Evaluating the improper integral $ \int_{0}^{\infty}{\frac{x^2}{x^{10} + 1}\mathrm dx} $

Question

I am trying to solve the following integral, but I don't have a solution, and the answer I am getting doesn't seem correct.

So I am trying to integrate this:

$$ \int_{0}^{\infty}{\frac{x^2}{x^{10} + 1}\,\mathrm dx} $$

To integrate this, I want to use a contour that looks like a pizza slice, out of a pie of radius R. One edge of this pizza slice is along the positive x-axis, if that makes sense. Since $ z^{10} + 1 $ has 10 zeroes, the slice should only be one tenth of a whole circle. So let's call this contour $ C $. Then:

$$ \int_{C}{\frac{z^2}{z^{10} + 1}\,\mathrm dz} = 2 \pi i\,\operatorname{Res}(\frac{x^2}{x^{10} + 1}, e^{i \pi/10}) $$ This is because this slice contains only one singularity. Furthermore:

$$ \int_{C}{\frac{z^2}{z^{10} + 1}\,\mathrm dz} = \int_0^R{\frac{z^2}{z^{10} + 1}\,\mathrm dz} + \int_\Gamma{\frac{z^2}{z^{10} + 1}\,\mathrm dz} $$

And then, by the M-L Formula, we can say that $ \int_\Gamma{\frac{z^2}{z^{10} + 1}\,\mathrm dz} $ goes to $0$ as $R$ goes to infinity. Evaluating $ 2 \pi i\ \operatorname{Res}(\frac{x^2}{x^{10} + 1}, e^{i \pi/10}) $ I get $ \dfrac{\pi}{e^{i \pi/5}} $. Since this answer isn't real, I don't think this could be correct. What did I do wrong?

It is the arc of the circle, plus the diagonal line connecting the arc to the origin. — Max, May 15 '13 at 10:27
You should get the answer $ {\frac {\pi }{10\sin \left( \frac{3\pi}{10} \right) }} .$ — Mhenni Benghorbal, May 15 '13 at 10:27
Your Contour should consist of three pieces. The third piece is the diagonal line. — Mhenni Benghorbal, May 15 '13 at 10:29
Ok, I see the problem now. But the third part converges to infinity as R goes to infinity, so how can I fix this? — Max, May 15 '13 at 10:38
I've enhanced the typesetting of your question. Here are some tips and guides to help you do that yourself on future questions. — Lord_Farin, May 15 '13 at 10:49
Letting $t=\frac{1}{x^{10}+1}$ should yield something in terns of the Beta function. — L. F., May 15 '13 at 10:58

score 1 · Answer 1 · edited Apr 13 '17 at 12:20

1

This answer, applied with $n=2$ and $m=10$, yields $$ \int_0^\infty\frac{x^2}{1+x^{10}}\,\mathrm{d}x=\frac\pi{10}\csc\left(\frac{3\pi}{10}\right) $$

edited Apr 13 '17 at 12:20

Community

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answered May 15 '13 at 10:53

robjohn

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Evaluating the improper integral $ \int_{0}^{\infty}{\frac{x^2}{x^{10} + 1}\mathrm dx} $

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