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Section 8.5 Indented Contour Integrals

If \(f\) is continuous on the interval \(b\lt x\le c\text{,}\) but discontinuous at \(b\text{,}\) then the improper integral of \(f\) over \([b,c]\) is defined by
\begin{equation*} \int_b^c f(x)\,dx = \lim_{r \to b^{+}}\int_{r}^{c}f(x)\,dx\text{,} \end{equation*}
provided the limit exists. Similarly, if \(f\) is continuous on the interval \(a \le x\lt b\text{,}\) but discontinuous at \(b\text{,}\) then the improper integral of \(f\) over \([a,b]\) is defined by
\begin{equation*} \int_{a}^{b}f(x)\,dx = \lim_{R \to b^{-}} \int_{a}^{R}f(x)\,dx\text{,} \end{equation*}
provided the limit exists. For example,
\begin{equation*} \int_0^{9}\frac{1}{2\sqrt{x}}\,dx = \lim_{r \to 0^{+}} \int_{r}^{9}\frac{1}{2\sqrt{x}}\,dx = \lim\limits_{r \to 0^{+}}(\sqrt{x})\Big|_{x=r}^{x=9} = 3-\lim\limits_{r \to 0^{+}}\sqrt{r} = 3\text{.} \end{equation*}
If \(f\) is continuous for all values of \(x\) in the interval \([a,c]\text{,}\) except at the value \(x=b\text{,}\) where \(a\lt b\lt c\text{,}\) then the Cauchy principal value of \(f\) over \([a,c]\) is defined by
\begin{equation*} \text{ P.V. } \int_{a}^{c}f(x)\,dx = \lim_{r \to 0^{+}}\left[\int_{a}^{b-r}f(x)\,dx+\int_{b+r}^{c}f(x)\,dx\right]\text{,} \end{equation*}
provided the limit exists.

Example 8.5.1.

\begin{equation*} \text{ P.V. } .\int_{-1}^{8}\frac{1}{x^\frac{1}{3}}\,dx = \lim_{r \to 0^{+}}\left[\int_{-1}^{-r}\frac{1}{x^\frac{1}{3}}\,dx+\int_{r}^{8}\frac{1}{x^\frac{1}{3}}\,dx\right]\text{.} \end{equation*}
Evaluating the integrals and computing limits gives
\begin{equation*} \lim_{r \to 0^{+}}\left[\frac{3}{2}r^\frac{2}{3} - \frac{3}{2} + 6 - \frac{3}{2}r^\frac{2}{3}\right] = \frac{9}{2}\text{.} \end{equation*}
In this section we show how to use residues to evaluate the Cauchy principal value of the integral of \(f\) over \((-\infty, \infty)\) when the integrand \(f\) has simple poles on the \(x\)-axis. We state our main results and then look at some examples before giving proofs.

Remark 8.5.4.

The formulas in these theorems give the Cauchy principal value of the integral, which pays special attention to the manner in which any limits are taken. They are similar to those in Sections 8.3 and Section 8.4, except here we add one-half the value of each residue at the points \(t_1, \, t_2, \, \ldots, \, t_l\) on the \(x\)-axis.

Example 8.5.5.

Evaluate P.V. \(\int_{-\infty}^{\infty}\frac{x}{x^3-8}\,dx\) by using complex analysis.
Solution.
The integrand
\begin{equation*} f(z) = \frac{z}{z^3-8}=\frac{z}{(z-2)(z+1+i\sqrt{3})(z+1-i\sqrt{3})} \end{equation*}
has simple poles at the points \(t_1=2\) on the \(x\)-axis and \(z_1=-1+i \sqrt{3}\) in the upper half-plane. By Theorem 8.5.2,
\begin{align*} \text{ P.V. } \int_{-\infty}^{\infty}\frac{x}{x^3-8}\,dx \amp = 2\pi i\mathrm{Res}[f,z_1] +\pi i\mathrm{Res}[f,t_1]\\ \amp = 2\pi i\frac{-1-i\sqrt{3}}{12}+\pi i\frac{1}{6}=\frac{\pi \sqrt{3}}{6}\text{.} \end{align*}

Example 8.5.6.

