To evaluate $I = \int_{-\infty}^{\infty} e^{-x^2}dx$, consider the contour integral of $e^{-z^2}$ around a rectangle with vertices at $\pm R$ and $\pm R + ia$. Since $e^{-z^2}$ is entire (analytic everywhere), the integral around the closed contour is zero. As $R \to \infty$, the vertical sides vanish, giving:

$\int_{-\infty}^{\infty} e^{-x^2}dx = e^{a^2}\int_{-\infty}^{\infty} e^{-(x+ia)^2}dx = e^{a^2}\int_{-\infty}^{\infty} e^{-x^2}dx$

Setting $I^2 = \int\int e^{-(x^2+y^2)}dx\,dy$ and converting to polar coordinates gives $I = \sqrt{\pi}$.

The integral $I = \int_0^{\infty} \frac{\sin x}{x}dx$ can be evaluated using $\oint_C \frac{e^{iz}}{z}dz$ with a semicircular contour in the upper half-plane. The residue at $z = 0$ contributes:

$\lim_{\epsilon \to 0}\int_{|z|=\epsilon} \frac{e^{iz}}{z}dz = i\pi$

The large semicircle vanishes by Jordan's lemma, so:

$\int_{-\infty}^{\infty} \frac{e^{ix}}{x}dx = i\pi \implies \int_0^{\infty} \frac{\sin x}{x}dx = \frac{\pi}{2}$

In quantum field theory, the propagator integral:

$D(x) = \int \frac{e^{ikx}}{k^2 - m^2 + i\epsilon}dk$

requires careful contour choice. The $i\epsilon$ prescription shifts the poles off the real axis:

$k = \pm(m - i\epsilon)$

For $x > 0$, close the contour in the upper half-plane (where $e^{ikx} \to 0$), enclosing $k = m - i\epsilon$:

$D(x) = -2\pi i \cdot \frac{e^{imx}}{2m} = -\frac{\pi i}{m}e^{imx} \quad (x > 0)$

This encodes causality: signals propagate forward in time only.

Consider potential flow in the $z$-plane around a cylinder of radius $a$. The complex potential for uniform flow $U$ in the $x$-direction is:

$w(z) = U\left(z + \frac{a^2}{z}\right)$

The velocity field is $\mathbf{v} = \nabla\phi = (\partial u/\partial x, \partial u/\partial y)$ where $w = \phi + i\psi$. At $z = ae^{i\theta}$ (cylinder surface):

$w = Ua(e^{i\theta} + e^{-i\theta}) = 2Ua\cos\theta$

showing that $\psi = 0$ on the boundary (the cylinder is a streamline). The pressure distribution from Bernoulli's equation is:

$p - p_{\infty} = \frac{1}{2}\rho(U^2 - v^2) = -2\rho U^2\sin^2\theta$

giving the d'Alembert paradox: zero net force on the cylinder in ideal flow.

The Schwarz-Christoffel transformation maps the upper half-plane to polygonal regions:

$z = A + B\int^w \prod_{k=1}^{n}(\zeta - w_k)^{\alpha_k - 1}d\zeta$

For a parallel-plate capacitor, this maps to a strip, giving the potential:

$\phi(x,y) = \frac{V}{d}y$

For more complex geometries (e.g., comb-drive actuators in MEMS), the mapping provides exact solutions.

The equation $\ddot{x} + 2\gamma\dot{x} + \omega_0^2 x = F(t)/m$ with $x(0) = \dot{x}(0) = 0$ transforms under $\mathcal{L}\{x(t)\} = \tilde{x}(s)$:

$(s^2 + 2\gamma s + \omega_0^2)\tilde{x}(s) = \frac{\tilde{F}(s)}{m}$

For $F(t) = \delta(t)$, $\tilde{F} = 1$, giving the Green's function:

$\tilde{G}(s) = \frac{1}{m(s^2 + 2\gamma s + \omega_0^2)} = \frac{1}{m(s - s_+)(s - s_-)}$

where $s_{\pm} = -\gamma \pm \sqrt{\gamma^2 - \omega_0^2}$. Inverting by partial fractions:

$G(t) = \frac{1}{m\omega}\sin(\omega t)e^{-\gamma t}\theta(t), \quad \omega = \sqrt{\omega_0^2 - \gamma^2}$

for underdamped motion ($\gamma < \omega_0$).