Section 6.1 Complex Integrals
We introduce the integral of a complex function by defining the integral of a complex-valued function of a real variable.
Definition 6.1.1. Integral of $f(t)$.
Let \(f(t)=u(t)+iv(t)\text{,}\) where \(u\) and \(v\) are real-valued functions of the real variable \(t\) for \(a \le t \le b\text{.}\) Then
\begin{equation}
\int_a^b f(t)\,dt=\int_a^b u(t)\, dt+i\int_a^b v(t)\, dt\text{.}\tag{6.1.1}
\end{equation}
We generally evaluate integrals of this type by finding the antiderivatives of
\(u\) and
\(v\) and evaluating the definite integrals on the right side of Equation
(6.1.1). That is, if
\(U\,'(t)=u(t)\text{,}\) and
\(V\,'(t)=v(t)\text{,}\) for
\(a \le t \le b\text{,}\) we have
\begin{equation}
\int_a^b f(t)\,dt = \big[U(t) +iV(t)\big] \Big|_{t=a}^{t=b}=U(b)-U(a) +i\big[V(b) -V(a)\big]\text{.}\tag{6.1.2}
\end{equation}
Example 6.1.2.
Show that
\begin{equation*}
\int_0^1(t-i)^3\,dt=-\frac{5}{4}\text{.}
\end{equation*}
Solution.
We write the integrand in terms of its real and imaginary parts, i.e., \(f(t)=(t-i)^3=t^3-3t+i(-3t^2+1)\text{.}\) Here, \(u(t)=t^3-3t\) and \(v(t)=-3t^2+1\text{.}\) The integrals of \(u\) and \(v\) are
\begin{equation*}
\int_0^1(t^3-3t)\,dt=-\frac{5}{4}, \text{ and } \int_0^1(-3t^2+1) \,dt=0\text{.}
\end{equation*}
\begin{equation*}
\int_0^1(t-i)^3\,dt = \int_0^1u(t)\,dt+i\int_0^1v(t)\,dt = -\frac{5}{4}
\end{equation*}
Example 6.1.3.
Show that
\begin{equation*}
\int_0^{\frac{\pi}{2}}\exp(t+it)\,dt = \frac{1}{2} (e^{\frac{\pi}{2}}-1) + \frac{i}{2}(e^{\frac{\pi}{2}}+1)\text{.}
\end{equation*}
Solution.
\begin{align*}
\int_0^{\frac{\pi}{2}}\exp (t+it)\,dt \amp = \int_0^{\frac{\pi}{2}}e^te^{it}dt\\
\amp = \int_0^{\frac{\pi}{2}}e^t(\cos t+i\sin t \,dt\\
\amp = \int_0^{\frac{\pi}{2}}e^t\cos t \, dt + i\int_0^{\frac{\pi}{2}}e^t\sin t \,dt\text{.}
\end{align*}
We can evaluate each of the integrals via integration by parts. For example,
\begin{align*}
\int_0^{\frac{\pi}{2}}\underset{u}{\underbrace{e^t}} \; \underset{dv}{\underbrace{\cos t\,dt}} \amp = (\underset{u}{\underbrace{e^t}} \; \underset{v}{\underbrace{\sin t}}) \Big|_{t=0}^{t=\frac{\pi}{2}} -\int_0^{\frac{\pi}{2}}\underset{v}{\underbrace{\sin t}} \; \underset{du}{\underbrace{e^tdt}}\\
\amp = (e^{\frac{\pi}{2}}\sin \frac{\pi}{2}-e^0\sin 0) -\int_0^{\frac{\pi}{2}}\underset{u}{\underbrace{e^t}} \; \underset{dv}{\underbrace{\sin t \,dt}}\\
\amp = e^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}}\underset{u}{\underbrace{e^t}} \; \underset{dv}{\underbrace{\sin t \,dt}}\\
\amp = e^{\frac{\pi}{2}} - (\underset{u}{\underbrace{e^t}} \; [\underset{v}{\underbrace{-\cos t}}] ) \Big|_{t=0}^{t=\frac{\pi}{2}} +\int_0^{\frac{\pi}{2}}\underset{v}{\underbrace{-\cos t}} \; \underset{du}{\underbrace{e^t\,dt}}\\
\amp = e^{\frac{\pi}{2}}-1-\int_0^{\frac{\pi}{2}}e^t\cos t\,dt\text{.}
\end{align*}
