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 $R &gt; 1$  $$\gamma_1(t) = t, t \in [0,R]$$ $$\gamma_2(t) = Re^{it}, t \in [0,\frac{2\pi}{5}]$$ $$\gamma_3(t) = -te^{i\frac{2\pi}{5}}, t \in [0,R]$$  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 ?
I've been looking at  $$\int\limits_0^\infty  {\frac{{{x^n}}}{{1 + {x^m}}}dx }$$  It seems that it always evaluates in terms of $\sin X$ and $\pi$, where $X$ is to be determined. For example:  $$\displaystyle \int\limits_0^\infty  {\frac{{{x^1}}}{{1 + {x^3}}}dx = } \frac{\pi }{3}\frac{1}{{\sin \frac{\pi }{3}}} = \frac{{2\pi }}{{3\sqrt 3 }}$$  $$\int\limits_0^\infty  {\frac{{{x^1}}}{{1 + {x^4}}}dx = } \frac{\pi }{4}$$  $$\int\limits_0^\infty  {\frac{{{x^2}}}{{1 + {x^5}}}dx = } \frac{\pi }{5}\frac{1}{{\sin \frac{{2\pi }}{5}}}$$  So I guess there must be a closed form - the use of $\Gamma(x)\Gamma(1-x)$ first comess to my mind because of the $\dfrac{{\pi x}}{{\sin \pi x}}$ appearing. Note that the arguments are always the ratio of the exponents, like $\dfrac{1}{4}$, $\dfrac{1}{3}$ and $\dfrac{2}{5}$. Is there any way of finding it? I'll work on it and update with any ideas.    UPDATE:  The integral reduces to finding   $$\int\limits_{ - \infty }^\infty  {\frac{{{e^{a t}}}}{{{e^t} + 1}}dt} $$  With $a =\dfrac{n+1}{m}$ which converges only if   $$0 &lt; a &lt; 1$$  Using series I find the solution is     $$\sum\limits_{k =  - \infty }^\infty  {\frac{{{{\left( { - 1} \right)}^k}}}{{a + k}}} $$
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