Section 8.4 Improper Integrals of Trigonometric Functions
Let \(P\) and \(Q\) be polynomials of degree \(m\) and \(n\text{,}\) respectively, where \(n\ge m+1\text{.}\) We can show (but omit the proof) that if \(Q(x) \ne 0\) for all real \(x\text{,}\) then
\begin{equation*}
\text{ P.V. } \int_{-\infty}^{\infty}\frac{P(x)}{Q(x)}\cos x\,dx \text{ and } \text{ P.V. } \int_{-\infty}^{\infty}\frac{P(x)}{Q(x)}\sin x\,dx
\end{equation*}
are convergent improper integrals. You may encounter integrals of this type in the study of Fourier transforms and Fourier integrals. We now show how to evaluate them.
Particularly important is our use of the identities
\begin{equation*}
\cos(\alpha x) =\mathrm{Re}[\exp(i\alpha x)] \text{ and } \sin (\alpha x) =\mathrm{Im}[\exp (i\alpha x)]\text{,}
\end{equation*}
where
\(\alpha\) is a positive real number. The crucial step in the proof of the following Theorem wouldn’t hold if we were to use
\(\cos(\alpha z)\) and
\(\sin (\alpha z)\) instead of
\(\exp (i\alpha z)\text{,}\) as you will see when you get to
Lemma 8.4.4.
Theorem 8.4.1.
Let \(P\) and \(Q\) be polynomials with real coefficients of degree \(m\) and \(n\text{,}\) respectively, where \(n \ge m+1\) and \(Q(x) \ne 0\text{,}\) for all real \(x\text{.}\) If \(\alpha>0\) and
\begin{equation}
f(z) = \frac{\exp(i\alpha z)P(z)}{Q(z)}\text{.}\tag{8.4.1}
\end{equation}
then
\begin{align}
\text{ P.V. } \int_{-\infty}^{\infty}\frac{P(x)}{Q(x)}\cos(\alpha x)\,dx \amp = -2\pi\sum \limits_{j=1}^k \mathrm{Im}(\mathrm{Res}[f,z_j]), \text{ and }\tag{8.4.2}\\
\text{ P.V. } \int_{-\infty}^{\infty}\frac{P(x)}{Q(x)}\sin(\alpha x)\,dx \amp = 2\pi\sum \limits_{j=1}^k\mathrm{Re}(\mathrm{Res}[f,z_j])\text{,}\tag{8.4.3}
\end{align}
where \(z_1, \, z_2, \, \ldots, \, z_{k-1}, \, z_k\) are the poles of \(f\) that lie in the upper half-plane, and \(\mathrm{Re}(\mathrm{Res}[f,z_j])\) and \(\mathrm{Im}(\mathrm{Res}[f,z_j])\) are the real and imaginary parts of \(\mathrm{Res}[f,z_j]\text{,}\) respectively.
Example 8.4.2.
Evaluate P.V. \(\int_{-\infty}^{\infty}\frac{x\sin x}{x^2+4}\,dx\text{.}\)
Solution.
The function
\(f\) in Equation
(8.4.1) is
\(f(z) = \frac{\exp (iz) z}{z^2+4}\text{,}\) which has a simple pole at the point
\(2i\) in the upper half-plane. Calculating the residue yields
\begin{equation*}
\mathrm{Res}[f,2i] =\lim\limits_{z \to 2i}\frac{\exp(iz)z}{z+2i} = \frac{2ie^{-2}}{4i}=\frac{1}{2e^2}\text{.}
\end{equation*}
\begin{equation*}
\text{ P.V. } \int_{-\infty}^{\infty}\frac{x\sin x}{x^2+4}\,dx = 2\pi \mathrm{Re}(\mathrm{Res}[f,2i]) = \frac{\pi}{e^2}
\end{equation*}
Example 8.4.3.
Evaluate P.V. \(\int_{-\infty}^{\infty}\frac{\cos x\,dx}{x^4+4}\text{.}\)
Solution.