Adding \(\int_0^{\frac{\pi}{2}}e^t\cos t\,dt\) to both sides of this equation and then dividing by 2 gives \(\int_0^{\frac{\pi}{2}}e^t\cos t)\,dt=\frac{1}{2}(e^{\frac{\pi}{2}}-1)\text{.}\) A similar computation procedure yields \(i\int_0^{\frac{\pi}{2}}e^t\sin t\,dt=\frac{i}{2}(e^{\frac{\pi}{2}}+1)\text{.}\) Therefore,
\begin{equation*}
\int_0^{\frac{\pi}{2}}\exp (t+it)\,dt=\frac{1}{2}(e^{\frac{\pi}{2}}-1) + \frac{i}{2}(e^{\frac{\pi}{2}}+1)
\end{equation*}
Complex integrals have properties that are similar to those of real integrals. We now trace through several commonalities. Let \(f(t)=u(t)+iv(t)\) and \(g(t)=p(t)+iq(t)\) be continuous on \(a \le t \le b\text{.}\)
Using
Definition (6.1.1), we can easily show that the integral of their sum is the sum of their integrals, that is
\begin{equation}
\int_a^b[f(t)+g(t)]\,dt = \int_a^bf(t)\,dt + \int_a^b g(t)\,dt\text{.}\tag{6.1.3}
\end{equation}
If we divide the interval
\(a \le t \le b\) into
\(a \le t \le c\) and
\(c \le t \le b\) and integrate
\(f(t)\) over these subintervals by using
Definition (6.1.1), then we get
\begin{equation}
\int_a^bf(t)\,dt=\int_{a}^{c}f(t)\,dt+\int_c^{b}f(t)\,dt\text{.}\tag{6.1.4}
\end{equation}
Similarly, if \(c+id\) denotes a complex constant, then
\begin{equation}
\int_a^b(c+id) f(t)\,dt=(c+id) \int_a^bf(t)\,dt\text{.}\tag{6.1.5}
\end{equation}
If the limits of integration are reversed, then
\begin{equation}
\int_a^bf(t)\,dt=-\int_{b}^{a}f(t)\,dt\text{.}\tag{6.1.6}
\end{equation}
The integral of the product \(fg\) becomes
\begin{align}
\int_a^bf(t) g(t)\,dt \amp = \int_a^b \big[u(t) p(t) -v(t) q(t)\big] \, dt\notag\\
\amp + i\int_a^b \big[u(t)q(t) + v(t)p(t)\big] \, dt\text{.}\tag{6.1.7}
\end{align}
Example 6.1.4.
Let us verify Property
(6.1.5). We start by writing
\begin{equation*}
(c+id) f(t)= (c+id)\big(u(t) + iv(t)\big)= cu(t) -dv(t) +i \big[cv(t) +du(t)\big]\text{.}
\end{equation*}
\begin{equation*}
c\int_a^bu(t)\,dt-d\int_a^bv(t)\,dt + ic\int_a^bv(t) \, dt + id\int_a^bu(t)\,dt\text{.}
\end{equation*}
which is equivalent to
\begin{equation*}
(c+id) \left[ \int_a^bu(t)\,dt + i\int_a^bv(t)\,dt\right]
\end{equation*}
It is worthwhile to point out the similarity between Equation
(6.1.2) and its counterpart in calculus. Suppose that
\(U\) and
\(V\) are differentiable on
\(a\lt t\lt b\) and
\(F(t)=U(t)+iV(t)\text{.}\) Since
\(F\,'(t)=U\,'(t) +iV\,'(t) =u(t) +iv(t)=f(t)\text{,}\) Equation
(6.1.2) takes on the familiar form
\begin{equation}
\int_a^bf(t)\,dt = F(t) \Big|_{t=a}^{t=b}=F(b) -F(a)\text{.}\tag{6.1.8}
\end{equation}
where
\(F\,'(t)=f(t)\text{.}\) We can view Equation
(6.1.8) as an extension of the fundamental theorem of calculus. In
Section 6.5 we show how to generalize this extension to analytic functions of a complex variable. For now, we simply note an important case of Equation
(6.1.8):
\begin{equation}
\int_a^bf\,'(t)\,dt=f(b)-f(a)\text{.}\tag{6.1.9}
\end{equation}
Example 6.1.5.
\begin{equation*}
\int_0^{\frac{\pi}{2}}\exp(t+it)\,dt=\frac{1}{2} (e^{\frac{\pi}{2}}-1) + \frac{i}{2}(e^{\frac{\pi}{2}}+1)
\end{equation*}
Solution.
