ML estimate

Bound on the magnitude of a contour integral. For a contour $C$ of length $L$ and a continuous function $f$ with $|f(z)|\le M$ on $C$,

$\left|{\int }_{C}f(z)dz\right|\le M\cdot L.$

Hence the name: M(aximum of $|f|$) times L(ength of $C$).

Proof

$\left|{\int }_{C}f(z)dz\right|=\left|{\int }_a^{b}f(z(t))z^{\prime }(t)dt\right|\le {\int }_a^b|f(z(t))||z^{\prime }(t)|dt\le M{\int }_a^b|z^{\prime }(t)|dt=M\cdot L.$

First step: triangle inequality for integrals. Second: bound the integrand. Third: $\int |z^{\prime }(t)|dt$ is the Arc length of the curve, which is $L$. ∎

Standard use: showing semicircle contributions vanish

Mostly used to estimate the contribution of a large semicircle when computing real improper integrals by contour integration.

Example. Let $C_R$ be the upper semicircle $|z|=R$, $\mathrm{Im}(z)\ge 0$. Estimate $\left|{\int }_{C_R}\frac{dz}{z^2+1}\right|$ as $R\to \infty$.

Length $L=\pi R$.

On $C_R$, $|z|=R$. By the reverse Triangle inequality, $|z^2+1|\ge |z{|}^2-1=R^2-1$, so

$\left|\frac{1}{z^2+1}\right|\le \frac{1}{R^2-1}.$

Apply ML:

$\left|{\int }_{C_R}\frac{dz}{z^2+1}\right|\le \frac{\pi R}{R^2-1}\to 0\text{ as }R\to \infty .$

So the semicircle contribution vanishes. This is the “vanishing semicircle” estimate that shows up everywhere in computing real improper integrals by contour integration. The trick: close a real integral ${\int }_{-\infty }^{\infty }f(x)dx$ with a large semicircle in the upper half-plane; the semicircle contribution dies by ML, and the real integral equals $2\pi i\sum$(residues) by the Residue theorem.

When ML is sharp

ML gives an upper bound, not the exact value. Sharp (equality) only when $|f(z)|$ is constant on $C$ and $z^{\prime }(t)$ has constant phase relative to $f$, which is rare in practice. Usually it overestimates by a constant factor, fine for arguing limits.

Jordan’s lemma (sharpening)

For integrands of the form $f(z)e^{iaz}$ with $a>0$ and $|f(z)|\to 0$ as $|z|\to \infty$ in the upper half-plane, the semicircle contribution dies even when ML alone wouldn’t do it: the exponential decay $|e^{iaz}|=e^{-ay}$ for $y>0$ helps. This is Jordan’s lemma, a finer version of ML for $\int f(x)\cos (ax)dx$-type integrals.

Where it shows up

• Bounding contour integrals when exact computation is hard.
• Showing semicircle / quarter-circle / rectangular-corner contributions vanish in limits.
• Convergence of Power series: the radius-of-convergence proof bounds Taylor coefficients with an ML-style estimate via Cauchy’s estimate.
• The proof of the Cauchy integral formula, where the error term from deforming a contour to a small circle has to vanish.

It’s the contour-integral version of $\left|{\int }_a^{b}f(x)dx\right|\le M(b-a)$ from real analysis, dressed up for complex paths.