Theorem 3.9 . Let \(D\) be any simply-connected domain of analyticity of \(f(z)\) . If \(C\) is any simple closed curve in \(D\) , \(\zeta\) any point of its interior then \[\tag{3.00} f(\zeta)=\frac{1}{2 \pi i} \int_{C} \frac{f(z)}{z-\zeta} \,d z\]
This is a very remarkable statement. It demonstrates the strong inter-relation among the values of a regular function. In order to know the values of the function in the interior of \(C\) we have only to specify the values on the boundary.
Proof. For the proof we employ any circle with \(\zeta\) at its center and interior in the interior of \(C\) . The function \(\frac{f(z)}{z-\zeta}\) is clearly analytic everywhere in the interior of \(C\) except at \(\zeta\) . By the proof of the Cauchy theorem for multiply-connected domains it follows that the integral about any such circle is independent of the radius and equal to the integral around \(C\) . We have \(f(z)=f(\zeta)+\eta(z, \zeta)\) where \(|\eta(z, \zeta)|<\varepsilon\) whenever \(|z-\zeta|<\delta\) . We choose a circle \(C_{\rho}\) of radius \(\rho<\delta\) . \begin{align} \int_{C} \frac{f(z)}{z-\zeta} \,d z & =\int_{C_{\rho}} \frac{f(z)}{z-\zeta} \,d z \\ & =f(\zeta) \int_{C_{\rho}} \frac{d z}{z-\zeta}+\int_{C_{\rho}} \frac{\eta \,d z}{z-\zeta} \\ & =f(\zeta) \int_{0}^{2 \pi} \frac{\rho i e^{i \theta}}{\rho e^{i\theta}} \,d \theta+\int_{C_{\rho}} \frac{\eta}{z-\zeta}\,dz \\ & =2 \pi i f(\zeta)+\int_{C_{\rho}} \frac{\eta}{z-\zeta} \,d z. \end{align} Since \(|z-\zeta|=\rho\) the absolute value of the integrand cannot exceed \(\varepsilon / \rho\) . By Lemma 3.1.4 we have \[\left|\int_{C_{\rho}} \frac{\eta}{z-\zeta} \,d z\right| \leq 2 \pi \varepsilon.\] But the integral does not depend on \(\rho\) . Hence \(\int_{C_{\rho}} \frac{\eta \,d z}{z-\zeta}=0\) and \[\frac{1}{2 \pi i} \int_{C} \frac{f(z)}{z-\zeta} \,d z=f(\zeta).\] ◻
Corollary 3.1 . If \(f(z)\) , \(g(z)\) are analytic in a domain containing \(C\) and \(f(z)=g(z)\) on \(C\) then \(f(z)=g(z)\) in the interior of \(C\) .
