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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.

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    $\begingroup$ $1+z+z^2+\cdots+z^{n-1}$ is a geometric series, maybe there is a way to rewrite this... $\endgroup$
    – PC1
    Commented Oct 31, 2021 at 3:18
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    $\begingroup$ It is not $1-x$ but $1-x^n$ that causes the problem. $\endgroup$
    – Gary
    Commented Oct 31, 2021 at 11:14
  • $\begingroup$ Several solutions, including the complex integration one - math.stackexchange.com/questions/4138297/… $\endgroup$
    – Svyatoslav
    Commented Oct 31, 2021 at 11:15

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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 $\psi(1-x)-\psi(x)=\pi \cot\pi x$ $$I(n)=\frac{\pi}{n}\Big(\cot\frac{\pi}{n}-\cot\frac{2\pi}{n}\Big)=\frac{\pi}{n}\frac{1}{\sin\frac{2\pi}{n}}$$

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