Find $\int_0^\infty \frac{dx}{x^{n-1}+x^{n-2}+\cdots +x+1}$ using contour integral I have solved the problem of integration
$$\int_0^\infty \frac{1}{x^n+1}dz$$
using the contour as an arc of a circle. But I don't know how to  approach this problem:
$$I=\int_0^\infty\frac{1}{x^{n-1}+x^{n-2}+\cdots +x+1}dz$$
Please Help me with this problem
Edit:
With the help of geometric series (suggested in comment), I can reduce the integral to the evaluation of
$$\int_0^\infty\frac{1-x}{1-x^n}dx$$
Now it's $1-x$ that's causing the problem if I choose the contour to be arc with angle $2\pi /n$. I still need little help.
 A: The problem can be solved via contour integration (making substitution $z=x^n$, using the keyhole contour with the cut $[0;\infty)$ and adding additional arches around $z=1$).
Bit the problem can also be solved via direct integration.
$$I(n)=\int_0^1\frac{1-x}{1-x^n}dx+\int_1^\infty\frac{1-x}{1-x^n}dx$$
Making change in the second integral ($x=\frac{1}{t}$)
$$I=\int_0^1\frac{1-x}{1-x^n}dx+\int_0^1\frac{x^{n-3}-x^{n-2}}{1-x^n}dx$$
Making another substitution ($t=x^n$)
$$I=\frac{1}{n}\int_0^1 t^{1/n-1}(1-t^{1/n})\frac{dt}{1-t}+\frac{1}{n}\int_0^1 (t^{-2/n}-t^{-1/n})\frac{dt}{1-t}$$
To evaluate these integrals we make regularisation ($\epsilon\to 0$) and consider $$I=\lim_{\epsilon \to0} \,I_{1\epsilon}+I_{2\epsilon}$$
where
$$I_{1\epsilon}=\frac{1}{n}\int_0^1 t^{1/n-1}(1-t^{1/n})\frac{dt}{(1-t)^{1-\epsilon}}=\frac{1}{n}\Big(B(1/n;\epsilon)-B(2/n;\epsilon)\Big)$$
$$=\frac{\Gamma(\epsilon)}{n}\Bigg(\frac{\Gamma(1/n)}{\Gamma(1/n+\epsilon)}-\frac{\Gamma(2/n)}{\Gamma(2/n+\epsilon)}\Bigg)$$
Given that at $\epsilon\to 0\, \Gamma(\epsilon)\to \frac{1}{\epsilon}$ and $\Gamma(a+\epsilon)=\Gamma(a)+\Gamma'(a)\epsilon +O(\epsilon^2)$
$$I_{1\epsilon}=\frac{1}{n}\frac{1}{\epsilon}\Bigg(\frac{\Gamma(1/n)}{\Gamma(1/n)+\Gamma'(1/n)\epsilon +...}-\frac{\Gamma(2/n)}{\Gamma(2/n)+\Gamma'(2/n)\epsilon +...}\Bigg)$$
$$=\frac{1}{n}\Bigg(\frac{\Gamma'(2/n)}{\Gamma(2/n)}-\frac{\Gamma'(1/n)}{\Gamma(1/n)}\Bigg)+O(\epsilon)=\frac{1}{n}\Big(\psi(2/n)-\psi(1/n)\Big)+O(\epsilon)$$
where $\psi(a)=\frac{\Gamma'(a)}{\Gamma(a)}$ - digamma function https://en.wikipedia.org/wiki/Digamma_function .
In the same way we evaluate the second integral:
$$I_{2\epsilon}=\frac{1}{n}\Big(\psi(1-1/n)-\psi(1-2/n)\Big)+O(\epsilon)$$
Taking the limit $\epsilon\to 0$
$$I(n)=\frac{1}{n}\Big(\psi(1-1/n)-\psi(1/n)+\psi(2/n)-\psi(1-2/n)\Big)$$
Using the property of digamma function $\psi(1-x)-\psi(x)=\pi \cot\pi x$
$$I(n)=\frac{\pi}{n}\Big(\cot\frac{\pi}{n}-\cot\frac{2\pi}{n}\Big)=\frac{\pi}{n}\frac{1}{\sin\frac{2\pi}{n}}$$