The representation of an analytic function by means of the Cauchy integral enables us to obtain an integral representation for its derivative. In fact we have in general:
Lemma 3.9 . Let \(C\) be any finite curve (not necessarily closed) and let \(\phi(z)\) be any continuous function on \(C\) . The function \[\tag{3.01} F(\zeta)=\int_{C} \frac{\phi(z)}{z-\zeta}\,dz\] is analytic in the entire complex plane except on \(C\) itself and has the derivative \[\tag{3.02} F'(\zeta)=\int_{C} \frac{\phi(z)}{(z-\zeta)^{2}} \,dz\]
Proof. Consider the difference quotient \[\frac{F(\zeta+\Delta \zeta)-F(\zeta)}{\Delta \zeta}=\int_{C} \frac{\phi(z)}{(z-\zeta)(z-\zeta-\Delta \zeta)} \,d z\] where \(\Delta \zeta\) is kept smaller than the distance between \(\zeta\) and the nearest point of \({C}\) . We have \begin{align} \frac{\Delta F}{\Delta \zeta}-\int_{C} \frac{\phi(z)}{(z-\zeta)^{2}}\,dz & =\int_{C} \phi(z)\left[\frac{1}{(z-\zeta)(z-\zeta-\Delta \zeta)}-\frac{1}{(z-\zeta)^{2}}\right]d z \\ & =\int_{C} \frac{\phi(z) \Delta \zeta}{(z-\zeta)^{2}(z-\zeta-\Delta \zeta)} \,d z. \end{align}
Since \(\phi(z)\) is continuous on \(C\) it is bounded \(|\phi(z)| \leq M\) . Let \(d\) be the minimum of the distances of \(\zeta\) and \(\zeta+\Delta \zeta\) from \(C\) . We then have \[\left|\frac{\Delta F}{\Delta \zeta}-\int_{C} \frac{\phi(z)}{(z-\zeta)^{2}}\,dz\right| \leq \frac{|\Delta \zeta| M L}{d^{3}}\] where \(L\) is the length of \(C\) . It follows that \(F(\zeta)\) has the derivative (3.02) . ◻
Since \(\phi(z)\) is any continuous function whatever we cannot hope that \(F(\zeta)\) will be analytic on \(C\) . However, it might be supposed that \(F(\zeta)\) tends to \(2 \pi i \phi(z)\) as \(\zeta\) approaches a point \(z\) on \(C\) . But this will not generally be true. For example, let \(C\) be the unit circle \(|z|=1\) and take \(\phi(z)=\frac{1}{z}\) . We have, for a point \(\zeta\) inside the circle,
\begin{align} F(\zeta) =\int_{C} \frac{d z}{z(z-\zeta)}&=\int \bigg(\frac{1}{\zeta(z-\zeta)}-\frac{1}{\zeta z} \bigg)d z \\ & =\frac{1}{\zeta} \int \frac{d z}{z-\zeta}-\frac{1}{\zeta} \int \frac{d z}{z} \\ & =2 \pi i\left[\frac{1}{\zeta}-\frac{1}{\zeta}\right]\\ &=0. \end{align} Hence the limiting value of \(F(\zeta)\) on the boundary is zero.
If \({C}\) is a closed path in a simply connected domain of analyticity of \(f(z)\) it follows by this lemma that \[\tag{3.03} f'(\zeta)=\frac{1}{2 \pi i} \int_{C} \frac{f(z)}{(z-\zeta)^{2}} \,d z,\] for all points \(\zeta\) in the interior of \(C\) . The function (3.03) is differentiable once again and, in fact, any number of times.
Theorem 3.10 . A complex function \(f(z)\) having a derivative of first order has derivatives of all orders. The \(n^\text{th}\) derivative is given by \[\tag{3.04} f^{(n)}(\zeta)=\frac{n !}{2 \pi i} \int_{C} \frac{f(z)}{(z-\zeta)^{n+1}} \,d z.\]
Proof. The proof goes by induction. The formula (3.04) certainly holds for \(n=1\) . If the theorem is true for \(n-1\) it is certainly true for \(n\) . For suppose \[f^{(n-1)}(\zeta)=\frac{(n-1) !}{2\pi i}\int_{C} \frac{f(z) \,d z}{(z-\zeta)^{n}}.\] We have \begin{align} & \left|\frac{\Delta f^{(n-1)}(\zeta)}{\Delta \zeta}-\frac{n !}{2 \pi i} \int_{C} \frac{f(z) \,d z}{(z-\zeta)^{n+1}}\right|\\ & \qquad =\frac{(n-1) !}{2 \pi}\left|\int_{C} \frac{f(z)}{\Delta \zeta}\left(\frac{(z-\zeta)^{n}-(z-\zeta-\Delta \zeta)^{n}}{(z-\zeta)^{n}(z-\zeta-\Delta \zeta)^{n}}-\frac{n}{(z-\zeta)^{n+1}}\right)d z\right| \\ & \qquad =\frac{(n-1)!}{2 \pi}\left|\int_{C} f(z)\left(\frac{\sum_{r=1}^{n}(z-\zeta)^{n-r+1}(z-\zeta-\Delta \zeta)^{r-1}-n(z-\zeta-\Delta \zeta)^{n}}{(z-\zeta)^{n+1}(z-\zeta-\Delta \zeta)^{n}}\right) d z\right| \\ & \qquad =\frac{(n-1) !}{2 \pi}\left|\int_{C} f(z)\left(\frac{\sum_{r=1}^{n}\left[\left(\frac{z-\zeta}{z-\zeta-\Delta \zeta}\right)^{n-r+1}-1\right]}{(z-\zeta)^{n+1}}\right) d z\right| \\ & \qquad <\frac{(n-1) !}{2 \pi} \frac{\varepsilon M L}{d^{n+1}} \end{align} where \(M\) , \(L\) and \(d\) are as before and \(\Delta \zeta\) is chosen so small that the absolute value of the sum is less than \(\varepsilon\) . ◻
We have proved an important result:
Theorem 3.11 . An analytic function \(f(z)\) has derivatives of all orders in the domain of analyticity of \(f(z)\) . Thus the derivatives of an analytic function are analytic.
