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I have problem with integration $I = \int_0^\infty \frac{x^{1/3}}{1+x^2} dx$ using residue theory.

Define $\log z$ on the complex plane except the positive real line so that its imaginary part is in $(0, 2\pi)$.

Consider a counterclockwise path $C$ with a small circle around $0$, a real line from $0$ to $\infty$ on the upper half plane, a big circle around $0$, and a real line from $\infty$ to $0$ on the lower half plane.

The integral $\int_0^\infty \frac{z^{1/3}}{1+z^2} dz$ converges to $(1-e^{2\pi i /3})I$ as on the lower real line $z^{1/3} = e^{2\pi i /3}x^{1/3}$.

Using residues I have $\frac{1}{2\pi i} \int_0^\infty \frac{z^{1/3}}{1+z^2} dz = \frac{1}{2i} (e^{\pi i /6} - e^{\pi i /2})$.

But this leads to $I = \frac{\pi i}{\sqrt 3}$.

Where did this argument go wrong? Is it not OK to apply residue theorem in this situation?

Another User
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  • Also see here . Exactly same question.If this post doesn't answer you question, then mention in your post and add proper link. – SoG Aug 12 '22 at 15:36

1 Answers1

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Residue equals:

$$Res=\frac{e^{\pi i/6}}{2i}+\frac{e^{\pi i/2}}{-2i}$$

The integral goes to:

$$(1-e^{2\pi i/3})I=2\pi i\cdot Res=\pi\left( e^{\pi i/6}-i\right)\Rightarrow I=\frac{\pi}{\sqrt{3}}$$

MathFail
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    Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here.Instead of answering it, it would be better to look for a good duplicate target, or help the user by posting comments.You can use this – SoG Aug 12 '22 at 15:30