# Improper Integral using Residue Theory.

I want to evaluate the improper integral $\int\limits_{0}^{\infty}\frac{x^{1/4}}{1+x^3}\, dx$ via residue theorem but something odd is happening.

When I use the key-hole contour where I integrate above/below the postive real axis, I end up getting that the real and imaginary part of the integral is $-\int\limits_{0}^{\infty}\frac{t^{1/4}}{1+t^3}dt + \int\limits_{0}^{\infty}\frac{t^{1/4}}{1+t^3}dt*i$

When I compute the contour via residues I get answers that not only do not match up to numerical calculation but I but have different real and imaginary scaler values.

The 3 roots of $1+z^3$ are $-1, 1/2+\frac{\sqrt{3}}{2i}, 1/2-\frac{\sqrt{3}}{2}*i$

And residue values computed at each are:

for $-1$, $\frac{\sqrt{2}}{6}(1+i)$

for $1/2 + \sqrt(3)/2i$, $\frac{-(\sqrt{3}-1)\sqrt{2}}{12} - \frac{(\sqrt{3}+1)\sqrt{2}}{12}i$

for $1/2 - \sqrt(3)/2i$, $\frac{-(\sqrt{3}-1)\sqrt{2}}{12} + \frac{(\sqrt{3}+1)\sqrt{2}}{12}i$

Now clearly the sum of these multiplied by $2\pi*i$ will not have real and imaginary parts which are scaler multiples of each other.

What did I do wrong?

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Did you take into account that when you go around the big almost-circle in this countour your add\substract $\,2\pi\,$ to the argument of $\,z^{1/4}=e^{\frac{1}{4}Log(z)}\,$? – DonAntonio Oct 15 '12 at 12:15
I believe so, as on the contour below the postive real axis you add $2\pi*i$ to $Log(z)$ to get $e^{\frac{1}{4}*Log(z) + 1/2\pi*i}$ which equals $z^{\frac{1}{4}}$ – Mike Oct 15 '12 at 18:20
oops I kept trying to edit as my TeX was wrong, I meant $e^{\frac{1}{4}*Log(z) + 1/2\pi*i}$ equals $z^{\frac{1}{4}}*e^{\frac{1}{2}\pi*i}$ – Mike Oct 15 '12 at 18:26

Just as a remark, this kind of integrals can be evaluated in terms of the Euler Beta function:

$$I=\int_{0}^{+\infty}\frac{x^{1/4}}{1+x^3}dx = \frac{1}{3}\int_{0}^{+\infty}\frac{1}{y^{7/12}(1+y)}dy,$$

use now the substitution $\frac{1}{1+y}=1-z$ and the identity $\Gamma(z+1)=z\,\Gamma(z)$ to have:

$$I = \frac{1}{3}\int_{0}^{1} z^{-7/12}(1-x)^{-5/12} dz = \frac{\Gamma(5/12)\,\Gamma(7/12)}{3},$$

but since $\Gamma(z)\,\Gamma(1-z)=\frac{\pi}{\sin(\pi z)}$ we have:

$$I = \frac{1}{3}\cdot\frac{\pi}{\sin\frac{5\pi}{12}} = \frac{\pi(1+\sqrt{3})}{6\sqrt{2}}.$$

The result suggests that, in the computation of $I$ through residues, you have to consider the behaviour of the integrand in the neighbourhood of any twelth roots of unity.

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Pretty development. +1 – DonAntonio Oct 15 '12 at 12:09
I was able to find the solution, I had calculated the residues wrong and it numerically matches wolfram and ti89 but your answer does not match up, I'm not sure if it was rounding error or not. – Mike Oct 15 '12 at 19:22