In these proofs we have avoided the assumption that the first partial derivatives of the real and imaginary parts of \(f(z)=u+i v\) are continuous. The continuity of these derivatives – and even the existence of derivatives of higher order – is assured if only they exist and satisfy the Cauchy-Riemann equations.
The converse of the Cauchy Integral Theorem , the second half of Theorem 3.2.3 , is generally known as Morera’s Integral Theorem .
Theorem 3.12 (Morera’s Integral Theorem). If \(f(z)\) is continuous in a domain \(D\) and \[\int_{C} f(z) \,d z=0\] for every closed curve \(C\) in \(D\) then \(f(z)\) is analytic in \(D\) .
The hypothesis of the theorem is equivalent to the assertion that \(F(z)=\int_{z_{0}}^{z} f(z) \,d z\) takes on the same value on all possible paths joining \(z_{0}\) to \(z\) . If a function \(f(z)\) has this property it is said to be integrable . We have proved ( Theorem 3.2.1 ) that an integrable function is the derivative of an analytic function. We conclude that an integrable function must in turn be analytic.
We see now that we could have defined an analytic function in terms of integrability instead of differentiability, the one property implying the other in contrast with the more complicated situation in the theory of real functions.
Table of Contents
3.3.1 Consequences of the Cauchy Integral Formula
An immediate and interesting result is the Mean Value Theorem .
Theorem 3.13 (Mean Value Theorem). For any circle in the domain of analyticity of \(f(z)\) the function value at the center is equal to the average of its values on the circumference.
Proof. We have \begin{align} f(\zeta)&=\frac{1}{2 \pi i} \int_{C_{\rho}} \frac{f(z)}{z-\zeta} \,d z\\ &=\frac{1}{2 \pi i} \int_{0}^{2 \pi} \frac{i \rho e^{i \theta}}{\rho e^{i \theta}} f(z) \,d\theta. \end{align} Denoting the arclength by \(s\) we have \(d s=\rho \,d \theta\) whence \[\tag{3.10} f(\zeta)=\frac{1}{2 \pi \rho} \int_{C_{\rho}} f(z)\,ds.\] ◻
An important consequence of the formula (3.04) is the fact that the derivatives of \(f(z)\) are bounded in a very definite way in terms of the bound of the original function. There is of course no analogous statement for real functions.