Evaluate P.V. \(\int_{-\infty}^{\infty}\frac{t\,dx}{t^3-8}\) with a computer algebra system.
Solution.
Computer algebra systems such as Mathematica or MAPLE give the indefinite integral
\begin{align*} \int\frac{t}{t^3-8}\,dt \amp = \frac{\mathrm{Arctan}\left(\frac{1+t}{\sqrt{3}}\right)}{2\sqrt{3}} + \frac{\mathrm{Log}(t-2)}{6} + \frac{\mathrm{Log}(t^2+2t+4)}{12}\\ \amp =g(t)\text{.} \end{align*}
However, for real numbers, we should write the second term as \(\frac{\mathrm{ Log}[(t-2)^2] }{12}\) and use the equivalent formula:
\begin{equation*} g(t) = \frac{\mathrm{Arctan}\left(\frac{1+t}{\sqrt{3}}\right)}{2\sqrt{3}} + \frac{\mathrm{Log}[(t-2)^2]}{12}+\frac{\mathrm{Log}(t^2+2t+4)}{12}\text{.} \end{equation*}
This antiderivative has the property \(\lim\limits_{t \to 2}g(t) = -\infty\text{,}\) as Figure 8.5.7 shows.
Figure 8.5.7. Graph of \(s=g(t) =\int \frac{t}{t^3-8}\ dt\)
We also compute
\begin{equation*} \lim_{t\to \infty}g(t) = \frac{\pi \sqrt{3}}{12} \quad \text{and} \quad \lim_{t \to -\infty}g(t) = -\frac{\pi \sqrt{3}}{12}\text{,} \end{equation*}
and the Cauchy principal limit at \(t=2\) as \(r \to 0\) is
\begin{equation*} \lim_{r \to 0^{+}}[g(2+r) -g(2-r)] = 0\text{.} \end{equation*}
Therefore, the Cauchy principal value of the improper integral is
\begin{align*} \text{ P.V. } \int_{-\infty}^{\infty} \frac{t}{t^3-8}\,dt \amp = \lim_{r \to 0^{+}}\left[\int\limits_{-\infty}^{2-r} \frac{t}{t^3-8}\,dt + \int\limits_{2+r}^{\infty} \frac{t}{t^3-8}\,dt\right]\\ \amp = \left(\lim_{r \to 0^{+}}g(2-r) - \lim_{t \to -\infty}g(t)\right)\\ \amp \qquad \qquad + \left(\lim_{t \to \infty}g(t) - \lim_{r \to 0^{+}}g(2+r)\right)\\ \amp = \left(\lim_{r \to 0^{+}}g(2-r) - \lim_{r \to 0^{+}}g(2+r)\right)\\ \amp \qquad \qquad + \left(\lim_{t \to \infty}g(t) - \lim_{t \to -\infty}g(t)\right)\\ \amp = 0 + \frac{\pi \sqrt{3}}{12} + \frac{\pi \sqrt{3}}{12}\\ \amp = \frac{\pi \sqrt{3}}{6}\text{.} \end{align*}

Example 8.5.8.

Evaluate P.V. \(\displaystyle\int_{-\infty}^{\infty}\frac{\sin x}{(x-1)(x^2+4)}\,dx\text{.}\)
Solution.
The integrand \(f(z) = \frac{\exp(iz)}{(z-1)(z^2+4)}\) has simple poles at the points \(t_1=1\) on the \(x\)-axis and \(z_1=2i\) in the upper half-plane. By Theorem 8.5.3,
\begin{align*} \text{ P.V. } \int_{-\infty}^{\infty}\frac{\sin x}{(x-1)(x^2+4)}\,dx \amp = 2\pi \mathrm{Re}\left(\mathrm{Res}[f,z_1]\right)\\ \amp \qquad \qquad +\pi \mathrm{Re}\left(\mathrm{Res}[f,t_1]\right)\\ \amp = 2\pi \mathrm{Re}\left(\frac{-2+i}{20e^2}\right)\\ \amp \qquad \qquad +\pi \mathrm{Re}\left(\frac{\cos 1+i\sin 1}{5}\right)\\ \amp = \frac{\pi}{5}\left(\cos 1-\frac{1}{e^2}\right)\text{.} \end{align*}
We are almost ready for the proofs of Theorems 8.5.2 and Theorem 8.5.3. First, we need the following lemma.

