Im trying to solve $$\int_{0}^{\infty}\dfrac{x^2}{x^6+1} dx$$

So I begin by considering the function $f(z) = \frac{z^2}{z^6+1}$, I consider the isolated singularities, which are indeed simple poles given by $z_k = e^{i\left(\frac{\pi}{6}+\frac{2\pi k}{6}\right)}$ for $k$ from 0 to 5. None of these poles lie on the real axis, and furthermore, I consider the first 3 poles that are in the upper half-plane, $z_0$, $z_1$, and $z_2$. Therefore, I can calculate the integral using the residue theorem by integrating over the semicircle centered at 0 with a radius of $R$. Thus, $$ \int_{0}^{\infty} \frac{x^2}{x^6+1} \, dx = \frac{1}{2}\int_{-R}^{R}f(x) \, dx +\int_{\gamma} f(z) \, dz = 2\pi i \left(\text{Res}(f,z_0)+\text{Res}(f,z_1)+\text{Res}(f,z_2)\right). $$ I have difficulties finding the residues as evaluating the limit becomes complicated and tedious. Is there a more elegant way to solve these residues?

  • 2
    $\begingroup$ If $Q$ is a polynomial with a zero of order $1$ at $r$, and $P$ is another polynomial, then the residue of $P/Q$ at $r$ is $P(r)/Q'(r)$. In your case that gives $\frac{z_i^{-3}}{6}=-\frac{z_i^3}{6}$ $\endgroup$
    – NDB
    Jul 12, 2023 at 19:23
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    $\begingroup$ With the substitution $x = \sqrt{u}$, you can write $\int_{0}^{\infty}\frac{x^{2}}{x^{6}+1}dx = \frac{1}{2}\int_{0}^{\infty}\frac{\sqrt{u}}{u^{3}+1}du$ and proceed with a keyhole contour as an alternative way to use residues. $\endgroup$ Jul 12, 2023 at 21:44

2 Answers 2



Another possible approach to compare with:

\begin{align*} \int_{0}^{\infty}\frac{x^{2}}{x^{6} + 1}\mathrm{d}x & = \frac{1}{3}\int_{0}^{\infty}\frac{3x^{2}}{x^{6} + 1}\mathrm{d}x\\\\ & = \frac{1}{3}\int_{0}^{\infty}\frac{\mathrm{d}(x^{3})}{(x^{3})^{2} + 1}\\\\ & = \frac{1}{3}\int_{0}^{\infty}\frac{\mathrm{d}u}{u^{2} + 1} \end{align*}


We are interested in computing $$I=\int_0^\infty \frac{x^2}{x^6+1} \, dx.$$

Perhaps a simpler approach is to take the contour $\Gamma$ to be the positive real axis from $0$ to $R$, a sixth-of-a-circle of radius $R$ and then back down to the origin from $Re^{i\pi/3}$ to $0.$

Then you need only compute one residue:

Consider $\displaystyle \oint_\Gamma f(z) \, dz$ where $\displaystyle f(z)=\frac{z^2}{z^6+1}.$

By Jordan's lemma, the integral goes to zero along the arc.

Now, along the third piece, $z=e^{i\pi/3}x,$ so that $z^2\,dz = -x^2\,dx$ and $z^6=x^6$:

$$2 I = 2\pi i \,\text{Res} \{ f(z); z=e^{i \pi/6 }\} $$

$$\text{Res} \{ f(z); z=e^{i \pi/6 }\} = \left. \frac{z^2}{6z^5} \right|_{z=e^{i\pi/6}} =\frac{1}{6i}.$$

Putting it all together:


If you insist on calculating three residues in the upper half plane:

$$\displaystyle z_k \in \{ e^{i\pi/6}, e^{i\pi/2}, e^{5i\pi/6} \}, \quad {k=1,2,3}$$

$$\sum_{k=1}^3 \text{Res}\{f(z);z_k\} = \frac{1}{6} \left[ e^{-i\pi/2}+e^{i\pi/2}+e^{-i\pi/2} \right]=\frac{1}{6i}$$

$$2I=\int_{-\infty}^\infty \frac{x^2 \, dx}{x^6+1} = \frac{2\pi i }{6i}$$


$$I = \frac{\pi}{6}.$$

UPDATE in response to a question that appeared for a short while in the comments (Thank you):

To compute the residue at a simple pole for a rational function, we take the numerator divided by the derivative of the denominator, and evaluate this at the pole.

$$\displaystyle \text{Res} \{ f(z); z=e^{i \phi_0 }\} = \left. \frac{z^2}{\frac{d}{dz} (z^6+1)} \right|_{z=e^{i\phi_0}} = \left. \frac{z^2}{6z^5} \right|_{z=e^{i\phi_0}} =\frac{1}{6}e^{-3i\phi_0}.$$

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    $\begingroup$ (+1) for the efficient way of proceeding $\endgroup$
    – Mark Viola
    Jul 12, 2023 at 22:05
  • 2
    $\begingroup$ It was my pleasure $\endgroup$
    – Mark Viola
    Jul 12, 2023 at 22:06
  • $\begingroup$ @MarkViola, Thank you very much! [accidentally deleted with some other comments, reposting] $\endgroup$
    – mjw
    Jul 13, 2023 at 20:25

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