Let \(R\) be a region of analyticity of \(f(z)\) bounded by a simple closed curve \(C\) . If \(\zeta\) is any point in the interior of \(C\) , \(\rho\) the nearest distance any point of \(C\) approaches \(\zeta\) , then we have by (3.04) \[|f^{(n)}(\zeta)| = \frac{n!}{2\pi} \left|\int_C \frac{f(z)}{(z - \zeta)^{n+1}}\,dz\right| \leq \frac{n!}{2\pi}\frac{ML}{\rho^{n+1}}.\] Note that this bound does not depend on \(\zeta\) but only on its distance from the boundary. More specially, let \(C\) be the circle of radius \(\rho\) about \(\zeta\) . We then have \[\tag{3.11} \left|f^{(n)}(\zeta)\right| \leq n ! \frac{M(\rho)}{\rho^{n}}\] where \(M(\rho)\) is now the maximum value of \(|f(z)|\) on \(C_{\rho}\) . In particular, for \(f(\zeta)\) itself we have \[\tag{3.12} |f(\zeta)| \leq M(p).\]
This last result could have been obtained as a direct consequence of the mean value theorem. From the special result (3.12) it is easy to prove the Maximum Modulus Theorem .
Theorem 3.14 (Maximum Modulus Theorem). If \(f(z)\) is analytic in a region \(R\) the maximum value of \(|f(z)|\) is attained on the boundary. 1 Further, if the maximum is attained at any interior point then \(f(z)= \textit{constant}\) .
Proof. Let \(\zeta_{0}\) be any interior point of \(R\) . We may then find a circle \(C_{\rho}\) about \(\zeta_{0}\) which lies completely in the interior of \(R\) . Now, in (3.12) , unless \(|f(z)|=M(\rho)\) everywhere on the circle, we must have the strict inequality \[\left|f(\zeta_{0})\right|
First, if \(\left|f(\zeta_{0})\right|=M\) then the above argument insures that \(|f(z)|=M\) on every circle in \(R\) about \(\zeta_{0}\) . In other words, \(|f(z)|=M\) in the largest circle that can be drawn about \(\zeta_{0}\) in \(R\) .
Now, let \(\zeta'\) be any point of \(R\) . We can find a path in the interior of \(R\) which connects \(\zeta_{0}\) to \(\zeta_{1}\) . Suppose that \(\left|f(\zeta_{1})\right|< M\) . By the continuity of \(|f(z)|\) there must be a first point \(\zeta'\) on the path for which \(\left|f(\zeta')\right|=M\) and immediately following, \(|f(z)|
Another consequence of the formula (3.11) is Liouville’s Theorem .
Theorem 3.15 (Liouville’s Theorem). If \(f(z)\) is analytic and bounded in the entire plane, \(|f(z)| \leq M\) , then \(f(z)\) is constant.
Proof. The proof consists in showing that \(f'(z) \equiv 0\) . At any point \(\zeta\) of the \(z\) -plane we have by (3.11)
\[\left|f'(\zeta)\right| \leq \frac{M}{\rho}\] where \(\rho\) is now the radius of any circle about \(\zeta\) . Since \(\rho\) can be chosen arbitrarily large it follows that \(f'(z) \equiv 0\) . This can be true only if \(f(z)\) is constant. ◻
Liouville’s Theorem figures importantly in many ways. Consider, for example, the following proof of The Fundamental Theorem of Algebra :
Theorem 3.16 (The Fundamental Theorem of Algebra). Every polynomial \[P(z)=a_{n} z^{n}+a_{n-1} z^{n-1}+\cdots+a_{0}, \quad a_{n} \neq 0\] of positive degree \({n}\) has a root in the complex domain.
Proof. For proof suppose on the contrary that \(P(z)\) has no root. It follows that the function \[Q(z)=\frac{1}{P(z)}\] is analytic in the entire plane. But \(Q(z)\) must be bounded since \({P}({z}) \rightarrow \infty\) as \({z} \rightarrow \infty\) . Hence, by Liouville’s Theorem \(Q(z)\) must be constant. But this contradicts the hypothesis that \(n>0\) . ◻
3.3.2 Residues, Evaluations of Integrals
Let \(f(z)\) be a regular function in a simply-connected domain \(D\) with the possible exception of a single point \(\zeta\) . Then if \(C\) is any simple closed curve containing \(\zeta\) we define the residue of \(f(z)\) at \(\zeta\) to be the integral \[\tag{3.21} \frac{1}{2 \pi i} \oint_{C} f(z) \,d z\] This integral is certainly independent of the curve containing \(\zeta\) .