We seek a function \(F\) with the property that \(F\,'(t)= \exp(t+it)\text{.}\) We note that \(F(t)=\frac{1}{1+i} e^{t(1+i)}\) satisfies this requirement, so
\begin{align*}
\int_0^{\frac{\pi}{2}}\exp (t+it)\,dt \amp = \frac{ 1}{1+i}e^{t(1+i)}\Big|_{t=0}^{t=\frac{\pi}{2}}=\frac{1}{ 1+i}(ie^{\frac{\pi}{2}}-1)\\
\amp = \frac{1}{2}(1-i) (ie^{\frac{\pi}{2}}-1)\\
\amp \frac{1}{2}(e^{\frac{\pi}{2}}-1) +\frac{i}{2}(e^{\frac{ \pi}{2}}+1)\text{,}
\end{align*}
which is the same result we obtained in
Example 6.1.3, but with a lot less work!
Exercises Exercises
1.
(a)
\(\int_0^1(3t-i)^2dt\text{.}\)
Solution.
(b)
\(\int_0^1(t+2i)^3dt\text{.}\)
(c)
\(\int_0^{\frac{\pi}{2}}\cosh (it)\,dt\text{.}\)
Solution.
(d)
\(\int_0^2\frac{t}{t+i}dt\text{.}\)
(e)
\(\int_0^{\frac{\pi}{4}}t\exp (it)\,dt\text{.}\)
Solution.
\(\frac{\sqrt{2}\pi}{8} + \frac{\sqrt{2}}{2} - 1 + i(\frac{\sqrt{2}}{2} - \frac{\sqrt{2}\pi}{8})\text{.}\)
2.
Let \(m\) and \(n\) be integers. Show that
\begin{equation*}
\int_0^{2\pi}e^{imt}e^{-int}\,dt = \begin{cases}\; 0, \amp \text{ when } \; m \ne n;\\ 2\pi, \amp \text{ when } \; m = n. \end{cases}
\end{equation*}
3.
Show that \(\int_0^{\infty}e^{-zt}dt=\frac{1}{z}\) provided \(\mathrm{Re}(z) >0\text{.}\)
Solution.
Using Equation
(6.1.8) we get
\(\int_0^{\infty}e^{-zt}\,dt = \lim\limits_{T \to \infty}\int_0^{T}e^{-zt}\,dt = \lim\limits_{T \to \infty}(-\frac{1}{z}e^{-zT} + \frac{1}{z}e^{-z(0)}) = \frac{1}{z} + \lim\limits_{T \to \infty}(-\frac{1}{z}e^{-zT})\text{.}\) Show that
\(\mathrm{Re}(z)>0\) implies this last limit equals zero.
4.
Establish the following identities .
(a)
(b)
(c)
(d)
5.
Let \(f(t)=u(t) +iv(t)\text{,}\) where \(u\) and \(v\) are differentiable. Show that
\begin{equation}
\int_a^bf(t) f\,'(t)\,dt=\frac{1}{2}[f(b)]^2-\frac{1}{2}\left[ f(a) \right] ^2\tag{6.1.10}
\end{equation}
Solution.
This follows from Equation
(6.1.8), and the fact that if
\(u\) and
\(v\) are differentiable, then
\(f\) is differentiable, and
\(\frac{d}{dt}\left[\frac{1}{2}\big(f(t)\big)^2\right] = f(t)f\,'(t)\text{.}\)
6.
Use integration by parts to verify that \(i\int_0^{\frac{\pi}{2}}e^t\sin t)\,dt=\frac{i}{2}(e^{\frac{\pi}{2}}+1)\text{.}\) \label {6.1.6}
