Calculating a real integral using complex integration

$$\int^\infty_0 \frac{dx}{x^6 + 1}$$

Does someone know how to calculate this integral using complex integrals? I don't know how to deal with the $x^6$ in the denominator.

Answer

Thankfully the integrand is even, so we have

$$\int^\infty_0 \frac{dx}{x^6 + 1} = \frac{1}{2}\int^\infty_{-\infty} \frac{dx}{x^6 + 1}. \tag{1}$$

To find this, we will calculate the integral

$$\int_{\Gamma_R} \frac{dz}{z^6+1},$$

where $\Gamma_R$ is the semicircle of radius $R$ in the upper half-plane, $C_R$, together with the line segment between $z=-R$ and $z=R$ on the real axis.

(Image courtesy of Paul Scott.)

Then

$$\int_{\Gamma_R} \frac{dz}{z^6+1} = \int_{-R}^{R} \frac{dx}{x^6+1} + \int_{C_R} \frac{dz}{z^6+1}.$$

We need to show that the integral over $C_R$ vanishes as $R \to \infty$. Indeed, the triangle inequality gives

$$\begin{align} \left| \int_{C_R} \frac{dz}{z^6+1} \right| &\leq L(C_R) \cdot \max_{C_R} \left| \frac{1}{z^6+1} \right| \\ &\leq \frac{\pi R}{R^6 - 1}, \end{align}$$

where $L(C_R)$ is the length of $C_R$. From this we may conclude that

$$\lim_{R \to \infty} \int_{\Gamma_R} \frac{dz}{z^6+1} = \int_{-\infty}^{\infty} \frac{dx}{x^6+1}. \tag{2}$$

The integral on the left is evaluated by the residue theorem. For $R > 1$ we have

$$\int_{\Gamma_R} \frac{dz}{z^6+1} = 2\pi i \sum_{k=0}^{2} \operatorname{Res}\left(\frac{1}{z^6+1},\zeta^k \omega\right),$$

where $\zeta$ is the primitive sixth root of unity and $\omega = e^{i\pi/6}$. Note that this is because $\omega$, $\zeta\omega$, and $\zeta^2 \omega$ are the only poles of the integrand inside $\Gamma_R$. The sum of the residues can be calculated directly, and we find that

$$\int_{\Gamma_R} \frac{dz}{z^6+1} = 2\pi i \sum_{k=0}^{2} \operatorname{Res}\left(\frac{1}{z^6+1},\zeta^k \omega\right) = \frac{\pi}{3 \sin(\pi/6)} = \frac{2\pi}{3}.$$

Thus, from $(1)$ and $(2)$ we conclude that

$$\int_{0}^{\infty} \frac{dx}{x^6+1} = \frac{\pi}{3}.$$

In general,

$$\int_{0}^{\infty} \frac{dx}{x^{2n}+1} = \frac{\pi}{2 n \sin\left(\frac{\pi}{2n}\right)}$$

for $n \geq 1$.