# Evaluating $\int_0^{\infty} \frac{\sqrt{x}}{x^3+1} \mathrm{d}x$ with contour integration

This integral has stumped me for quite a bit.

$$\int_0^{\infty} \frac{\sqrt{x}}{x^3+1} \mathrm{d}x$$

I have identified poles at $$x=e^{-\frac{i\pi}{3}}, e^{\frac{i\pi}{3}}, -1$$.

Edit: I have changed my approach to this question thanks to Hans Lundmark's comment.

Since $$\sqrt{x}$$ is a multivalued function, I'm using a keyhole contour with a branch cut at $$[0,\infty)$$.

However, I have issues computing the residue at $$x=-1$$.

By L'Hopital's Rule: $$\text{Res}_{x=-1}=\lim_{x \to -1}\frac{\sqrt{x}}{3x^2}$$

But $$\sqrt{-1}=\pm i$$ since $$-1=e^{\pm i\pi}$$.

Which value should I be taking in cases like these?

• – Hans Lundmark Sep 19 '18 at 12:26
• @HansLundmark dang I completely forgot $z$ is multivalued. I haven't done it before but I'll attempt using a keyhole contour. – fysh Sep 19 '18 at 12:53
• take also a look here: math.stackexchange.com/questions/110457/… – Masacroso Sep 20 '18 at 11:05