Let \(a>0\) and calculate \[ \int_0^\infty {x^2 \over (x^2 + a^2)^2} \, dx .\]
Solution
1. The integrand is even, so \(\int_0^\infty \frac{x^2}{(x^2 + a^2)^2}\,dx = \frac{1}{2}\int_{-\infty}^{\infty} \frac{x^2}{(x^2 + a^2)^2}\,dx\).
evenness.
2. Consider \(f(z) = \frac{z^2}{(z^2 + a^2)^2} = \frac{z^2}{(z - ia)^2 (z + ia)^2}\).
definition.
3. \(f\) has a double pole at \(z = ia\) (in the upper half-plane) and a double pole at \(z = -ia\) (in the lower half-plane).
factor the denominator.
4. Integrate over a semicircular contour in the upper half-plane; the integral over the arc tends to \(0\) as the radius \(\to \infty\).
\(|f(z)| \le C/R^2\) on the arc of radius \(R\), and the arc length is \(\pi R\), so the arc integral is \(O(1/R) \to 0\).
5. \(\operatorname{Res}(f, ia) = \frac{1}{4ia}\).
computing the residue at the double pole \(z = ia\): \(\operatorname{Res} = \frac{d}{dz}\left[\frac{z^2}{(z + ia)^2}\right]_{z = ia} = \frac{1}{4ia}\).
6. By the residue theorem, \(\int_{-\infty}^{\infty} f(x)\,dx = 2\pi i \cdot \frac{1}{4ia} = \frac{\pi}{2a}\).
4 and 5.
7. Hence \(\int_0^\infty \frac{x^2}{(x^2 + a^2)^2}\,dx = \frac{1}{2} \cdot \frac{\pi}{2a} = \frac{\pi}{4a}\).
1 and 6.
8. Q.E.D.
7.