# Evaluation of $\int_0^\infty \frac{x^2}{1+x^5} \mathrm{d} x$ by contour integration

Consider the following integral: $$\int_{0}^{\infty} \frac{x^{2}}{1+x^{5}} \mathrm{d} x \>.$$

I did the following: Since $-1$ is a pole on the real axis, I took $z_{1}=e^{3\pi/5}$ then constructed an arc between $Rz_{1}$ and $R$ :

$f(e^{2\pi i /5}z)=f(z)$ it follows that : $(1-e^{2pi i/5})\int_{\alpha}f(z)dz + \int_{\beta}f(z)dz = 2\pi i \text{ Res } z_{2} f$ ,
where $\alpha$ is the way along the eral axis, $\beta$ is the way along the arc and $z_{2}=e^{i\pi/5}$

This gives: $\int_{0}^{\infty}...dx = 2\pi i (\frac{e^{2i\pi/5}}{-5})\cdot \frac{1}{(1-e^{2\pi i/5})}$

The real part of the right side in wolframalpha does not give the same result as the left side.

Does anybody see why? Please do tell.

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What is your $f$? $1+(e^{3\pi i/5}z)^5=1-z^5$ so I am not seeing $f(e^{3\pi i/5}z)= f(z)$. –  robjohn Nov 19 '11 at 23:50
Imaginary part of $2\pi i \frac{e^{2\pi i/5}}{-5}\frac{1}{1-e^{3\pi i/5}}$ is not zero. –  Norbert Nov 19 '11 at 23:54

I think you want to use the curve $e^{2\pi i/5}x$ for the inbound curve circling the poie at $e^{\pi i/5}$. In that case we get $$(1-e^{6\pi i/5})\int_{0}^{\infty} \frac{x^{2}}{1+x^{5}}\;\mathrm{d}x=2\pi i\frac{1}{5e^{2\pi i/5}}$$ Which gives $$\int_{0}^{\infty} \frac{x^{2}}{1+x^{5}}\;\mathrm{d}x=\frac{2\pi i}{5(e^{2\pi i/5}-e^{-2\pi i/5})}=\frac{\pi}{5\sin(2\pi/5)}$$

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$\csc (2\pi /5)=\sqrt{2-2/\sqrt{5}}$, in case the OP was comparing this answer with the output from WolframAlpha. –  Jonathan Gleason Nov 20 '11 at 0:31
@Jonathan: Thanks. I should have checked Wolfram's form of the answer. –  robjohn Nov 20 '11 at 0:37
robjohn, why $(1-e^{6\pi i /5})$ and not $(1-e^{2\pi i /5})$ ? Why for the residue : $\frac{1}{5e^{2\pi i/5}}$ and not : $\frac{(z_{2}^{2}}{5(z_{2})^{4}}$ with $z_{2}=e^{\pi i/5}$ ? –  VVV Nov 20 '11 at 0:42
@VVV: we get 2 factors of $e^{2\pi i/5}$ from the $x^2$ and 1 factor from the $\mathrm{d}x$; that gives $e^{6\pi i/5}$. The residue is $2\pi i$ times \begin{align}\lim_{z\to e^{\pi i/5}}\frac{z^2(z-e^{\pi i/5})}{1+z^5}&=\lim_{z\to e^{\pi i/5}}\frac{z^2}{5z^4}\\&=\frac{1}{5e^{2\pi i/5}}\end{align} –  robjohn Nov 20 '11 at 0:53
When you use the path $z=x$ and the path $z=e^{2pi i/5}x$ you substitute $x$ and $e^{2pi i/5}x$ for every occurrence of $z$. Thus, $\int_C\frac{z^2}{1+z^5}\;\mathrm{d}z$ becomes $\int_0^\infty\frac{x^2}{1+x^5}\;\mathrm{d}x$ on the curve $z=x$ and $\int_0^\infty\frac{e^{4\pi i/5}x^2}{1+x^5}\;e^{2\pi i/5}\;\mathrm{d}x$ on the curve $z=e^{2\pi i/5}x$ (but then we subtract since the curve is inbound). –  robjohn Nov 20 '11 at 1:29
I ran out of patience, but the partial fraction approach can give an indefinite integral in closed form, as the tenth roots of 1 have real and imaginary parts that can be written with nothing worse than square root signs. I got $$\frac{x^2}{1 + x^5} \; = \; \; \frac{A}{x + 1} + \frac{B_1 x + B_0}{x^2 - \frac{1}{2}(\sqrt 5 + 1)x + 1} + \frac{C_1 x + C_0}{x^2 + \frac{1}{2}(\sqrt 5 - 1)x + 1}.$$ Then I got $$A = \frac{1}{3}, \; B_0 = B_1 = \frac{1}{6}(\sqrt 5 - 1), \; \; C_0 = C_1 = -\frac{1}{6}(\sqrt 5 + 1).$$ The two quadratic denominators terms are positive over the reals. The rest of the task is also cookbook, you rewrite, for example $B_1 x + B_0$ as $\frac{B_1}{2}$ times the derivative of $x^2 - \frac{1}{2}(\sqrt 5 + 1)x + 1$ plus a constant term , call it $B_3.$ So, for this part, you get the indefinite integral being $$\frac{B_1}{2} \log(x^2 - \frac{1}{2}(\sqrt 5 + 1)x + 1) + B_3 \arctan(linear)$$ I imagine you can force your computer algebra system to show all the steps in this calculation while getting the specific numbers right. The beginning is to define $$\omega = e^{\pi i / 5} = \frac{1}{4}(1 + \sqrt 5) + \frac{i}{4} \sqrt{10 - 2 \sqrt 5}$$ with $$1 + x^5 = (x + 1)(x - \omega)(x - \omega^3)(x - \omega^7)(x - \omega^9)$$