More generally, if \({C}\) is contained in a domain of analyticity of \(f(z)\) and \(f(z)\) is regular everywhere in the interior of \(C\) except at a finite number of points \(z_{1}, z_{2}, \ldots, z_{n}\) , then \(\frac{1}{2 \pi i} \int_{C} f(z) \,d z\) is equal to the sum of the residues at these points. The proof is evident. The notion of residue is a very useful tool for the evaluation of complex integrals.
As an example, let us determine the residue of the function \(f(z)=z^{-n}\) , \(n\) an integer. Setting \(z=r e^{i \theta}\) and integrating around a circle of radius \(\rho\) about the origin we obtain \[\frac{1}{2 \pi i} \int_{C_{\rho}} \frac{d z}{z^{n}}=\frac{1}{2 \pi \rho^{n-1}} \int_{0}^{2 \pi} e^{-i(n-1) \theta} \,d \theta\] Since \(e^{2 \pi i}=e^{0}=1\) we conclude that the residue is zero if \(n>1\) and if \(n=1\) the residue is \(1\) .
From the fact that \(\frac{{d} \log z}{{dz}}=\frac{1}{{z}}\) , we have \[\tag{3.22} \log {z}=\int_{1}^{{z}} \frac{{d} \zeta}{\zeta}\] This characterization displays the multivalued character of the logarithms. If we join \(1\) to the point \(z\) by means of two non-intersecting curves which enclose the origin then the two values obtained by (3.22) will differ by \(2 \pi i\) . Clearly there are an infinite number of values of \(\log{z}\) , the difference of any two of them being a multiple of \(2 \pi i\) .
By means of residues we can evaluate a large class of infinite integrals in the real domain:
Theorem 3.17 . Suppose \(f(z)\) to be regular in the upper half plane \(\operatorname{Im}(z) \geq 0\) except for a finite number of non-real values \(z_{1}, z_{2}, \ldots, z_{n}\) . Then if \(f(z)\) tends to zero faster than \(\frac{1}{z}\) , \[\tag{3.23} \lim _{z \rightarrow \infty} z f(z)=0,\] we have \[\tag{3.24} \int_{-\infty}^{\infty} f(x) \,d x=2 \pi i \sum r_{\nu}\] where \(r_{\nu}\) is the residue of \(f(z)\) at \(z_{\nu}\) .
Proof. Take \({R}>0\) so large that the \(z_{i}\) are all contained in the circle \(|z|
Consider the integral around the semicircle in the upper half-plane: \begin{align} \int_{\text{\tiny{semicircle}}} f(z) \,d z&=\int_{-R}^{R} f(x) \,d x+\int_{0}^{\pi} i R e^{i \theta} f(R e^{i \theta}) \,d \theta\\ &=2 \pi i \sum r_{\nu}. \end{align} The integral on the curved part of the path can be made arbitrarily small by increasing \(R\) sufficiently. For we have \[\left|\int_{0}^{\pi} R e^{i \theta} f(R e^{i \theta}) \,d \theta\right| \leq \pi \max_{|z|=R}\big\{|z f(z)|\big\}\] and by (3.23) the right hand side must go to zero. We conclude that \[\lim _{R \rightarrow \infty} \int_{-R}^{R} f(x) \,d x=\int_{-\infty}^{+\infty} f(x) \,d x=2 \pi i \sum r_{\nu}.\] ◻
This result may easily be generalized to the case where there is a denumerable infinity of points \(z_{\nu}\) for which \(f(z)\) is not regular, provided only that the \(z_{\nu}\) have no point of accumulation. The summation is taken in the order of increasing \(\left|z_{\nu}\right|\) .
