What’s the connection between these integrals and complex analysis? We signed up for the square root of -1 and that’s what we have fun with.
Suppose f is even and “nice” on \mathbb R and we want to evaluate \int_{-\infty}^\infty f(x)\,dx.
Suppose f is analytic in and on C = \Gamma_1 + \Gamma_2, possibly except for isolated singularities in \operatorname{Int}C. We know that \int_Cf=\int_{\Gamma_1}f+\int_{\Gamma_2}f. Hopefully, we can evaluate the left-hand side with the residue theorem (sum of residues at singularities). Then, if we let R \to \infty, the integral over \Gamma_2 is what we want: \operatorname{PV}\int_{-\infty}^\infty f=\int_{-\infty}^\infty f because we assumed f is even.
It remains to deal with \lim_{R\to\infty}\int_{\Gamma_1}f. Hopefully, we can estimate this to show this goes to zero, for example via M-\ell.
Example: Evaluate I = \int_0^\infty x^2/(x^6+1)\,dx. Note that f(z) = x^2/(x^6+1) is even and continuous. As x \to \pm \infty, f \sim 1/x^4 so I converges by the p-test since p>1. Moreover, the complex function f(z)=z^2/(z^6+1) is analytic on \mathbb C except for 6 zeros of z^6+1, i.e. (-1)^{1/6}.
f is analytic in and on C=\Gamma_1+\Gamma_2 except for the 3 zeros in the upper half-plane. These singularities are z_1=e^{\pi i/6}, z_2=i, and z_3=e^{5\pi i/6}. The residue theorem implies that \int_C f(z)\,dz = 2\pi i\sum_{j=1}^3 \operatorname{Res}_{z=z_j}f. We can see that f has the form p/q and at each z_j, p(z_j)\ne 0, q(z_j)=0, and q'(z_j)\ne 0. Thus, each singularity is a simple pole. From the theorem last lecture of \operatorname{Res}_{z_0}p/q=p(z_0)/q'(z_0), we have \int_Cf(z)\,dz=2\pi i \sum_{j=1}^3 \left.\frac{z^2}{(z^6+1)'}\right|_{z=z_j}=2\pi i\sum_{j=1}^3 \frac{z_j^2}{6z_j^5}=2\pi i\left(\frac 1 {6i}-\frac 1 {6i}+\frac 1 {6i}\right)=\frac \pi 3. Remember that we’re looking for \int_C f=\int_{\Gamma_1}f+\int_{\Gamma_2}f. We’ve got the left-hand side now. As the radius R \to \infty, \int_{\Gamma_2}f \to 2I because we’re looking for the integral from 0 to \infty. Now, we want to show that \int_{\Gamma_1}f \to 0 as R \to \infty. We claim that |\int_{\Gamma_1}f|\le M_R \ell_R where \ell_R is the length of \Gamma_1, which is \pi R. Also, \begin{aligned} M_R &= \max_{z\in\Gamma_1}|f(z)| \le \max_{|z|=R}\left|\frac{z^2}{z^6+1}\right|\le\max_{|z|=R}\frac{|z|^2}{|z|^6-|1|}\le\frac{R^2}{R^6-1} \end{aligned} where the denominator comes from the inverse triangle inequality. Therefore, \lim_{R \to \infty}M_R\ell_R = \lim_{R\to\infty}\frac{\pi R\cdot R^2}{R^6-1}=0. Finally, \int_C f=\int_{\Gamma_1}f+\int_{\Gamma_2}f \implies\frac \pi 3=2I+0\implies I=\frac \pi 6. Example: I=\int_0^\infty \sin x/x\,dx. Firstly, notice that this is an improper integral because the integrand is undefined at 0 and the bound is infinity. Near 0, the integrand approaches 1. Approaching \infty is an absolute pain to estimate (via real analysis).
The good news is \sin x/x is even. If I exists, then I=\frac 1 2\int_{-\infty}^\infty \frac{\sin x}x\,dx=\frac 1 2 \operatorname{PV}\int_{-\infty}^\infty \frac{\sin x}x\,dx. We need to be careful because our theorems don’t work across singularities, despite 0 being a removable singularity. This gives is a 4-part contour.
A standard trick when working with trig functions is take an exponential and use the real or imaginary part as needed. Take f(z)=e^{iz}/z which is analytic on \mathbb C_*. Cauchy tells us that the integral across the whole contour is 0. Thus, 0=\int_{\gamma_1+C_\rho+\gamma_2+C_R}f=\int_{\gamma_1}f+\int_{C_\rho}f+\int_{\gamma_2}f+\int_{C_R}f = I_1 + I_2 + I_3 + I_4. Let the integrals of the right-hand side be I_1, \ldots, I_4 respectively. Looking at I_1 and I_3, I_1 = \int_{-R}^{-\rho}\frac{e^{ix}}x\,dx=-\int_\rho^R\frac{e^{-iw}(-w)}{-1}\,dw \quad \text{and} \quad I_3 = \int_\rho^R\frac{e^{ix}}x\,dx. Combining these, I_1+I_3=\int_\rho^R\frac{e^{ix}-e^{-ix}}{x}\,dx=2i\int_\rho^R\frac{\sin x}x\,dx \quad\longrightarrow\quad 2i\int_0^\infty\frac {\sin x}{x}\,dx after taking the limits \rho\to0 and R \to \infty. Therefore, \int_0^\infty \frac {\sin x}{x}\,dx = -\frac 1{2i}\left(\lim_{\rho \downarrow 0}I_2 + \lim_{R \to \infty}I_4 \right) assuming these limits exist. Looking at I_2, we substitute z=\rho e^{i\theta} and \theta : \pi \to 0 (direction) with appropriate change of variables formula. \begin{aligned} I_2 &= \int_{C_\rho}\frac {e^{iz}}{z}\,dz = \int_\pi^0\frac {e^{i\rho e^{i\theta}}}{\rho e^{i\theta}} i\rho e^{i\theta}\,d\theta = -i\int_0^\pi e^{i\rho e^{i\theta}}\,d\theta. \end{aligned} Since |\rho e^{i\theta}| = \rho, the expression e^{i\rho e^{i\theta}}\to 1 as \rho \to 0 uniformly for \theta \in [0,\pi]. From real analysis, this means that the convergence radius \delta is independent of \theta, depending on \epsilon. This means we can write \lim_{\rho \downarrow 0}I_2 = -i\int_0^\pi \lim_{\rho \downarrow 0}\left(e^{i\rho e^{i\theta}}\right)\,d\theta=-i\int_0^\pi d\theta=-i\pi.
Finally, I_4 \to 0 as R \to \infty by Jordan’s lemma (next lecture) because we cannot use M-\ell in the usual way. Therefore, \int_0^\infty \frac {\sin x}{x}\,dx = -\frac 1{2i}\left(\lim_{\rho \downarrow 0}I_2 + \lim_{R \to \infty}I_4 \right)=-\frac 1 {2i }\left[-i\pi+0\right]=\frac\pi 2.