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

Asked Sep 19 '18 at 12:22

Active Sep 23 '18 at 06:07

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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?

edited Sep 23 '18 at 06:07

asked Sep 19 '18 at 12:22

calquathater

See here: https://math.stackexchange.com/questions/117619/another-residue-theory-integral – 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. – calquathater Sep 19 '18 at 12:53
take also a look here: https://math.stackexchange.com/questions/110457/closed-form-for-int-0-infty-fracxn1-xmdx – Sep 20 '18 at 11:05

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