4. *Applications of Residue Theorem

The following are applications of 3. Theorem 2.4.1 - Residue Theorem

APPLICATION 1
Suppose we want to compute the real integral

\int_{- \infty}^{\infty} R (x) d x

Where $R (x) = \frac{P (x)}{Q (x)}$ such that $Q (x)$ has no zeros on the real line, and $d e g (Q) \geq d e g (P) + 2$ (Guarantees absolute convergence of the integral). Then we use semicircles with radius $r$ , which correspond to integrals 1;; $\int_{- r}^{r} R (x) d x$ to see that:

\int_{- \infty}^{\infty} R (x) d x = 2 π i \sum_{w, Q (w) = 0, i m (w) > 0} \underset{z = w}{R e s} R (z) = - 2 π i \sum_{w, Q (w) = 0, i m (w) < 0} \underset{z = w}{R e s} R (z)

In particular, notice that as a corollary, we see that the sum of the residues is $0$ .

Example

Compute the integrals $\int_{- \infty}^{\infty} \frac{d x}{x^{2} + 1}$ and $\int_{- \infty}^{\infty} \frac{d x}{(x^{2} + 1)^{2}}$ using the above formula

APPLICATION 2:
Suppose $R (x)$ is defined the same as above. This time we want to compute the integral:

\int_{- \infty}^{\infty} R (x) \cos (x) d x

The trick here is to compute

\oint_{γ} R (z) e^{i} z d z

And then take the real part of the integral, where now our $γ$ is a rectangle along the real axis with height $r$ .

Example

Compute the integral $\int_{- \infty}^{\infty} \frac{\cos (b x)}{x^{2} + a^{2}} d x$

Our computations yield:

\int_{- \infty}^{\infty} R (x) \cos (x) d x = R e (2 π i \sum_{z_{k} sing of R(z)} \underset{z = z_{k}}{R e s} R (z) e^{i z})

APPLICATION 3:
Suppose that $R (x, y)$ is a rational function such that $R (\sin (t), \cos (t))$ has no singularities on the interval $[0, 2 π]$ . We would like to compute:

\int_{0}^{2 π} R (\cos (t), \sin (t)) d t

We first make the transformation $γ (t) = e^{i t}$ so we are left with the integral:

\oint_{| z | = 1} R (\frac{z + z^{- 1}}{2}, \frac{z - z^{- 1}}{2 i}) \frac{d z}{i z}

We now compute the integral using the residue theorem and we get:

= 2 π i \sum_{z_{k} poles of integrand with | z_{k} | < 1} \underset{z = z_{k}}{R e s} R (\frac{z + z^{- 1}}{2}, \frac{z - z^{- 1}}{2 i})

Example

Compute the integral $\int_{0}^{2 π} \frac{d x}{3 + 4 \cos (x)}$

APPLICATION 4:
Now suppose that $0 < s < 1$ , and let $R (x)$ be a rational function with $d e g (Q) \geq d e g (P) + 2$ . Furthermore suppose that $R (x)$ has no poles on $(0, \infty)$ (These conditions are for nice behavior around $0$ ). We are now interested in computing:

\int_{0}^{\infty} x^{s} R (x) d x

The idea is to use pac-man to eat the positive real axis. After some computations, we get the following formula:

= \frac{2 π i}{1 - e^{2 π i s}} \sum_{sing z_{k} \neq 0 of R(z)} \underset{z = z_{k}}{R e s} z^{s} R (z)

Example

Compute the following integral $\int_{0}^{\infty} \frac{x^{s}}{x^{2} + x} d x$ .

Danger

One might be tempted to say that $e^{2 π i s} = 1$ for any value of $s$ by exponent rules, but these exponent rules don't hold generally for all complex numbers! Consider the complex formula for $e^{i z}$ , and try raising everything to some power.