Solve $\int_{0}^{\infty} \frac{x^{1/3}}{1+x^2}dx$ using residue calculus

Question

I need to find the value of the following integral using residue calculus - $$I = \int_{0}^{\infty} \frac{x^{1/3}}{1+x^2}dx$$

I have used substitution to start with. Let $x = t^3$. Then $I $ changes to $$I = \int_{0}^{\infty} 3\frac{t^3}{1+t^6}dt$$

On further substitution, $u = t^2$, I get $$I = \frac{3}{2} \int_{0}^{\infty} \frac{u}{1+u^3}du $$ At this point I am stuck. Also, I do not have any good idea about the choice of contour. Any kind of help is appreciated.

A good contour would be the (positively oriented) boundary of the sector of the circle centered at $0$ with radius $R>0$ whose circular arc is given by $R\exp(\text{i}t)$ for $t=0$ to $t=\frac{2\pi}{3}$. Then, take $R\to\infty$. — Batominovski, Apr 18 '16 at 09:48

score 1 · Answer 1 · answered Apr 18 '16 at 08:56

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Hint:$$\frac{u}{1+u^3}=\frac{1}{3}(\frac{u+1}{u^2-u+1}-\frac{1}{u+1})$$

answered Apr 18 '16 at 08:56

So, should I consider $f(z ) = \frac{z}{1+z^3}=\frac{1}{3}(\frac{z+1}{z^2-z+1}-\frac{1}{z+1})$? Its poles are coming at $z = \frac{1+3i}{2} , \frac{1-3i}{2}, -1$. – Dark_Knight Apr 18 '16 at 09:15
??? I would finish the complex partial fraction decomposition – reuns Apr 18 '16 at 09:15
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( @Dark_Knight : note how using the residue theorem is almost equivalent to performing the partial fraction decomposition) – reuns Apr 18 '16 at 09:17
What about the choice of contour? – Dark_Knight Apr 18 '16 at 09:27

Solve $\int_{0}^{\infty} \frac{x^{1/3}}{1+x^2}dx$ using residue calculus

1 Answers1