10.6: Integrals over portions of circles

We will need the following theorem in order to combine principal value and the residue theorem.

Theorem \(\PageIndex{1}\)

Suppose \(f(z)\) has a simple pole at \(z_0\). Let \(C_r\) be the semicircle \(\gamma (\theta) = z_0 + re^{i \theta}\), with \(0 \le \theta \le \pi\). Then

\[\lim_{r \to 0} \int_{C_r} f(z) \ dz = \pi i \text{Res} (f, z_0) \nonumber \]

Figure \(\PageIndex{1}\): Small semicircle of radius \(r\) around \(z_0\). (CC BY-NC; Ümit Kaya)

Proof

Since we take the limit as \(r\) goes to 0, we can assume \(r\) is small enough that \(f(z)\) has a Laurent expansion of the punctured disk of radius \(r\) centered at \(z_0\). That is, since the pole is simple,

\[f(z) = \dfrac{b_1}{z - z_0} + a_0 + a_1 (z - z_0) + \ ... \ \ \ \ \text{for } 0 < |z - z_0| \le r. \nonumber \]

Thus,

\[\int_{C_r} f(z)\ dz = \int_{0}^{\pi} f(z_0 + re^{i \theta}) rie^{i \theta} \ d \theta = \int_{0}^{\pi} (b_1 i + a_0 ire^{i \theta} + a_1 ir^2 e^{i 2 \theta} + \ ...)\ d \theta \nonumber \]

The \(b_1\) term gives \(\pi i b_1\). Clearly all the other terms go to 0 as \(r \to 0\). \(\text{QED}\)

If the pole is not simple the theorem doesn’t hold and, in fact, the limit does not exist.

The same proof gives a slightly more general theorem.

Theorem \(\PageIndex{2}\)

Suppose \(f(z)\) has a simple pole at \(z_0\). Let \(C_r\) be the circular \(\text{arc } \gamma (\theta) = z_0 + re^{i \theta}\), with \(\theta_0 \le \theta \le \theta_0 + \alpha\). Then

\[\lim_{r \to 0} \int_{C_r} f(z)\ dz = \alpha i \text{Res} (f, z_0) \nonumber \]

Figure \(\PageIndex{1}\): Small circular arc of radius \(r\) around \(z_0\). (CC BY-NC; Ümit Kaya)

Example \(\PageIndex{1}\) Return to Example 10.6.1

A long time ago we left off Example 10.6.1 to define principal value. Let’s now use the principal value to compute

\[\tilde{I} = \text{p.v.} \int_{-\infty}^{\infty} \dfrac{e^{ix}}{x} \ dx. \nonumber \]

Solution

We use the indented contour shown below. The indentation is the little semicircle the goes around \(z = 0\). There are no poles inside the contour so the residue theorem implies

\[\int_{C_1 - C_r + C_2 + C_R} \dfrac{e^{iz}}{z} \ dz = 0. \nonumber \]

Figure \(\PageIndex{3}\): (CC BY-NC; Ümit Kaya)

Next we break the contour into pieces.

\[\lim_{R \to \infty, r \to 0} \int_{C_1 + C_2} \dfrac{e^{iz}}{z}\ dz = \tilde{I}. \nonumber \]

Theorem 10.2.2(a) implies

\[\lim_{R \to \infty} \int_{C_R} \dfrac{e^{iz}}{z} \ dz = 0. \nonumber \]

Equation 10.7.1 in Theorem 10.7.1 tells us that

\[\lim_{r \to 0} \int_{C_r} \dfrac{e^{iz}}{z} \ dz = \pi i \text{Res} (\dfrac{e^{iz}}{z}, 0) = \pi i \nonumber \]

Combining all this together we have

\[\lim_{R \to \infty, r \to 0} \int_{C_1 - C_r + C_2 + C_R} \dfrac{e^{iz}}{z} \ dz = \tilde{I} - \pi i = 0, \nonumber \]

so \(\tilde{I} = \pi i\). Thus, looking back at Example 10.3.3, where \(I = \int_{0}^{\infty} \dfrac{\sin (x)}{x} \ dx\), we have

\[I = \dfrac{1}{2} \text{Im} (\tilde{I}) = \dfrac{\pi}{2}, \nonumber \]

There is a subtlety about convergence we alluded to above. That is, \(I\) is a genuine (conditionally) convergent integral, but \(\tilde{I}\) only exists as a principal value. However since \(I\) is a convergent integral we know that computing the principle value as we just did is sufficient to give the value of the convergent integral.