Worked: the bell-shaped integral
Evaluate \displaystyle\int_{-\infty}^{\infty} \frac{dx}{1 + x^2}
by residues, treating it as a complex integral of
f(z) = \dfrac{1}{1 + z^2} = \dfrac{1}{(z - i)(z + i)}.
Step 1 — close the contour. Integrate
f around the half-disk: the segment from
-R to R plus the upper arc
\Gamma_R. Splitting the closed loop,
\oint = \int_{-R}^{R} \frac{dx}{1 + x^2} + \int_{\Gamma_R} \frac{dz}{1 + z^2}.
Step 2 — kill the arc with the ML inequality. On
\Gamma_R we have |z| = R, so for large
R the integrand is bounded by
M = \dfrac{1}{R^2 - 1}, and the arc has length
L = \pi R. By the
ML
inequality,
\left|\int_{\Gamma_R} \frac{dz}{1 + z^2}\right| \le M\,L = \frac{\pi R}{R^2 - 1} \;\xrightarrow{R \to \infty}\; 0.
Step 3 — pick up the enclosed residue. Inside the upper half-plane the only
pole is z = i. By the quotient rule with
h(z) = 1 + z^2, h'(z) = 2z:
\operatorname{Res}(f, i) = \frac{1}{h'(i)} = \frac{1}{2i}.
Step 4 — assemble. The arc vanishes, so the real integral equals the closed
contour integral, which the residue theorem evaluates:
\int_{-\infty}^{\infty} \frac{dx}{1 + x^2} = 2\pi i\,\operatorname{Res}(f, i) = 2\pi i \cdot \frac{1}{2i} = \pi.
Step 5 — sanity check. This is an
improper
integral we can also do directly:
\int_{-\infty}^{\infty} \frac{dx}{1 + x^2} = \bigl[\arctan x\bigr]_{-\infty}^{\infty} = \tfrac{\pi}{2} - (-\tfrac{\pi}{2}) = \pi.
The two methods agree.
A harder one: a fourth-degree denominator
The same machine evaluates
\displaystyle\int_{-\infty}^{\infty} \frac{dx}{1 + x^4}, which has
no friendly antiderivative at all. The denominator vanishes at the four fourth roots of
-1; the two in the upper half-plane are
z_1 = e^{i\pi/4}, \qquad z_2 = e^{i 3\pi/4}.
Each is a simple pole, and the quotient rule gives
\operatorname{Res}(f, z_k) = \dfrac{1}{4 z_k^3} = -\dfrac{z_k}{4}
(using z_k^4 = -1). Summing the two upper residues and multiplying
by 2\pi i,
\int_{-\infty}^{\infty} \frac{dx}{1 + x^4} = 2\pi i\bigl(\operatorname{Res}(f, z_1) + \operatorname{Res}(f, z_2)\bigr) = \frac{\pi}{\sqrt{2}}.
The astonishing thing is that the answer is real, yet the route runs entirely through the
complex plane and through a pole the real integral never even meets. The decay of
f guarantees the arc contributes nothing, so the line integral
is fully determined by the residues hiding above it. Whole families of integrals — rational
functions, products with e^{iax} (Fourier integrals), and more
— fall to exactly this contour-closing technique. The line was never the whole story; the
upper half-plane was.