Integration in complex analysis using Residue Theorem

Question

Prove that for $a>0$, $$ \int_{-\infty}^{\infty} \frac{1}{x^4+a^4}dx=\frac{\pi}{a^3\sqrt{2}} $$ I think I'm supposed to use Cauchy's Residue Theorem somehow, but I don't know what closed path to use or even where to begin really! Any help would be appreciated, thanks.

Answer 1

Define $C$ by a quarter-circle of radius $R$ in the first quadrant. Then consider

$$\oint_C \frac{dz}{z^4+a^4} = \int_0^R \frac{dx}{x^4+a^4} + i R \int_0^{\pi/2} d\theta \, e^{i \theta} \frac{1}{R^4 e^{i 4 \theta}+a^4}+i\int_R^0 \frac{dy}{y^4+a^4}$$

Note that in the third integral, I used the fact that $i^4=1$. Now consider the limit as $R \to \infty$; the second integral vanishes in this limit. This is because that integral has a magnitude bounded by

$$\frac{\pi}{2} \frac{R}{R^4-a^4}$$

which clearly vanishes as $R\to\infty$. Thus, in this limit, the contour integral is

$$(1-i) \int_0^{\infty} \frac{dx}{x^4+a^4}$$

The contour integral is also, by the residue theorem, equal to $i 2 \pi$ times the residue at the pole $z=a e^{i \pi/4}$. Thus,

$$\sqrt{2} e^{-i \pi/4} \int_0^{\infty} \frac{dx}{x^4+a^4} = i 2 \pi \frac{1}{4 a^3 e^{i 3 \pi/4}}$$

or

$$\int_0^{\infty} \frac{dx}{x^4+a^4} = \frac{\pi}{2 \sqrt{2} a^3}$$

The integral over the entire real line is twice this, as expected.