Show that $\large\int_{0}^{\infty} \frac{x^{m-1}}{x^n+1} dx= \frac{\frac{\pi}{n}}{\sin(\frac{m}{n}\pi)}$; $n>m$
Given $z \in \mathbb{C}; z^n\neq-1$, I can define $f(z)= \frac{z^{m-1}}{z^n+1}$ and considering the simple loop $\gamma_R$:$[0,R] \lor \alpha_{R} \lor [e^{\frac{2\pi i}{n}},0]; \alpha_{R}(t)=Re^{it}; t \in [0, \frac{2\pi }{n}]; R>1$;
$$\int_{\gamma_R}f(z)dz=\int_{0}^Rf(x)dx+ \int_{\alpha_R}f(z)dz+\int_{[e^{\frac{2\pi i}{n}},0]}f(z)dz=2\pi i (\mathit{res}(e^{\frac{i\pi}{n}},f))$$
And this lead to:
$$\large \int_{0}^{\infty}f(x)dx=2\pi i\Big(\frac{e^{i\frac{\pi}{n}(m-1)}}{ne^{i\frac{\pi}{n}(n-1)}}\Big)$$ $(\ast)$ $$\begin{align*} &\quad\frac{2 \pi i}{n} \frac{1}{e^{i\frac{n-m}{n}}} \\[9 pt]&=\quad\frac{2 \pi i}{n} \frac{1}{\cos(\pi\frac{n-m}{n})+i\sin(\pi\frac{n-m}{n})} \\ &= \frac{2 \pi i}{n} \frac{1}{\cos(\pi)\cos(-\pi\frac{m}{n})-\sin(\pi)\sin(-\pi\frac{m}{n})+i(\sin(\pi)\cos(-\pi\frac{m}{n})+\cos(\pi)\sin(-\pi\frac{m}{n}))} \\ &=\frac{2 \pi i}{n} \frac{1}{-\cos(\pi\frac{m}{n})+i\sin(\pi\frac{m}{n})} \end{align*}$$
But I can't relate this with the given result; any suggestions?
EDIT:I forgot to account for $\int_{[e^{\frac{2\pi i}{n}},0]}f(z)dz$:
$\large \int_{[e^{\frac{2\pi i}{n}},0]}f(z)dz=-e^{\frac{i 2 \pi m}{n}}\int_{0}^{\infty}\frac{x^{m-1}}{x^n+1}dx$
And then, $$(1-e^{\frac{2\pi i m}{n}})\int_{0}^{\infty}\frac{x^{m-1}}{x^n+1}dx=\frac{2 \pi i }{n}e^{i\frac{\pi(m-n)}{n}} \iff \int_{0}^{\infty}\frac{x^{m-1}}{x^n+1}dx=\frac{\frac{\pi}{n}}{\sin(\frac{m}{n}\pi)}$$