Examples
The function \[f(z)=\frac{1}{a z^{2}+b z+c}\] with \(a>0\) , \(a\) , \(b\) , \(c\) real and \(b^{2}-4 a c<0\) is regular for \(\operatorname{Im}(z) \geq 0\) with the exception of the point \(z_{1}=\frac{1}{2a}(-b+i \sqrt{4 a c-b^{2}})\) , where the sign in front of the radical is to be the same as that of \(a\) . Clearly, \(f(z)\) satisfies the condition (3.23) . Hence by the preceding theorem \(\int_{-\infty}^{+\infty} \frac{1}{a x^{2}+b x+c} \,d x\) is equal to \(2 \pi i\) times the residue \(r_{1}\) of \(f(z)\) at \(z_{1}\) . But, setting \(z=z_{1}+\rho e^{i \phi}\) , we have \begin{align} r_{1} & =\frac{1}{2 \pi i} \int_{C_{\rho}} f(z) \,d z\\ &=\lim_{\rho \rightarrow 0} \frac{1}{2 \pi i} \int_{C_{\rho}} \frac{d z}{a(z-z_{1})(z-\bar{z}_{1})} \\ & =\lim_{\rho \rightarrow 0} \frac{1}{2 \pi i} \int_{0}^{2 \pi} \frac{i \rho e^{i\phi} \,d \phi}{a \rho e^{i\phi}(\rho e^{i\phi}+z_{1}-\bar{z}_{1})} \\ & =\frac{1}{a(z_{1}-\bar{z}_{1})}\\ &=\frac{1}{i \sqrt{4 a c-b^{2}}}. \end{align} We conclude that \[\int_{-\infty}^{\infty} f(x) \,d x=\frac{2 \pi}{\sqrt{4 a c-b^{2}}}.\]
The method used in the above theorem may be adapted to other improper integrals. Take, for example, \[\int_{-\infty}^{\infty} \frac{\sin z}{z} \,d z\] To evaluate this integral we consider \[\int \frac{e^{i z}}{z} \,d z.\] We choose a semicircular domain of integration as before but we cut out a small piece of radius \(\varepsilon\) about the singular point at the origin.
Denote the integral over the large semicircle \(S_{R}\) by \(-I_{R}\) , over the small semicircle \({S}_{\varepsilon}\) by \({I}_{\varepsilon}\) , and the part of the integral over the real axis exclusive of the interval \(-\varepsilon<{x}<\varepsilon\) by \({I}_{{H}}\) . Then, by Cauchy’s Integral Theorem , \[I_{H}=I_{R}-I_{\varepsilon}.\] It is easily verified that \[I_{H}=\int_{-R}^{-\varepsilon}+\int_{\varepsilon}^{R} \frac{e^{i x}}{x} \,d x=2 i \int_{\varepsilon}^{R} \frac{\sin x}{x} \,d x.\] We have only to evaluate the integrals over \(S_{\varepsilon}\) and \(S_{R}\) as \(\varepsilon \rightarrow 0\) and \({R} \rightarrow \infty\) . \[I_{R}=\int_{S_{R}} \frac{e^{i z}}{z} \,d z=-i \int_{0}^{\pi} e^{-R \sin \theta} e^{i R \cos \theta} \,d \theta.\] Hence \[\left|I_{R}\right| \leq \int_{0}^{\pi} e^{-R \sin \theta} \,d \theta=2 \int_{0}^{\pi / 2} e^{-R \sin \theta} \,d \theta.\] By choosing \(R\) sufficiently large we can make \(\left|I_{R}\right|\) as small as we please. We consider the integral in two parts \[\left|I_{R}\right| \leq 2\left(\int_{0}^{\delta}+\int_{\delta}^{\pi / 2} e^{-R \sin \theta} \,d \theta\right).\] We have \(\left|e^{-R \sin \theta}\right| \leq 1\) for \(\theta \leq \delta\) and by taking \(R\) sufficiently large we may be sure that \(\left|e^{-R \sin \theta}\right|<\delta\) in the interval \(\delta \leq \theta \leq \frac{\pi}{2}\) . Thus we have the estimate \[\left|I_{R}\right|<2 \delta+\frac{\pi}{2} \delta=\frac{4+\pi}{2} \delta.\]