The function
\(f\) in Equation
(8.4.1) is
\(f(z) = \frac{\exp (iz)}{z^4+4}\text{,}\) which has simple poles at the points
\(z_1=1+i\) and
\(z_2=-1+i\) in the upper half-plane. We get the residues with the aid of L’Hôpital’s rule:
\begin{align*}
\mathrm{Res}[f, \, 1+i] \amp = \lim_{z \to 1+i}\frac{(z-1-i) \exp (iz)}{z^4+4}\\
\amp = \lim_{z \to 1+i}\frac{[1+i(z-1-i)]\exp (iz)}{4z^3}\\
\amp = \frac{\exp (-1+i)}{4(1+i)^3}\\
\amp = \frac{\sin 1-\cos 1-i(\cos 1+\sin 1)}{16e}\text{.}
\end{align*}
Similarly,
\begin{equation*}
\mathrm{Res}[f, \, -1+i] = \frac{\cos 1 -\sin 1 -i(\cos 1+\sin 1)}{16e}\text{.}
\end{equation*}
\begin{align*}
\int_{-\infty}^{\infty}\frac{\cos x}{x^4+4}\,dx \amp = -2\pi [\mathrm{Im}(\mathrm{Res}[f,1+i] ) + \mathrm{Im}(\mathrm{Res}[f,-1+i])]\\
\amp = \frac{\pi (\cos 1+\sin 1)}{4e}\text{.}
\end{align*}
We are almost ready to give the proof of
Theorem 8.4.1, but first we need one preliminary result.
Lemma 8.4.4. Jordan’s lemma.
Suppose that \(P\) and \(Q\) are polynomials of degree \(m\) and \(n\text{,}\) respectively, where \(n \ge m+1\text{.}\) If \(C_R\) is the upper semicircle \(z=Re^{i\theta}\text{,}\) for \(0 \le \theta \le \pi\text{,}\) then
\begin{equation*}
\lim_{R \to \infty}\int_{c_{R}}\frac{\exp(iz)P(z)}{Q(z)}\,dz=0\text{.}
\end{equation*}
Proof.
From \(n \ge m+1\text{,}\) it follows that \(|\frac{P(z)}{Q(z)}| \to 0\) as \(|z| \to \infty\text{.}\) Therefore, for any \(\varepsilon>0\text{,}\) there exists \(R_{\varepsilon}>0\) such that
\begin{equation}
|\frac{P(z)}{Q(z)}| \lt \frac{\varepsilon}{\pi}\tag{8.4.4}
\end{equation}
\begin{equation}
\left|\int_{C_R}\frac{\exp(iz)P(z)}{Q(z)}\,dz\right| \le \int_{C_R}\frac{\varepsilon}{\pi}|e^{iz}|\,|dz|\text{.}\tag{8.4.5}
\end{equation}
provided \(R\ge R_{\varepsilon}\text{.}\) The parametrization of \(C_R\) leads to the equation
\begin{equation}
|dz| = R\,d\theta, \text{ and } |e^{iz}|=e^{-y}=e^{-R\sin \theta}\text{.}\tag{8.4.6}
\end{equation}
Using the trigonometric identity
\(\sin (\pi -\theta) = \sin \theta\) and Equations
(8.4.6), we express the integral on the right side of
Inequality (8.4.5) as
\begin{equation}
\int_{C_R}\frac{\varepsilon}{\pi}|e^{iz}|\,|dz| = \frac{\varepsilon}{\pi}\int_0^{\pi}e^{-R\sin \theta}R\,d\theta = \frac{2\varepsilon}{\pi}\int_0^{\pi /2}e^{-R\sin \theta}R\,d\theta\text{.}\tag{8.4.7}
\end{equation}
On the interval \(0 \le \theta \le \pi /2\) we can use the inequality
\begin{equation*}
0 \le \frac{2\theta}{\pi}\le \sin \theta\text{.}
\end{equation*}
\begin{align*}
\left|\int_{C_R}\frac{\exp(iz)P(z)}{Q(z)}\,dz\right| \amp \le \frac{2\varepsilon}{\pi} \int_0^{\pi /2}e^{-\frac{2R\theta}{\pi}}R\,d\theta\\
\amp = \left. -\varepsilon e^{-\frac{2R\theta}{\pi}}\right|_{\theta=0}^{\theta = \pi /2}\\
\amp = \varepsilon (1-e^{-R})\\
\amp \lt \varepsilon\text{.}
\end{align*}
Because \(\varepsilon >0\) is arbitrary, the proof of Jordan’s lemma is complete.
We now turn to the proof of our main Theorem.
Proof.
