Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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
add comment

1 Answer

up vote 2 down vote accepted

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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.