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Complex Analysis

Russell W. Howell, John H. Mathews

Section 8.6 Integrands with Branch Points

We now show how to evaluate certain improper real integrals involving the integrand \(x^{\alpha}\frac{P(x)}{Q x)}\text{.}\) The complex function \(z^{\alpha}\) is multivalued, so we must first specify the branch to be used.
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Let \(\alpha\) be a real number with \(0\lt \alpha \lt 1\text{.}\) In this section we use the branch of \(z^{\alpha}\) corresponding to the branch of the logarithm \(\log_0\) (see Equation (5.2.11)) as follows:
\begin{align} z^{\alpha} \amp = e^{\alpha [\log_0(z) ] } =e^{\alpha(\ln |z| +i\arg_0z)}\notag\\ \amp=e^{\alpha (\ln r+i\theta)} = r^{\alpha}(\cos \alpha \theta +i\sin \alpha \theta)\text{,}\tag{8.6.1} \end{align}
where \(z=re^{i\theta}\ne 0\) and \(0\lt \theta \le 2\pi\text{.}\) Note that this is not the traditional principal branch of \(z^{a}\) and that, as defined, the function \(z^{a}\) is analytic in the domain \(\{re^{i\theta} : r>0, \; 0 \lt \theta \lt 2\pi\}\text{.}\)
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Proof.

Let \(C\) denote the simple closed positively oriented contour that consists of the portions of the circles \(C_r(0)\) and \(C_R(0)\) and the horizontal segments joining them, per Figure 8.6.2.
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Figure 8.6.2. Contour \(C\) encloses the nonzero poles \(z_, \ z_2, \ \ldots, \ z_k\) of \(\frac{P}Q\)
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We select a small value of \(r\) and a large value of \(R\) so that the nonzero poles \(z_1, \, z_2, \, \ldots, \, z_k\) of \(\frac{P}{Q}\) lie inside \(C\text{.}\) Using the residue theorem, we write
\begin{equation} \int_{C}f(z)\,dz = 2\pi i\sum\limits_{j=1}^k\mathrm{Res}[f,z_j]\text{.}\tag{8.6.2} \end{equation}
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If we let \(r \to 0\) in Equation (8.6.2), the integrand \(f(z)\) on the upper horizontal line of Figure 8.6.2 approaches \(\frac{x^{\alpha}P(x)}{Q(x)}\text{,}\) where \(x\) is a real number; however, because of the branch we chose for \(z^{a}\) (see Equation (8.6.1)), the integrand \(f(z)\) on the lower horizontal line approaches \(\frac{x^{\alpha}e^{i\alpha 2\pi}P(x)}Q(x)\text{.}\) Therefore,
\begin{align} \lim_{r \to 0}\int_{C}f(z)\,dz = \int_0^R\frac{x^{\alpha}P(x)}{Q(x)}\,dx \amp + \int_R^0 \frac{x^{\alpha}e^{i\alpha 2\pi}P(x)}{Q(x)}\,dx\notag\\ \amp +\int_{C_R^{+}(0)}f(z)\,dz\text{.}\tag{8.6.3} \end{align}
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It is here that we need the function \(Q\) to have a zero of order at most 1 at the origin. Otherwise, the first two integrals on the right side of Equation (8.6.3) would not necessarily converge. Combining this result with Equation (8.6.2) gives
\begin{align*} \int_0^{R}\frac{x^{\alpha}P(x)}{Q(x)}\,dx \amp -\int_0^{R}\frac{x^{\alpha}e^{i\alpha 2\pi}P(x)}{Q(x)}\,dx\\ \amp \qquad \qquad = 2\pi i\sum\limits_{j=1}^k\mathrm{Res}[f,z_j] -\int_{C_R^{+}(0)}f(z)\,dz\text{,} \end{align*}
so
\begin{equation*} \left(\int_0^{R}\frac{x^{a}P(x)}{Q(x)}\,dx\right)\big(1-e^{ia2\pi}\big) = 2\pi i\sum\limits_{j=1}^kRes[f,z_j] -\int_{C_R^{+}(0)}f(z)\,dz\text{,} \end{equation*}
which we rewrite as
\begin{align} \int_0^{R}\frac{x^{\alpha}P(x)}{Q(x)}\,dx \amp = \frac{2\pi i}{1-e^{i\alpha 2\pi}}\sum\limits_{j=1}^k\mathrm{Res}[f,z_j]\notag\\ \amp \qquad \qquad- \frac{1}{1-e^{i\alpha 2\pi}}\int_{C_R^{+}(0)}f(z)\,dz\text{.}\tag{8.6.4} \end{align}
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Using the ML inequality (Theorem 6.2.19) gives
\begin{equation} \lim_{R \to \infty}\int_{C_R^{+}(0)}f(z)\,dz=0\text{.}\tag{8.6.5} \end{equation}
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The argument is essentially the same as that used to establish Equation (8.3.5), and we omit the details. If we combine Equations (8.6.4) and (8.6.5) and let \(R \to \infty\text{,}\) we arrive at the desired result.
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Example 8.6.3.

