Exercise. Evaluate \begin{equation*} \int_{-\infty}^{+\infty}\frac{x^{2k}}{1+x^{2n}}dx \end{equation*} , where $0\leq k < n$ and $n,k \in \mathbb{N_0}$
My attempt. I have tried to solve the integral through the residues method,i.e., completing the segment $[-R,R]$ with a curve $C_R$ such that $|z| = R, Im(z)\geq 0$ and then declaring that \begin{equation*} \int_D \dots = \int_{-R}^{R}\dots + \int_{C_R} \dots \end{equation*} , where \begin{equation*} \int_D f(x)dx= 2\pi i \sum_{a_j \in int(D)}Res(f,a_j) \end{equation*} By calculating $f$ singularities I've that $x=\sqrt[n]{i}$ is a pole of order $n$.
But how I am stuck calculating the associated residues. Is there some alternative method to solve this? Thanks for ali the helpin advance.
I got nerdsniped and ended up solving the problem. Hope it helps you.
Let $f(z) = \dfrac{z^{2k}}{z^{2n}+1}$. If $\omega$ is any $2n$th root of $-1$, then $f$ has simple pole at $\omega$ and \begin{align*} \operatorname{Res}(f,\omega) &= \lim_{z\to\omega} \frac{z^{2k}(z-\omega)}{z^{2n} - \omega^{2n}} \\&= \frac{\omega^{2k}}{2n\omega^{2n-1}} = - \frac{\omega^{2k+1}}{2n} \end{align*} Set $\omega = \exp\left(\frac{\pi i}{2n}\right)$. The $2n$th roots of $-1$ that are in the upper half-plane are $\omega, \omega^3, \omega^5, \dots, \omega^{2n-1}$. Using your contour notation and assuming $R > 1$, \begin{align*} \int_D f(z)\,dz &= 2\pi i \sum_{j=0}^{n-1} \operatorname{Res}(f,\omega^{2j+1}) \\&= -\frac{\pi i}{n} \sum_{j=0}^{n-1} (\omega^{2j+1})^{2k+1} \\&= -\frac{\pi i}{n} \sum_{j=0}^{n-1} (\omega^{2k+1})^{2j+1} \end{align*} Let $\zeta = \omega^{2k+1}$. Then \begin{align*} \int_D f(z)\,dz &= -\frac{\pi i}{n} \sum_{j=0}^{n-1} \zeta^{2j+1} = -\frac{\pi i \zeta}{n} \sum_{j=0}^{n-1} (\zeta^2)^j \\&= - \frac{\pi i\zeta}{n} \left(\frac{1 - \zeta^{2n}}{1-\zeta^2}\right) = -\frac{2\pi i}{n} \cdot \frac{\zeta}{1-\zeta^2} \end{align*} (since $\zeta^{2n} = (\omega^{2k+1})^{2n} = (\omega^{2n})^{2k+1} = 1$).
Since $|\zeta| = 1$, $\zeta^{-1} = \bar\zeta$. So \begin{align*} \frac{\zeta}{1-\zeta^2} &= \frac{\zeta}{1-\zeta^2} \cdot \frac{\bar\zeta}{\bar\zeta} \\&= \frac{1}{\bar\zeta - \zeta} = \frac{i}{2}\cdot \frac{1}{\Im(\zeta)} \end{align*} Therefore, $$ \int_D f(z)\,dz = \frac{\pi}{n \Im(\zeta)} $$ The integral over $C_R$ is bounded by $\pi R^{2(k-n)+1}$, which (since $0 \leq k < n$) tends to zero as $R\to\infty$. Also, by Euler's formula, $$ \zeta = \exp\left(\frac{(2k+1)\pi i}{2n}\right) = \cos \left(\frac{(2k+1)\pi}{2n}\right) + i \sin \left(\frac{(2k+1)\pi}{2n}\right) $$ Therefore, $$ \int_{-\infty}^\infty \frac{x^{2k}\,dx}{1+x^{2n}} = \frac{\pi}{n} \csc \left(\frac{(2k+1)\pi}{2n}\right) $$