10.1: Integrals of functions that decay

The theorems in this section will guide us in choosing the closed contour \(C\) described in the introduction.

The first theorem is for functions that decay faster than \(1/z\).

Theorem \(\PageIndex{1}\)

(a) Suppose \(f(z)\) is defined in the upper half-plane. If there is an \(a > 1\) and \(M > 0\) such that

\[|f(z)| < \dfrac{M}{|z|^a} \nonumber \]

for \(|z|\) large then

\[\lim_{R \to \infty} \int_{C_R} f(z)\ dz = 0, \nonumber \]

where \(C_R\) is the semicircle shown below on the left.

Semicircles: left: \(Re^{i \theta}, 0 < \theta < \pi\) right: \(Re^{i \theta}, \pi < \theta < 2\pi\). (CC BY-NC; Ümit Kaya)

(b) If \(f(z)\) is defined in the lower half-plane and

\[|f(z)| < \dfrac{M}{|z|^a}, \nonumber \]

where \(a > 1\) then

\[\lim_{R \to \infty} \int_{C_R} f(z)\ dz = 0, \nonumber \]

where \(C_R\) is the semicircle shown above on the right.

Proof

We prove (a), (b) is essentially the same. We use the triangle inequality for integrals and the estimate given in the hypothesis. For \(R\) large

\[|\int_{C_R} f(z)\ dz| \le \int_{C_R} |f(z)|\ |dz| \le \int_{C_R} \dfrac{M}{|z|^a} |dz| = \int_{0}^{\pi} \dfrac{M}{R^a} R \ d\theta = \dfrac{M \pi}{R^{a - 1}}. \nonumber \]

Since \(a > 1\) this clearly goes to 0 as \(R \to \infty\). \(\text{QED}\)

The next theorem is for functions that decay like \(1/z\). It requires some more care to state and prove.

Theorem \(\PageIndex{2}\)

(a) Suppose \(f(z)\) is defined in the upper half-plane. If there is an \(M > 0\) such that

\[|f(z)| < \dfrac{M}{|z|} \nonumber \]

for \(|z|\) large then for \(a > 0\)

\[\lim_{x_1 \to \infty, x_2 \to \infty} \int_{C_1 + C_2 + C_3} f(z) e^{iaz}\ dz = 0, \nonumber \]

where \(C_1 + C_2 + C_3\) is the rectangular path shown below on the left.

Rectangular paths of height and width \(x_1 + x_2\). (CC BY-NC; Ümit Kaya)

(b) Similarly, if \(a < 0\) then

\[\lim_{x_1 \to \infty, x_2 \to \infty} \int_{C_1 + C_2 + C_3} f(z) e^{iaz}\ dz = 0, \nonumber \]

where \(C_1 + C_2 + C_3\) is the rectangular path shown above on the right.

Note: In contrast to Theorem 10.2.1 this theorem needs to include the factor \(e^{iaz}\).

Proof

(a) We start by parametrizing \(C_1, C_2, C_3\).

\(C_1: \gamma_1 (t) = x_1 + it\), \(t\) from 0 to \(x_1 + x_2\)

\(C_2: \gamma_2 (t) = t + i(x_1 + x_2)\), \(t\) from \(x_1\) to \(-x_2\)

\(C_3: \gamma_3 (t) = -x_2 + it\), \(t\) from \(x_1 + x_2\) to 0.

Next we look at each integral in turn. We assume \(x_1\) and \(x_2\) are large enough that

\[|f(z)| < \dfrac{M}{|z|} \nonumber \]

on each of the curves \(C_j\).

\[\begin{array} {rcl} {|\int_{C_1} f(z) e^{iaz}\ dz|} & \le & {\int_{C_1} |f(z) e^{iaz}|\ |dz| \le \int_{C_1} \dfrac{M}{|z|} |e^{iaz}|\ |dz|} \\ {} & = & {\int_{0}^{x_1 + x_2} \dfrac{M}{\sqrt{x_1^2 + t^2}} |e^{iax_1 - at}|\ dt} \\ {} & \le & {\dfrac{M}{x_1} \int_{0}^{x_1 + x_2} e^{-at}\ dt} \\ {} & = & {\dfrac{M}{x_1} (1 - e^{-a(x_1 + x_2)})/a.} \end{array} \nonumber \]

Since \(a > 0\), it is clear that this last expression goes to 0 as \(x_1\) and \(x_2\) go to \(\infty\).

\[\begin{array} {rcl} {|\int_{C_2} f(z) e^{iaz}\ dz|} & \le & {\int_{C_2} |f(z) e^{iaz}|\ |dz| \le \int_{C_2} \dfrac{M}{|z|} |e^{iaz}|\ |dz|} \\ {} & = & {\int_{-x_2}^{x_1} \dfrac{M}{\sqrt{t^2 + (x_1 + x_2)^2}} |e^{iat - a(x_1 + x_2)}|\ dt} \\ {} & \le & {\dfrac{Me^{-a(x_1 + x_2)}}{x_1 + x_2} \int_{0}^{x_1 + x_2} \ dt} \\ {} & \le & {Me^{-a(x_1 + x_2)}} \end{array} \nonumber \]

Again, clearly this last expression goes to 0 as \(x_1\) and \(x_2\) go to \(\infty\).

The argument for \(C_3\) is essentially the same as for \(C_1\), so we leave it to the reader.

The proof for part (b) is the same. You need to keep track of the sign in the exponentials and make sure it is negative.

Example \(\PageIndex{1}\)

See Example 10.8.1 below for an example using Theorem 10.2.2.