Evaluate P.V. \(\int_0^{\infty}\frac{x^{\alpha}}{x(x+1)}\,dx\text{,}\) where \(0\lt a\lt 1\text{.}\)
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Solution.
The function \(f(z) =\frac{z^{a}}{z(z+1)}\) has a nonzero pole at the point \(-1\text{,}\) and the denominator has a zero of order at most 1 (in fact, exactly 1) at the origin. Using Theorem 8.6.1, we compute
\begin{align*} \int_0^{\infty}\frac{x^{a}}{x(x+1)}\,dx \amp = \frac{2\pi i}{1-e^{ia2\pi}}\mathrm{Res}[f,-1]\\ \amp = \frac{2\pi i}{1-e^{ia2\pi}}\left(\frac{e^{ia\pi}}{-1}\right)\\ \amp =\frac{\pi}{\frac{e^{ia\pi}-e^{-ia\pi}}{2i}}\\ \amp = \frac{\pi}{\sin a\pi} \text{.} \end{align*}
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We can apply the preceding ideas to other multivalued functions.
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Example 8.6.4.

Evaluate P.V. \(\int_0^{\infty}\frac{\ln x}{x^2+a^2}dx\text{,}\) where \(a>0\text{.}\) To get started, refer to Figure 8.6.5.
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Solution.
We use the function \(f(z) =\frac{\mathrm{\log }_{-\frac{\pi}{2}}z}{z^2+a^2}\text{.}\) Recall that
\begin{equation*} \log_{-\frac{\pi}{2}}z = \ln |z| + i\arg_{-\frac{\pi}{2}}z = \ln r+i\theta\text{,} \end{equation*}
where \(z=re^{i\theta}\ne 0\) and \(-\frac{\pi}{2}\lt \theta \le \frac{3\pi}{2}\text{.}\) The path \(C\) of integration will consist of the segments \([-R,-r ]\) and \([r,R]\) of the \(x\)-axis together with the upper semicircles \(C_{r}:z=re^{i\theta}\) and \(C_R:z=Re^{i\theta}\text{,}\) for \(0 \le \theta \le \pi\text{,}\) as shown in Figure 8.6.5.
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Figure 8.6.5. The contour \(C\) for the integrand \(f(z) = \frac{\mathrm{\log}_{-\frac{\pi}{2}}z}{z^2+a^2}\)
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We chose the branch \(\log_{-\frac{\pi}{2}}\) because it is analytic on \(C\) and its interior, hence so is the function \(f\text{.}\) This choice enables us to apply the residue theorem properly (see the hypotheses of Theorem 8.1.5), and we get
\begin{equation*} \int_{C}f(z)\,dz=2\pi i\mathrm{Res}[f,ai] = \frac{\pi \ln a}{a}+i\frac{\pi^2}{2a}\text{.} \end{equation*}
Keeping in mind the branch of logarithm that we’re using, we then have
\begin{align} \int_{C}f(z)\,dz \amp = \int_{-R}^{-r}f(x)\,dx + \int_{-C_{r}}f(z)\,dz\notag\\ \amp \qquad \qquad + \int_{r}^{R}f(x)\,dx + \int_{C_R}f(z)\,dz\notag\\ \amp = \int_{-R}^{-r}\frac{\ln |x| +i\pi}{x^2+a^2}\,dx + \int_{-C_{r}}f(z)\,dz\notag\\ \amp \qquad \qquad + \int_{r}^{R}\frac{\ln x}{x^2+a^2}\,dx+\int_{C_R}f(z)\,dz\notag\\ \amp = \frac{\pi \ln a}{a}+i\frac{\pi^2}{2a}\text{.}\tag{8.6.6} \end{align}
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If \(R^2>a^2\text{,}\) then by the ML inequality (Theorem 6.2.19)
\begin{align*} \left|\int_{C_R}f(z)\,dz\right| \amp = \left|\int_0^{\pi}\frac{\ln R+i\theta}{R^2e^{i2\theta}+a^2} iRe^{i\theta}d\theta\right|\\ \amp \le \frac{R(\ln R+\pi ) \pi}{R^2-a^2}\text{,} \end{align*}
and L’Hôpital’s rule yields \(\lim\limits_{R \to \infty}\int_{C_R}f(z)r\,dz=0\text{.}\) Engaging in a similar computation shows that \(\lim\limits_{r \to 0^{+}}\int_{c_{r}}f(z)\,dz=0\text{.}\) We use these results when we take limits Equations (8.6.6) to get
\begin{equation*} \text{ P.V. } \left(\int_{-\infty}^{0}\frac{\ln|x| +i\pi}{x^2+a^2}\,dx + \int_0^{\infty}\frac{\ln x}{x^2+a^2}\,dx\right) = \frac{\pi \ln a}{a}+i\frac{\pi^2}{2a}\text{.} \end{equation*}
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Equating the real parts in this equation gives
\begin{equation*} \text{ P.V. } \int_0^{\infty}\frac{\ln x}{x^2+a^2}\,dx=\frac{\pi \ln a}{2a} \end{equation*}
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Remark 8.6.6.

