Show that $\int_{0}^{\infty}\frac{x^2}{1+x^5}dx = \frac{\pi}{5\sin(\frac{2\pi}{5})}$

Question

I have to show that

$$\int_{0}^{\infty}\frac{x^2}{1+x^5}dx = \frac{\frac{\pi}{5}}{\sin(\frac{2\pi}{5})}$$

I know how to solve almost every polynomial integral. However, this one is not even and I can not see how to prove this. I try different kinds of parametrization. I tried the semicircle from $[0,\frac{pi}{2}]$, from $[0,\frac{\pi}{4}]$ and so on. I still can not solve this integral. I know the residue theorem and also I know that I have to consider the roots of unity of the equation and the solutions are:

$$z_k = e^{i\frac{n\pi}{5}}, n = 1,3,5,7,9$$ for $k = 1,2,3,4,5$

If I use the following parametrization for $R > 1$

$$\gamma_1(t) = t, t \in [0,R]$$ $$\gamma_2(t) = Re^{it}, t \in [0,\frac{2\pi}{5}]$$ $$\gamma_3(t) = -te^{i\frac{2\pi}{5}}, t \in [0,R]$$

Then using L'hopital to simplify the calculation of the residues we can note that the only residue in the first quadrant is $e^{i\frac{\pi}{5}}$.

Then $$\int_{\gamma}f(z) = 2\pi i \mbox{ Res}(f,e^{i\frac{\pi}{5}}) = \frac{2\pi i}{5e^{\frac{2\pi i}{5}}}$$

Also,

$$\int_{\gamma}f(z) = \int_{0}^{R}f(t)dt + \int_{0}^{\frac{2\pi}{5}}f(Re^{it})iRe^{it}dt - \int_{0}^{R}f(te^{i\frac{2\pi}{5}})e^{i\frac{2\pi}{5}}dt$$

As $R \to \infty$,

$$ (1 - e^{\frac{6\pi i}{5}})\int_{0}^{\infty}f(t)dt = \frac{2\pi i}{5e^{\frac{2\pi i}{5}}} $$

Therefore $$ \int_{0}^{\infty}f(t)dt = \frac{2\pi i}{(1 - e^{\frac{6\pi i}{5}})5e^{\frac{2\pi i}{5}}} $$

Still I can not get the correct answer. Why ?

For the real approach, simply set $x=1/\sqrt{u^2-1}$ and the integral should reduce tremendously. — Simply Beautiful Art, Sep 03 '17 at 22:48
You mean the contour depends on the integral? If you wanted to be one-counter-does-all see robjohn's answer in the given link. — Simply Beautiful Art, Sep 03 '17 at 23:01
:P Well, different integrals, different contours (usually). What more could you expect from it all? — Simply Beautiful Art, Sep 03 '17 at 23:03
Your residue is indeed correct, but as you noted, the integral at the end is not suitable for any calculations. Highly recommended that you use wedges with angles that are multiples of $2\pi/5$. — Simply Beautiful Art, Sep 04 '17 at 00:12
Yes, the residue is correct. It appears I mistook the problem. You want to evaluate the other arm? Notice that$$\frac{(xe^{2\pi i/5})^2}{1+(xe^{2\pi i/5})^5}=\frac{x^2}{1+x^5}e^{4\pi i/5}$$Thus, you have something of the form...$$(1-e^{4\pi i/5})\int_0^\infty\frac{x^2}{1+x^5}~\mathrm dx=\frac{2\pi i}{5e^{2\pi i/5}}$$Hopefully that makes sense? — Simply Beautiful Art, Sep 04 '17 at 00:23
Note that $(e^{2\pi i/5})^2=e^{4\pi i/5}$. Where'd you get the $6$? — Simply Beautiful Art, Sep 04 '17 at 00:31
Ah yes, it should be a 6, my bad. Now all you have to do is show that $$\frac{2\pi i}{5e^{2\pi i}(1-e^{6\pi i})}=\frac\pi{5\sin(2\pi i/5)}$$ — Simply Beautiful Art, Sep 04 '17 at 00:56
@SimplyBeautifulArt I get it! Thank you! We have to make magic to obtain this kind of solutions. lol — Dr Richard Clare, Sep 04 '17 at 01:14

Show that $\int_{0}^{\infty}\frac{x^2}{1+x^5}dx = \frac{\pi}{5\sin(\frac{2\pi}{5})}$

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