I spent a few hours trying to solve this indefinite integral:

$$\int \frac { dx}{x^{2m}+1}, m \in \mathbb R $$

I tried to transform the fraction to partial fractions getting $\int \frac{\imath}{2(x^m+\imath)} {dx} - \int \frac{\imath}{2(x^m-\imath)} {dx}$, but this doesn't help me because we enter in complex analisys, a field that I've no knowledgement and searching for methods that can help me with this make me nuts. I think the solution should involve some sort of induction, so I tried to solve for $ m= 0, 1, 2...$, but this doesn't help me neither. I've no luck looking for some books in the university library that can help me.

Putting the integral in wolframalpha gives me the result:$$ x \space_2F_1(1,\frac{1}{2m};1 + \frac{1}{2m}; -x^{2m}) \color{silver} {+constant}$$where $_2F_1(1,\frac{1}{2m};1 + \frac{1}{2m}; -x^{2m})$ is the hypergeometric function (I'd never hear about this before) .

This reveals to me that this problem is beyond my capabilities. Can some kind soul make a detailed resolution for me? If possible, not going into deep complex analysis.


The Wolfram Alpha expression of the indefinite integral in terms of the Gauss hypergeometric function is more straightforward than it might seem. We formally expand the integrand in powers of $x$ and integrate term-by-term to obtain

$$\int \frac{dx}{1+x^{2m}} = \int\left[1-x^{2m}+x^{4m}-\cdots\right]dx = x-\frac{x^{2m+1}}{2m+1} +\frac{x^{4m+1}}{m+1}-\cdots.$$ This expression is of the form $x \,f(-x^{2m})$ with $$f(u)=1+\frac{u}{2m+1}+\frac{u^2}{4m+1}+\cdots =\sum_{k=0}^\infty \frac{u^k}{2km+1}.$$ To compare with Wolfram Alpha, we recall that the Gauss hypergeometric function is defined as $\displaystyle _2F_1(a,b;c;u)=\sum_{k=0}^\infty \frac{(a)_k(b)_k}{(c)_k }\frac{u^k}{k!}$ where $(a)_k = a(a+1)\cdots (a+k-1)$. The particular case of $(a,b,c)=(1,\frac{1}{2m},1+\frac{1}{2m})$ gives

$$\frac{(1)_k}{k!}\frac{(\frac{1}{2m})_k}{(1+\frac{1}{2m})_k }=\frac{k!}{k!}\frac{\frac{1}{2m}(\frac{1}{2m}+1)\cdots (\frac{1}{2m}+k-1)}{(\frac{1}{2m}+1)(\frac{1}{2m}+2)\cdots (\frac{1}{2m}+k)}=\frac{\frac{1}{2m}}{\frac{1}{2m}+k}=\frac{1}{1+2mk}$$ and we conclude that $f(u)={_2F}_1(1,\frac{1}{2m},1+\frac{1}{2m};u)$ in agreement with Wolfram Alpha.

Note: For the formal expansion in powers of $x$ to be justified, we need $|x^m|<1$. Hence the above reasoning will be valid near $x=0$ under the assumption $m>0$.

  • $\begingroup$ I upvoted you because you actually answered the question, so clearly your answer should lie above mine. Can't you use the connection formulas for the hypergeometric functions to produce expressions valid outside the interval of convergence of the series? $\endgroup$ May 14 '16 at 1:49

It only takes complex arithmetic to expand in partial fractions, not complex analysis. There is a huge difference. Let $\omega_k=e^{i\theta_k}=\cos\theta_k+i\sin\theta_k$, where $\theta_k=\frac{(2k+1)\pi}{2m}$. Then $\omega_k^{2m}=-1$ and $$\frac1{x^{2m}+1}=\frac1{\prod_{k=0}^{2m-1}}(x-\omega_k)=\sum_{k=0}^{2m-1}\frac{A_k}{x-\omega_k}$$ By L'Hopital's rule, $$\begin{align}\lim_{x\rightarrow\omega_k}\frac{(x-\omega_k)}{x^{2m}+1}&=\lim_{x\rightarrow\omega_k}\frac1{2mx^{2m-1}}=\frac1{2m\omega_k^{2m-1}}=-\frac{\omega_k}{2m}\\ &=\sum_{\ell=0}^{2m-1}A_{\ell}\lim_{x\rightarrow\omega_k}\frac{x-\omega_k}{x-\omega_{\ell}}=\sum_{\ell=0}^{2m-1}A_{\ell}\delta_{k\ell}=A_k\end{align}$$ And now since $\theta_{2m-1-k}=2\pi-\theta_k$ it follows that $$\begin{align}\frac1{x^{2m}+1}&=\sum_{k=0}^{2m-1}-\frac{\omega_k}{2m(x-\omega_k)}=-\frac1{2m}\sum_{k=0}^{m-1}\left(\frac{\omega_k}{x-\omega_k}+\frac{\omega_k^*}{x-\omega_k^*}\right)\\ &=-\frac1{2m}\sum_{k=0}^{m-1}\frac{x\omega_k-1+x\omega_k^*-1}{x^2-(\omega_k+\omega_k^*)x+1}=-\frac1{2m}\sum_{k=0}^{m-1}\frac{2x\cos\theta_k-2}{x^2-2x\cos\theta_k+1}\\ &=\frac1{2m}\sum_{k=0}^{m-1}\frac{-2(x-\cos\theta_k)\cos\theta_k+2\sin^2\theta_k}{(x-\cos\theta_k)^2+\sin^2\theta_k}\end{align}$$ So now we can integrate $$\begin{align}\int\frac{dx}{x^2m+1}&=\frac1{2m}\sum_{k=0}^{m-1}\int\frac{-2(x-\cos\theta_k)\cos\theta_k+2\sin^2\theta_k}{(x-\cos\theta_k)^2+\sin^2\theta_k}dx\\ &=\frac1{2m}\sum_{k=0}^{m-1}\left[-\cos\theta_k\ln\left|x^2-2x\cos\theta_k+1\right|+2\sin\theta_k\tan^{-1}\left(\frac{x-\cos\theta_k}{\sin\theta_k}\right)\right]+C\end{align}$$ Now, you can simplify the above a little bit because $\cos\theta_{m-1-k}=-\cos\theta_k$ and $\sin\theta_{m-1-k}=\sin\theta_k$ but you have to be a little careful because there is a special term when $\theta_k=\pi/2$ when $m$ is odd and combining the arctangents gets a little tricky because their individual ranges span a width of $\pi$ so combining them in pairs requires a range of $2\pi$ so you would need the $\text{atan2}$ function or a restricted domain for $x$.

  • $\begingroup$ The partial fraction decomposition in the above only works if $m$ is a positive integer. But the OP has $m\in\mathbb{R}$. (That said, the case of $m$ a positive integer is far more common in practice.) $\endgroup$ May 13 '16 at 18:37
  • $\begingroup$ Sorry, should have read more carefully. $\endgroup$ May 13 '16 at 20:04
  • $\begingroup$ @Semiclassical Also, note that the OP refers to trying induction on $m=0,1,2,\ldots$ so it is highly likely they will find value in this answer. $\endgroup$
    – Erick Wong
    May 14 '16 at 3:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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