The theory of this section is not purely esoteric. Many applications of contour integrals surface in government and industry worldwide. Several years ago, for example, a briefing was given at the Korean Institute for Defense Analysis (KIDA) in which a sophisticated problem was analyzed by means of a contour integral whose path of integration was virtually identical to that given in Figure 8.6.2.
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Exercises Exercises

Use residues to compute the following integrals.
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1.

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

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

\(\text{ P.V. } \, \int_0^{\infty}\frac{x^\frac{1}{2}}{(1+x)^2}\,dx\text{.}\)
Solution.
\(\frac{\pi}{2}\text{.}\)
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4.

\(\text{ P.V. } \, \int_0^{\infty}\frac{x^\frac{1}{2}}{1+x^2}\,dx\text{.}\)
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5.

\(\text{ P.V. } \, \int_0^{\infty}\frac{\ln (x^2+1)}{x^2+1}\,dx\text{.}\)
Hint.
Use the integrand \(\, f(z)=\frac{\log (z+i)}{z^2+1}.\)
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Solution.
\(\pi \ln 2\text{.}\)
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6.

\(\text{ P.V. } \, \int_0^{\infty}\frac{\ln x}{(1+x^2)^2}\,dx\text{.}\)
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7.

\(\text{ P.V. } \, \int_0^{\infty}\frac{(\ln x)^2}{x^2+1}\,dx\text{.}\)
Solution.
\(\frac{\pi^3}{8}\text{.}\)
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8.

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

\(\int_0^{\infty}\frac{\ln x}{x^2+2^2}\,dx\text{.}\)
Solution.
\(\frac{1}{4}\pi \ln 2\text{.}\)
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10.

Carry out the following computations:
(a)
\(\text{For } \, f(z)=\frac{z^\frac{1}{3}}{z^3(z+1)}, \, \text{ show that } \, \mathrm{Res}[f,-1]=-\frac{1}{2}-\frac{\sqrt{3}}{2}i\text{.}\)
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(b)
\(\text{Use part (a) and } \, \alpha=\frac{1}{3} \, \text{ to verify the computation}\)
\begin{equation*} \frac{2\pi i}{1-e^{i \alpha 2\pi}}\mathrm{Res}[f,-1] = \frac{2\sqrt{3}}{3}\pi\text{.} \end{equation*}
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(c)
\(\text{Can you conclude that P.V. } \, \int_0^{\infty}\frac{x^\frac{1}{3}}{x^3(x+1)}\,dx=\frac{2\sqrt{3}}{3}\pi? \, \text{ Justify your answer.}\)
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11.

