Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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


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?

share|cite|improve this question
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
up vote 4 down vote accepted

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 $$ \begin{align} \int_0^\infty\frac{\mathrm{d}t}{1+t^{2n}} &=-\frac{2\pi i}{4n}\sum_{k=0}^{n-1}e^{\frac{\pi i}{2n}(2k+1)}\\ &=-\frac{\pi i}{2n}e^{\frac{\pi i}{2n}}\frac{1-(-1)}{1-e^{\frac{\pi i}{n}}}\\ &=\frac{\pi}{2n}\csc\left(\frac{\pi}{2n}\right)\tag{5} \end{align} $$

share|cite|improve this answer
+1 .......... nice answer – Bhaskara-III Dec 13 '15 at 5:10

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.

share|cite|improve this answer

I realized after I wrote this up that this is given in one of the papers mentioned by Dave L. Renfro, but I did all this work and the approach is not exactly the same, so here goes.

We wish to evaluate $$ \int \frac{1}{1+x^n}\ dx. $$

We will do this by partial fraction decomposition. Note that the roots of $1+x^n$ are the $2n$-th roots of unity that are not $n$-th roots of unity. That is to say $x^{2n}-1=(x^n-1)(x^n+1)$. It follows that the set of roots of $1+x^n$ is $$ \left\{\exp\left(\frac{(2k-1)\pi i}{n} \right):0\leq k\leq n-1\right\}. $$ If we consider the roots (excluding -1 if $n$ is odd) we have that $$ \left(x-\exp\left(\frac{(2k+1)\pi i}{n} \right)\right)\left(x-\exp\left(\frac{(2(n-k)-1)\pi i}{n} \right)\right)=\left(x-\exp\left(\frac{(2k+1)\pi i}{n} \right)\right)\left(x-\exp\left(\frac{-(2k+1)\pi i}{n} \right)\right) $$ $$ =x^2-\left(\exp\left(\frac{(2k+1)\pi i}{n}\right)+\exp\left(\frac{-(2k+1)\pi i}{n} \right) \right)x+1=x^2-2\cos\left(\frac{(2k+1)\pi}{n}\right)x+1. $$ Let $x_k=\frac{(2k+1)\pi}{n}$ and $\alpha_k=\exp((2k+1)\pi i/n)$, then by partial fraction decomposition (for $n$ even) we have that $$ \frac{1}{1+x^n}=\sum_{k=0}^{n/2-1}\frac{a_kx+b_k}{x^2-2\cos(x_k)x+1}=\sum_{k=0}^{n/2-1}\frac{(a_kx+b_k)\prod_{\overset{j\neq k}{j\neq n-1-k}}(x-\alpha_j)}{1+x^n}=\sum_{k=0}^{n/2-1}\frac{\frac{a_kx+b_k}{x-\alpha_{n-1-k}}\prod_{j\neq k}(x-\alpha_j)}{1+x^n}. $$ Furthermore $$ 1=\sum_{k=0}^{n/2-1}\frac{a_kx+b_k}{x-\alpha_{k}^{-1}}\prod_{j\neq k}(x-\alpha_j). $$ If we set $x=\alpha_k$ for $0\leq k\leq n/2$ we obtain $$ \frac{a_k\alpha_k+b_k}{\alpha_k-\alpha_{k}^{-1}}\prod_{j\neq k}(\alpha_k-\alpha_j)=1. $$ Note that $$ \prod_{k=1}^{n-1}(x-\exp(k2\pi i/n))=(1+x+\cdots+x^{n-1}) $$ so $$ \prod_{k=1}^{n-1}(1-\exp(k2\pi i/n))=n. $$ Furthermore $$ \prod_{j\neq k}(\alpha_k-\alpha_j)=\prod_{j\neq k}\alpha_k(1-\frac{\alpha_j}{\alpha_k})=\alpha_k^{n-1}\prod_{k=1}^{n-1}(1-\exp(k2\pi i/n))=n\alpha_k^{n-1}=-n\alpha^{-1}. $$ So we are left with $$ \frac{(a_k\alpha_k+b_k)(-n\alpha_k^{-1})}{\alpha_k-\alpha_{k}^{-1}}=1. $$ and $$ -n(a_k+\alpha_k^{-1}b_k)=\alpha_k-\alpha_{k}^{-1}=2i\sin(x_k) $$ implying that $$ a_k+\cos(x_k)b_k-i\sin(x_k)b_k=-\frac{2i}{n}\sin(x_k). $$ Hence $b_k=\frac{2}{n}$ and $a_k=-\frac{2}{n}\cos(x_k)$. So for even $n$ we have. $$ \frac{1}{1+x^n}=-\frac{1}{n}\sum_{k=0}^{n/2-1}\frac{2\cos(x_k)x-2}{x^2-2\cos(x_k)x+1} $$ If $n$ is odd we have the additional term $$ \frac{a}{1+x} $$ and it follows that $a\prod_{\alpha_k\neq 1}(x-\alpha_k)=a(1-x+\cdots-x^{n-2}+x^{n-1})=1$. Setting $x=-1$ we obtain $a=\frac{1}{n}$.

Noticing that $$ \frac{2\cos(x_k)x-2}{x^2-2\cos(x_k)x+1}=\frac{\cos(x_k)(2x-2\cos(x_k))}{x^2-2\cos(x_k)x+1}+\frac{2\cos^2(x_k)-2}{(x-\cos(x_k))^2+1-\cos^2(x_k)} $$ $$ =\frac{\cos(x_k)(2x-2\cos(x_k))}{x^2-2\cos(x_k)x+1}-2\frac{\sin^2(x_k)}{(x-\cos(x_k))^2+\sin^{2}(x_k)} $$ $$ =\cos(x_k)\frac{(2x-2\cos(x_k))}{x^2-2\cos(x_k)x+1}-2\sin(x_k)\frac{\csc(x_k)}{(\frac{x-\cos(x_k)}{\sin(x_k)})^2+1}. $$ So we have for even $n$ $$ \int\frac{1}{1+x^n}\ dx=-\frac{1}{n}\sum_{k=0}^{n/2-1}\left\{\cos(x_k)\int\frac{(2x-2\cos(x_k))}{x^2-2\cos(x_k)x+1}\ dx-2\sin(x_k)\int\frac{\csc(x_k)}{(\frac{x-\cos(x_k)}{\sin(x_k)})^2+1}\ dx\right\} $$ $$ =-\frac{1}{n}\sum_{k=0}^{n/2-1}\left\{\cos(x_k)\log|x^2-2\cos(x_k)x+1|-2\sin(x_k)\arctan\left(\frac{x-\cos(x_k)}{\sin(x_k)}\right)\right\}, $$ and for odd $n$ $$ \int\frac{1}{1+x^n}\ dx=\frac{1}{n}\log|x+1|-\frac{1}{n}\sum_{k=0}^{(n-1)/2-1}\left\{\cos(x_k)\log|x^2-2\cos(x_k)x+1|-2\sin(x_k)\arctan\left(\frac{x-\cos(x_k)}{\sin(x_k)}\right)\right\} $$ where $x_k=(2k+1)\pi/n$, $n\in\mathbb{Z}_{>0}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.