Let \(C\) be the contour that consists of the segment \(-R \le x \le R\) of the real axis together with the upper semicircle \(C_R\) parametrized by \(z=Re^{i\theta}\text{,}\) for \(0 \le \theta \le \pi\text{.}\) Using properties of integrals, we have
\begin{equation*}
\int_{-R}^{R}\frac{\exp(i\alpha x)P(x)}{Q(x)}\,dx = \int_{C}\frac{\exp(i\alpha z)P(z)}{Q(z)}\,dz - \int_{C_R}\frac{\exp(i\alpha z)P(z)}{Q(z)}\,dz\text{.}
\end{equation*}
If \(R\) is sufficiently large, all the poles \(z_1, \, z_2, \, \ldots, \, z_k\) of \(f\) will lie inside \(C\text{,}\) and we can use the residue theorem to obtain
\begin{equation}
\int_{-R}^{R} \frac{\exp (i\alpha x)P(x)}{Q(x)}\,dx = 2 \pi i\sum_{j=1}^k\mathrm{Res}[f,z_j] - \int_{C_R}\frac{\exp(i\alpha z)P(z)}{Q(z)}\,dz\text{.}\tag{8.4.8}
\end{equation}
Since
\(\alpha\) is a positive real number, the change of variables
\(\zeta=\alpha z\) shows that the conclusion of Jordan’s lemma holds for the integrand
\(\frac{\exp(i\alpha z)P(z)}{Q(z)}\text{.}\) Hence we let
\(R \to \infty\) in Equation
(8.4.8) to obtain
\begin{align*}
\text{ P.V. } \int_{-\infty}^{\infty}\frac{[\cos(\alpha x)+i\sin(\alpha x)]P(x)}{Q(x)}\,dx \amp = 2\pi i\sum_{j=1}^k\mathrm{Res}[f,z_j]\\
\amp = -2\pi \sum\limits_{j=1}^k\mathrm{Im}(\mathrm{Res}[f,z_j])\\
\amp + 2\pi i\sum\limits_{j=1}^k\mathrm{Re}(\mathrm{Res}[f,z_j])\text{.}
\end{align*}
Equating the real and imaginary parts of this equation gives us Equations
(8.4.2) and
(8.4.3), which completes the proof.
Exercises Exercises
Use residues to find the Cauchy principal value of the following:
1.
\(\int_{-\infty}^{\infty}\frac{\cos x}{x^2+9}\,dx, \text{ and } \int_{-\infty}^{\infty}\frac{\sin x}{x^2+9}\,dx\text{.}\)
Solution.
\(\int_{-\infty}^{\infty}\frac{\cos x}{x^2+9}\,dx = \frac{\pi}{3e^3}\text{,}\) and \(\int_{-\infty}^{\infty}\frac{\sin x}{x^2+9}\,dx = 0\text{.}\)
2.
\(\int_{-\infty}^{\infty}\frac{x\cos x}{x^2+9}\,dx, \text{ and } \int_{-\infty}^{\infty}\frac{x\sin x}{x^2+9}\,dx\text{.}\)
3.
\(\int_{-\infty}^{\infty}\frac{x\sin x}{(x^2+4)^2}\,dx\text{.}\)
Solution.
\(\frac{\pi}{4e^2}\text{.}\)
4.
\(\int_{-\infty}^{\infty}\frac{\cos x}{(x^2+4)^2}\,dx\text{.}\)
5.
\(\int_{-\infty}^{\infty}\frac{\cos x}{(x^2+4)(x^2+9)}\,dx\text{.}\)
Solution.
\(\frac{\pi}{5}(\frac{1}{2e^2}-\frac{1}{3e^3})\text{.}\)
6.
\(\int_{-\infty}^{\infty}\frac{\cos x}{(x^2+1)(x^2+4)}\,dx\text{.}\)
7.
\(\int_{-\infty}^{\infty}\frac{\cos x}{x^2-2x+5}\,dx\text{.}\)
Solution.
\(\frac{\pi \cos 1}{2e^2}\text{.}\)
8.
\(\int_{-\infty}^{\infty}\frac{\cos x}{x^2-4x+5}\,dx\text{.}\)
9.
\(\int_{-\infty}^{\infty}\frac{x\sin x}{x^4+4}\,dx\text{.}\)
Solution.
\(\frac{\pi \sin 1}{2e}\text{.}\)
10.
\(\int_{-\infty}^{\infty}\frac{x^3\sin x}{x^4+4}\,dx\text{.}\)
11.
\(\int_{-\infty}^{\infty}\frac{\cos 2x}{x^2+2x+2}\,dx\text{.}\)
Solution.
\(\frac{\pi \cos 2}{e^2}\text{.}\)
12.
\(\int_{-\infty}^{\infty}\frac{x^3\sin 2x}{x^4+4}\,dx\text{.}\)
13.
Why do you need to use the exponential function when evaluating improper integrals involving the sine and cosine functions?
Solution.
The inequality \(\left|\int_{C_{R}}\frac{\exp(iz)P(z)}{Q(z)}\,dz\right| \le \frac{2\varepsilon}{\pi}\int\nolimits_0^\frac{\pi}{2}e^{\frac{-2R\theta}{\pi}}R\,d\theta \lt \varepsilon\) in Jordan’s lemma would not be possible to get if we replaced \(\exp(iz)\) by either the complex sine or cosine. Explain!
