I am reading "An Introduction to Complex Analysis" by Tomoki Kawahira.
There is the following problem in this book.
Calculate the following improper integral based on residues in complex analysis:
$$\int_{-\infty}^{\infty} \frac{x^4}{(x^2 + a^2)^4} dx,$$
where $a$ is a positive real number.
For example, we need to calculate the following integral:
$$\int_{C} \frac{z^4}{(z^2 + a^2)^4} dz,$$
where $C = J_R + H_R$, $J_R : z = x (-R \leq x \leq R)$, $H_R : z = R e^{i \theta} (0 \leq \theta \leq \pi), R > a$.
Let $g(z) := \frac{z^4}{(z + a i)^4}$.
$$\int_{C} \frac{z^4}{(z^2 + a^2)^4} dz = \int_{C} \frac{g(z)}{(z - a i)^4} dz = 2 \pi i \operatorname{Res}(\frac{g(z)}{(z - a i)^4}, a i) \\ = 2 \pi i \frac{1}{3!} g^{(3)}(a i).$$
We need to calculate $g^{(3)}(z)$.
First, I represented $g(z)$ as follows.
$$g(z) = 1 - 4 a i \frac{1}{z+a i} - 6 a^2 \frac{1}{(z+a i)^2} + 4 a^3 i \frac{1}{(z+a i)^3} + a^4 \frac{1}{(z+a i)^4}.$$
Then I calculated $g^{(3)}(z)$ , but the calculation was hard for me.
Is there an easier way to calculate $g^{(3)}(z)$?
By definition the residue at $z_0$ is the $c_{-1}$ coefficient in the Laurent series of the function. Thus you first calculate the Laurent Series of each piece centered at $ai$
\begin{align}z^4&=(z-ai+ai)^4=\sum_{n=0}^4{4\choose n}(ai)^n(z-ai)^n\\ \frac{1}{(z-ai)^4}&=(z-ai)^{-4}\\ \frac{1}{(z+ai)^4}&=\left(\frac{1}{z-ai+2ai}\right)^4=\left(\frac{1}{\frac{z-ai}{2ai}+1}\right)^4=\left(\sum \frac{(z-ai)^n}{(-2ai)^n}\right)^4 \end{align}
Then multiplying the three together and collecting only the terms whose power is $-1$ gives you the coefficient. Doing the same for $z+ai$ gives the answer.
In this case it's not fast, but sometimes it's a nice technique