Proof.
The Laurent series for \(f\) at \(z=t_0\) has the form
\begin{equation} f(z) = \frac{\mathrm{Res}[f,t_0] }{z-t_0}+g(z)\text{,}\tag{8.5.4} \end{equation}
where \(g\) is analytic at \(z=t_0\text{.}\) Using the parameterization of \(C_{r}\) and Equation (8.5.4), we get
\begin{align} \hspace{-0.5in} \int_{C_{r}}f(z)\,dz \amp = \mathrm{Res}[f,t_0] \int_0^{\pi}\frac{ire^{i\theta}}{re^{i\theta}}\,d\theta + ir\int_0^{\pi}g(t_0+re^{i\theta}) e^{i\theta}\,d\theta\notag\\ \amp = i\pi \mathrm{Res}[f,t_0] + ir\int_0^{\pi}g(t_0+re^{i\theta}) e^{i\theta}d\theta\text{.}\tag{8.5.5} \end{align}
Since \(g\) is continuous at \(t_0\text{,}\) there is an \(M>0\) so that \(|g(t_0+re^{i\theta})| \le M\text{,}\) and
\begin{equation*} \left|\lim_{r \to 0}ir\int_0^{\pi}g(t_0+re^{i\theta}) e^{i\theta}d\theta\right| \le \lim_{r \to 0}r\int_0^{\pi}M \,d\theta = \lim_{r \to 0}r\pi M=0\text{.} \end{equation*}
Combining this inequality with Equation (8.5.5) gives the desired result.
We are now ready to prove Theorems Theorem 8.5.2 and Theorem 8.5.3:

Proof of Theorems Theorem 8.5.2 and Theorem 8.5.3.
Since \(f\) has only a finite number of poles, we can choose \(r\) small enough that the semicircles
\begin{equation*} C_j : z = t_j+re^{i\theta}, \quad \text{for} \quad 0 \le \theta \le \pi \quad \text{and} \quad j=1,2,\ldots, l \text{,} \end{equation*}
are disjoint and the poles \(z_1, \, z_2, \, \ldots, \, z_k\) of \(f\) in the upper half-plane lie above them, as shown in Figure 8.5.10.
Figure 8.5.10. The poles \(t_1, t_2, \ldots, t_l\) of \(f\) that lie on the \(x\)- axis and the poles \(z_1, z_2, \ldots z_k\) that lie above the semicircles \(C_1, C_2, \ldots, C_{l}\)
Let \(R\) be large enough so that the poles of \(f\) in the upper half-plane lie under the semicircle \(C_R:z=Re^{i\theta}\text{,}\) for \(0 \le \theta \le \pi\text{,}\) and the poles of \(f\) on the \(x\)-axis lie in the interval \(-R \le x \le R\text{.}\) Let \(C\) be the simple closed positively oriented contour that consists of \(C_R\) and \(-C_1, \, -C_2, \, \ldots, \, -C_l\) and the segments of the real axis that lie between the semicircles shown in Figure 8.5.10. The residue theorem gives \(\int_Cf(z)\,dz = 2\pi i\sum\limits_{j=1}^k\mathrm{Res}[f,z_j]\text{,}\) which we rewrite as
\begin{equation} \hspace{-0.75in} \int_{I_{R}}f(x)\,dx = 2\pi i\sum\limits_{j=1}^k\mathrm{Res}[f,z_j] + \sum_{j=1}^{l}\int_{C_j}f(z)\,dz - \int_{C_R}f(z)\,dz\text{,}\tag{8.5.6} \end{equation}
where \(I_R\) is the portion of the interval \(-R \le x \le R\) that lies outside the intervals \((t_j-r,t_j+r)\) for \(j=1,2,\ldots,l\text{.}\) Using the same techniques that we used in Theorems 8.3.4 and Theorem 8.4.1 yields
\begin{equation} \lim_{R \to \infty}\int_{C_R}f(z)\,dz=0\text{.}\tag{8.5.7} \end{equation}
If we let \(R \to \infty\) and \(r \to 0\) in Equation (8.5.6) and use the results of Equations (8.5.7) and Lemma 8.5.9, we obtain
\begin{equation} \text{ P.V. } \int_{-\infty}^{\infty}f(x)\,dx = 2\pi i\sum\limits_{j=1}^k\mathrm{Res}[f,z_j] + \pi i\sum_{j=1}^{l}\mathrm{Res}[f,t_j]\text{.}\tag{8.5.8} \end{equation}
If \(f\) is the function given in Theorem 8.5.2, then Equation (8.5.8) becomes Equation (8.5.1). If \(f\) is the function given in Theorem 8.5.3, then equating the real and imaginary parts of Equation (8.5.8) results in Equations (8.5.2) and (8.5.3), respectively, which completes the proof.

