrt{2}} $$
Which eliminates the imaginary component in the sum of the residues, and therefore in the final answer.
You seem to be using the wrong branch for the square root. We have $-i=e^{3\pi i/2}$; note that because of how this integral is being calculated (using a "pacman contour" with pacman's mouth along the non-negative real axis), in the residue calculus we should use the branch where the argument lies between $0$ and $2\pi$. So, \begin{align} \text{Res}\left(\frac{\sqrt{z}}{z^2+1};-i\right) = \frac{\sqrt{-i}}{-2i}=\frac{e^{3\pi i/4}}{2e^{3\pi i /2}} = \frac{e^{-3\pi i/4}}{2}=\frac{-e^{i\pi/4}}{2}=\frac{-1-i}{2\sqrt{2}}, \end{align} which is the same thing as $\frac{1-i}{2i\sqrt{2}}$ as in the other answer.