I am trying to show that
$$\int_{-\infty}^\infty \frac{x^2}{\left(x^2 + a^2\right)^2} dx = \frac\pi{2a}$$
for $a > 0$ using the Residue Theorem.
The formula I am using says
$$\int_{-\infty}^\infty f(x)dx = 2\pi i\sum_{\nu=1}^k \operatorname{Res}\left(f ; a_\nu\right).$$
We have
$$\int_{-\infty}^\infty \frac{x^2}{\left(x^2 + a^2\right)^2} dx = \int_{-\infty}^\infty \frac{x^2}{(x + ai)^2(x – ai)^2} dx$$
which has the pole $x = ai$ in the upper half plane of order $2$. Now, we find
$$\frac{d^2}{dx^2}\frac{x^2}{(x + ai)^2} = \frac{6x^2}{(x + ia)^4} – \frac{8x}{(x + ia)^3} + \frac2{(x + ia)^2}$$
so
$$\operatorname{Res}(f;1)= \frac1{8a^2}.$$
Thus, we have that
$$\int_{-\infty}^\infty \frac{x^2}{\left(x^2 + a^2\right)^2} dx = 2\pi i\operatorname{Res}(f; ai) = 2\pi i \frac1{8a^2} = \frac{\pi i}{4a^2} \ne \frac\pi{2a}.$$
Where am I going wrong?
Best Answer
To compute the residue of a double pole, we need to compute the first derivative of the relevant factor of the integrand, not the second derivative.
Here's another method that builds on your approach: Applying the method of partial fractions to your complex factorization gives $$\frac{x^2}{(x^2 + a^2)^2} = \frac{1}{4} \frac{1}{(x - ai)^2} - \frac{i}{4a} \frac{1}{x - ai} + g(x)$$ for some function $g$ analytic in a neighborhood of $a i$. In particular, $$\operatorname{Res}(f; a i) = -\frac{i}{4 a} ,$$ and continuing with your computation as in the question statement gives the desired answer: $$\int_{-\infty}^\infty \frac{x^2 \,dx}{(x^2 + a^2)^2} = 2 \pi i \operatorname{Res}(f; a i) = 2 \pi i \left(-\frac{i}{4 a}\right) = \boxed{\frac{\pi}{2 a}} .$$
Alternatively, the real partial fractions decomposition of $f$ the integrand is $$\frac{x^2}{(x^2 + a^2)^2} = -\frac{a^2}{(x^2 + a^2)^2} + \frac{1}{x^2 + a^2} ,$$ so we can rewrite the original integral as $$-a^2 \int_{-\infty}^\infty \frac{dx}{(x^2 + a^2)^2} + \int_{-\infty}^\infty \frac{dx}{x^2 + a^2} .$$ Denote the second integral by $I(a)$; it is elementary, but we can also evaluate it using the Residue Theorem: $$I(a) = 2 \pi i \operatorname{Res}\left(\frac{1}{x^2 + a^2}; ai\right) = 2 \pi i \left(-\frac{i}{2 a}\right) = \frac{\pi}a .$$ Differentiating under the integral sign gives that $$-\frac{\pi}{a^2} = I'(a) = -2 a \int_{-\infty}^\infty \frac{dx}{(x^2 + a^2)^2},$$ so $$\int_{-\infty}^\infty \frac{dx}{(x^2 + a^2)^2} = \frac{\pi}{2 a^3},$$ hence the original integral is $$-a^2 \left(\frac{\pi}{2 a^3}\right) + \left(\frac{\pi}{a}\right) = \frac{\pi}{2 a} .$$