Exercises Exercises

Use residues to compute the following integrals.

1.

P.V. \(\int_{-\infty}^{\infty}\frac{1}{x(x-1)(x-2)}\,dx\text{.}\)
Solution.
\(0\text{.}\)

2.

P.V. \(\int_{-\infty}^{\infty}\frac{1}{x^3+x}\,dx\text{.}\)

3.

P.V. \(\int_{-\infty}^{\infty}\frac{x}{x^3+1}\,dx\text{.}\)
Solution.
\(\frac{\pi}{\sqrt{3}}\text{.}\)

4.

P.V\(.\int_{-\infty}^{\infty}\frac{1}{x^3+1}\,dx\text{.}\)

5.

P.V. \(\int_{-\infty}^{\infty}\frac{x^2}{x^4-1}\,dx\text{.}\)
Solution.
\(\frac{\pi}{2}\text{.}\)

6.

P.V. \(\int_{-\infty}^{\infty}\frac{x^4}{x^{6}-1}\,dx\text{.}\)

7.

P.V. \(\int_{-\infty}^{\infty}\frac{\sin x}{x}\,dx\text{.}\)
Solution.
\(\pi\text{.}\)

8.

P.V. \(\int_{-\infty}^{\infty}\frac{\cos x}{x^2-x}\,dx\text{.}\)

9.

P.V. \(\int_{-\infty}^{\infty}\frac{\sin x}{x(\pi^2-x^2)}\,dx\text{.}\)
Solution.
\(\frac{2}{\pi}\text{.}\)

10.

P.V. \(\int_{-\infty}^{\infty}\frac{\cos x}{\pi^2-4x^2}\,dx\text{.}\)

11.

P.V. \(\int_{-\infty}^{\infty}\frac{\sin x}{x(x^2+1)}\,dx\text{.}\)
Solution.
\(\pi (1-\frac{1}{e})\text{.}\)

12.

P.V. \(\int_{-\infty}^{\infty}\frac{x\cos x}{x^2+3x+2}\,dx\text{.}\)

13.

P.V. \(\int_{-\infty}^{\infty}\frac{\sin x}{x(1-x^2)}\,dx\text{.}\)
Solution.
\(\pi (1-\cos 1)\text{.}\)

14.

P.V. \(\int_{-\infty}^{\infty}\frac{\cos x}{a^2-x^2}\,dx\text{.}\)

15.

P.V. \(\int_{-\infty}^{\infty}\frac{\sin^2x}{x^2}\,dx\text{.}\)
Hint.
Use trigonometric identity \(\sin^2x=\frac{1}{2}-\frac{1}{2} \cos 2x.\)
Solution.
\(\pi\text{.}\)

16.

As a hint for the following exercises refer to Figure 8.5.11.
Figure 8.5.11. The contour \(C=L_1+C_r-L_2\) for Exercises 16 and 17

(a)

P.V. \(\int_0^{\infty}\frac{1}{x^3+1}\,dx\text{.}\)

(b)

P.V. \(\int_0^{\infty}\frac{x}{x^3+1}\,dx\text{.}\)
Solution.
\(\frac{2\sqrt{3}}{9}\pi\text{.}\)
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