# How to evaluate $\int 1/(1+x^{2n})\,dx$ for an arbitrary positive integer $n$?

How to find $$\int\dfrac{dx}{1+x^{2n}}$$ where $n \in \mathbb N$?

### Remark

When $n=1$, the antiderivative is $\tan^{-1}x+C$. But already with $n=2$ this is something much more complicated. Is there a general method?

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This feels very hypergeometric to me. Is that what you're looking for? Or perhaps you have bounds? –  mixedmath Jul 15 '12 at 7:59
If the integral ranges from $-\infty$ to $\infty$ there's a nice trick with the Residue theorem. –  Cocopuffs Jul 15 '12 at 8:20
@GerryMyerson : It's not always a good thing. –  Michael Hardy Jul 17 '12 at 0:30
@Michael, OP is zero-for-ten (and unfortunately unable to do anything about it, having been suspended for the next few weeks), I'm zero-for-one, and I try to make up for it in other ways. –  Gerry Myerson Jul 17 '12 at 5:17

If the integral is taken from $0$ to $\infty$, there is more than one way to evaluate this. One is \begin{align} \int_0^\infty\frac{\mathrm{d}t}{1+t^{2n}} &=\int_0^1\frac{\mathrm{d}t}{1+t^{2n}}+\int_0^1\frac{t^{2n-2}\,\mathrm{d}t}{1+t^{2n}}\\ &=\int_0^1(1-t^{2n}+t^{4n}-t^{6n}+\dots)\,\mathrm{d}t\\ &+\int_0^1(t^{2n-2}-t^{4n-2}+t^{6n-2}+\dots)\,\mathrm{d}t\\ &=1-\frac{1}{2n+1}+\frac{1}{4n+1}-\frac{1}{6n+1}+\dots\\ &+\frac{1}{2n-1}-\frac{1}{4n-1}+\frac{1}{6n-1}-\dots\\ &=\frac{1}{2n}\left(\frac{1}{0+\frac{1}{2n}}-\frac{1}{1+\frac{1}{2n}}+\frac{1}{2+\frac{1}{2n}}-\frac{1}{3+\frac{1}{2n}}+\dots\right)\\ &+\frac{1}{2n}\left(-\frac{1}{-1+\frac{1}{2n}}+\frac{1}{-2+\frac{1}{2n}}-\frac{1}{-3+\frac{1}{2n}}-\dots\right)\\ &=\frac{1}{2n}\sum_{k=-\infty}^\infty\frac{(-1)^k}{k+\frac{1}{2n}}\\ &=\frac{\pi}{2n}\csc\left(\frac{\pi}{2n}\right)\tag{1} \end{align} The last step uses the result from "An Infinite Alternating Harmonic Series" on this page.

Another method is to use contour integration to evaluate $$\frac12\int_{-\infty}^\infty\frac{\mathrm{d}t}{1+t^{2n}} =\frac12\oint_\gamma\frac{\mathrm{d}z}{1+z^{2n}}\tag{2}$$ where $\gamma$ is the path from $-\infty$ to $\infty$ along the real axis (which picks up the integral in question), then circling back counter-clockwise around the upper half-plane (which vanishes). The countour integral in $(2)$ is $2\pi i$ times the sum of the residues of $\frac{1}{1+z^{2n}}$ in the upper half-plane.

The poles of the integrand in $(2)$ are given by $$\zeta_k=e^{\frac{\pi i}{2n}(2k+1)}\tag{3}$$ where $k=0\dots n-1$ represent the roots in the upper half-plane. All the poles are simple, so the residues are \begin{align} \mathrm{Res}_{z=\zeta_k}\left(\frac{1}{1+z^{2n}}\right) &=\lim_{z\to\zeta_k}\frac{z-\zeta_k}{1+z^{2n}}\\ &=-\frac{1}{2n}\zeta_{k}\\ &=-\frac{1}{2n}e^{\frac{\pi i}{2n}(2k+1)}\tag{4} \end{align} Thus, we get \begin{align} \int_0^\infty\frac{\mathrm{d}t}{1+t^{2n}} &=-\frac{2\pi i}{4n}\sum_{k=0}^{n-1}e^{\frac{\pi i}{2n}(2k+1)}\\ &=-\frac{\pi i}{2n}e^{\frac{\pi i}{2n}}\frac{1-(-1)}{1-e^{\frac{\pi i}{n}}}\\ &=\frac{\pi}{2n}\csc\left(\frac{\pi}{2n}\right)\tag{5} \end{align}

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The following papers will be useful. Note that Gopalan/Ravichandran is freely available on the internet.

M. A. Gopalan and V. Ravichandran, Note on the evaluation of $\int \frac{1}{\;1\;+\;t^{2^{n}}\;}dt$, Mathematics Magazine 67 #1 (February 1994), 53-54.

Judith A. Palagallo and Thomas E. Price, Some remarks on the evaluation of $\int \frac{dt}{\;t^{m}\;+\;1\;}$, Mathematics Magazine 70 #1 (February 1997), 59-63.

V. Ravichandran, On a series considered by Srinivasa Ramanujan, Mathematical Gazette 88 #511 (March 2004), 105-110.

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