\(\text{Carry out the following computations:}\)
(a)
\(\text{For } \, f(z)=\frac{z^\frac{4}{3}}{z+1}, \, \text{ show that } \, \mathrm{Res}[f,-1]=-\frac{1}{2}-\frac{\sqrt{3}}{2}i\text{.}\)
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(b)
\(\text{Use part (a) and } \, \alpha=\frac{4}{3} \, \text{ to verify the computation}\)
\begin{equation*} \frac{2\pi i}{1-e^{i\alpha2\pi}}\mathrm{Res}[f,-1] = \frac{2\sqrt{3}}{3}\pi. \end{equation*}
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(c)
\(\text{Can you conclude that P.V. } \, \int_0^{\infty}\frac{x^\frac{4}{3}}{x+1}\,dx=\frac{2\sqrt{3}}{3}\pi ? \, \text{ Justify your answer.}\)
Solution.
\(\text{No. The hypotheses of} \, \text{ are not satisfied. Explain why they are not.}\)
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Again, use residues to compute the following integrals.
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12.

\(\text{ P.V. } \, \int_0^{\infty}\frac{1}{x^\frac{1}{2}(x+1)^2}\,dx\text{.}\)
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13.

\(\text{ P.V. } \, \int_0^{\infty}\frac{1}{x^\frac{1}{2}(1+x^2)}\,dx\text{.}\)
Solution.
\(\frac{\pi}{\sqrt{2}}\text{.}\)
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14.

\(\text{ P.V. } \, \int_0^{\infty}\frac{x^\frac{1}{3}}{(x+1)^2}\,dx\text{.}\)
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15.

\(\text{ P.V. } \, \int_0^{\infty}\frac{x^\frac{1}{3}}{x^2+1}\,dx\text{.}\)
Solution.
\(\frac{\pi\sqrt{2}}{1+\sqrt{3}}\text{.}\)
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16.

\(\text{ P.V. } \, \int_0^{\infty}\frac{x^\frac{1}{3}\ln x}{x^2+1}\,dx \, \) and P.V. \(\int_0^{\infty}\frac{x^\frac{1}{3}}{x^2+1}\,dx\text{.}\)
Hint.
Use the complex integrand \(\, f(z)=\frac{z^\frac{1}{3}\mathrm{Log} z}{z^2+1}.\)
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17.

\(\text{ P.V. } \, \int_0^{\infty}\frac{\ln (1+x)}{x^{1+a}}\,dx\text{,}\) where \(0\lt a\lt 1\text{.}\)
Solution.
\(\frac{\pi}{a\sin \pi a}\text{.}\)
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18.

\(\text{ P.V. } \, \int_0^{\infty}\frac{\ln x}{(x+a)^2}\,dx\text{,}\) where \(a>0\text{.}\)
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19.

\(\text{ P.V. } \, \int_{-\infty}^{\infty}\frac{\sin x}{x}\,dx\text{.}\)
Hint.
Use the integrand \(\, f(z) =\frac{\exp(iz)}{z} \,\) and the contour \(\, C \,\) in Figure 8.6.2. Then let \(\,r \to 0 \,\) and \(\, R \to \infty.\)
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Solution.
\(\pi\text{.}\)
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20.

\(\text{ P.V. } \, \int_{-\infty}^{\infty}\frac{\sin^2x}{x^2}\,dx.\)
Hint.
Use the integrand \(\, f(z) =\frac{1-\exp (i2z)}{z^2} \,\) and the contour \(\, C \,\) in Figure 8.6.2. Then let \(\, r \to 0 \,\) and \(\, R \to \infty.\)
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21.

The Fresnel integrals \(\int_0^{\infty}\cos(x^2)\,dx\) and \(\int_0^{\infty}\sin (x^2)\,dx\) are important in the study of optics. Use the integrand \(f(z)=\exp(-z^2)\) and the contour \(C\) shown in Figure 8.6.7. Then let \(R \to \infty\) to get the value of these integrals. Use the fact from calculus that \(\int_0^{\infty}e^{-x^2}dx=\frac{\sqrt{\pi}}{2}\text{.}\)
Figure 8.6.7. For Exercise Exercise 8.6.21
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Solution.
\(\frac{\sqrt{\pi}}{2\sqrt{2}}\text{.}\)
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