How to compute contour integrals using the residue theorem?

The **Residue Theorem** is a powerful tool in complex analysis used to evaluate real definite integrals by extending them into the complex plane. **1. Evaluation of $\int_{-\infty}^{\infty} \frac{dx}{x^2+1}$ Step-by-Step** We extend the function into the complex plane as $f(z) = \frac{1}{z^2+1} = \frac{1}{(z-i)(z+i)}$. **A. Choosing the Contour** - We use a closed semicircle $C$ in the Upper Half-Plane (UHP), consisting of a real segment $[-R, R]$ and a large arc $\Gamma_R$. - As $R \to \infty$, the integral along the arc $\Gamma_R$ vanishes to $0$ (by the Estimation Lemma, since the denominator is of order $R^2$ while the arc length is $\pi R$). **B. Computing the Residue at $z = i$** - The only pole inside our contour is the simple pole at **$z = i$** (since $z = -i$ lies in the lower half-plane). - Using the simple pole limit formula: $$\text{Res}(f, i) = \lim_{z \to i} (z - i) \frac{1}{(z-i)(z+i)} = \frac{1}{i+i} = \frac{1}{2i}$$ **C. Applying the Residue Theorem** $$\int_{-\infty}^{\infty} \frac{dx}{x^2+1} = 2\pi i \sum \text{Res} = 2\pi i \left( \frac{1}{2i} \right) = \pi$$ **2. Tricks for Higher-Order Poles (Order $m > 1$)** When a pole has a multiplicity greater than 1, standard limits fail. Use these two fast shortcuts: **- Trick 1: The Derivative Formula** Differentiate the isolated function $m-1$ times before taking the limit: $$\text{Res}(f, z_0) = \frac{1}{(m-1)!} \lim_{z \to z_0} \frac{d^{m-1}}{dz^{m-1}} \left[ (z-z_0)^m f(z) \right]$$ **- Trick 2: Series Expansion Shifting (Often Faster)** - Substitute $w = z - z_0$ to shift the pole to the origin ($0$). - Expand the remaining components into standard Taylor/Geometric series. - The residue is simply the **coefficient of the $\frac{1}{w}$ term** ($s_{-1}$).