# [SOLVED]Integral challenge

#### anemone

##### MHB POTW Director
Staff member
Evaluate $\displaystyle\int\limits_0^{\infty} \dfrac{x^2+2}{x^6+1} \, dx$.

#### MountEvariste

##### Well-known member
We have $$I= \int_0^\infty \frac{x^2+2}{x^6+1}\,\mathrm{dx}$$

Let $x \mapsto \frac{1}{x}$ then

$$I= \int_0^\infty \frac{x^2+2x^4}{x^6+1}\,\mathrm{dx}$$

So that $$2I = 2\int_0^\infty \frac{1+x^2+x^4}{1+x^6}\,\mathrm{dx}$$

i.e. $$I = \int_0^\infty \frac{1+x^2+x^4}{1+x^6}\,\mathrm{dx}$$

By partial fractions \begin{aligned} \frac{1+x^2+x^4}{1+x^6} &= \frac{1}{3}\cdot \frac{1}{1+x^2}+\frac{2}{3}\cdot \frac{x^2+1}{x^4-x^2+1} \\& = \frac{1}{3}\cdot \frac{1}{1+x^2} +\frac{2}{3}\cdot \frac{1+1/x^2}{x^2+\frac{1}{x^2}+1} \\& = \frac{1}{3}\cdot \frac{1}{1+x^2} +\frac{2}{3}\cdot \frac{1+1/x^2}{(x-1/x)^2+1}\end{aligned}

This along with the sub $u = x-\frac{1}{x}$ helps us finish off the integral,

\begin{aligned} I & = \frac{1}{3}\cdot \int_0^\infty \frac{1}{1+x^2}\, \mathrm{dx} +\frac{2}{3}\cdot \int_0^\infty\frac{1+1/x^2}{(x-1/x)^2+1}\,\mathrm{dx} \\& = \frac{1}{3}\cdot \frac{\pi}{2}+\frac{2}{3} \cdot \int_{-\infty}^{\infty} \frac{1}{u^2+1}\,\mathrm{du} \\& = \frac{\pi}{6}+\frac{2\pi}{3} \\& = \frac{5 \pi}{6}.\end{aligned}

Therefore $$\int_0^\infty \frac{x^2+2}{x^6+1}\,\mathrm{dx}= \frac{5\pi}{6}.$$

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• • anemone and topsquark

#### MountEvariste

##### Well-known member
Consider the function $\displaystyle f(z) = \frac{z^2+2}{z^6+1}$; $f$ has simple poles at $z = \pm i, \pm i^{\frac{1}{3}}, \pm i^{\frac{5}{3}}$, three of which lie in the upper half plane;
those are $z_1 = i, ~ z_2 = i^{\frac{1}{3}}, ~ z_3 = i^{\frac{5}{3}}$. The sum of the residues at these poles is given by

\displaystyle \begin{aligned} \sum_{1 \le i \le 3} \text{res} f(z_i) & = \frac{1}{6} (-2 i - \sqrt{3})+\frac{1}{6}(-2 i +\sqrt{3})-\frac{i}{6} \\& = -\frac{5i}{6}\end{aligned}​

Now, let $\Gamma$ be the upper half-plane semi-circle $|z| = a$ with radius $a$ in counterclockwise direction.

Since $z_1, z_2, z_3$ all lie in the upper half plane, by the residue theorem we have

\displaystyle \begin{aligned} \oint_\Gamma f(z)\,dz & =2i \pi \sum_{1 \le i \le 3} \text{res} f(z_i) \\& = 2i \pi \cdot \frac{-5i}{6} \\& = \frac{5\pi}{3} \end{aligned}​

But also
$\displaystyle \oint_\Gamma f(z)\,dz=\int_{-a}^{a} f(z) \,\mathrm{dz} +\int_{\text{Arc}} f(z)\,\mathrm{dz}$​

Therefore

$\displaystyle \int_{-a}^{a} f(z) \,\mathrm{dz} +\int_{\text{Arc}} f(z)\,\mathrm{dz} = \frac{5\pi}{3}$​

By taking $a \to \infty$, since

$\displaystyle \bigg| \int_{\text{Arc}} f(z) \,\mathrm{dz} \bigg| \le \frac{a^2+2}{a^6-1}~ a \pi \longrightarrow 0$​

we have $\displaystyle \int_{\text{Arc}} f(z)\,\mathrm{dz} \to 0$ as $a \to \infty$, therefore

$\displaystyle \int_{-\infty}^{\infty}\frac{z^2+2}{z^6+1}\,\mathrm{dz} = \frac{5\pi}{3}$​

Therefore (since $f$ is an even function),

$\displaystyle \int_{0}^{\infty}\frac{z^2+2}{z^6+1}\,\mathrm{dz} = \frac{5\pi}{6}.$​

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• anemone