On the other hand we have \[I_{\varepsilon}=\int_{S_{\varepsilon}} \frac{e^{i z}}{z} \,d z=-i \int_{0}^{\pi} e^{-\varepsilon \sin \theta} e^{i \varepsilon \cos \theta} \,d \theta.\] Since the integrand is uniformly continuous in \(\varepsilon\) and \(\theta\) we may take the limit as \(\varepsilon \rightarrow 0\) inside the integral sign. We conclude that \[\lim_{\varepsilon \rightarrow 0} I_{\varepsilon}=-\pi i.\] Hence we obtain \[\int_{-\infty}^{+\infty} \frac{\sin x}{x} \,d x=-i \lim_{\substack{R \rightarrow \infty \\ \varepsilon \rightarrow 0}} I_{H}=\pi.\]
As a second example we propose to evaluate the Fresnel integrals \(\int_{0}^{\infty} \cos x^{2} \,d x\) and \(\int_{0}^{\infty} \sin x^{2} \,d x\) . To this end we introduce the function \({e}^{-{z}^{2}}\) and integrate around a circular sector of radius \(R\) and central angle \(\frac{\pi}{4}\) .
Setting \(z=r e^{i \theta}\) we have \begin{align} \int_{0}^{R} e^{-x^{2}} \,d x+\int_{0}^{\pi / 4} e^{-R^{2}} e^{2 i \theta} R i e^{i \theta} \,d \theta+\int_{R}^{0} e^{-r^{2} e^{i\pi/2}} e^{i \pi/4} \,d r=0. \end{align} We shall now show that the second integral tends to zero as \({R} \rightarrow \infty\) . We have \begin{align} \left|I_{2}\right|&=\left|R i \int_{0}^{\pi / 4} e^{-R^{2}} e^{2 i \theta} e^{i \theta} \,d \theta\right|\\ &\leq R \int_{0}^{\pi / 4}\left|e^{-R^{2} \cos 2 \theta}\right| \,d \theta \\ &= \frac{R}{2} \int_{0}^{\pi / 2} e^{-R^{2} \sin \phi} \,d \phi. \end{align} Now \(\frac{d^{2} \sin \phi}{d \phi^{2}}=-\sin \phi\) , whence it follows that the sine function is convex in the interval \(0 \leq \phi \leq \pi / 2\) and therefore lies above the straight line between the endpoints; thus \[\sin \phi \geq \frac{2}{\pi} \phi.\] Thus \[\left|I_{2}\right| \leq \frac{R}{2} \int_{0}^{\pi / 2} e^{-\frac{2}{\pi} R^{2} \phi} \,d \phi=\frac{\pi}{4 R}\left(1-e^{-R^{2}}\right)\] and this clearly goes to zero as \({R}\) tends to infinity. Consequently we have \[\int_0^{\infty} e^{-x^{2}} \,d x-\int_0^{\infty} e^{-i r^{2}} \frac{\sqrt{2}}{2}(1+i) \,d r=0.\] Now \(\int_{0}^{\infty} e^{-x^{2}} \,d x\) is a well known integral. Its value is \(\sqrt{\pi} / 2\) . 2 Thus we have \[\int_{0}^{\infty} \cos r^{2} \,d r-i \int_{0}^{\infty} \sin r^{2} \,d r=\frac{\sqrt{\pi}}{2} \frac{(1-i)}{\sqrt{2}},\] whence \[\int_{0}^{\infty} \cos x^{2} \,d x=\int_{0}^{\infty} \sin x^{2} \,d x=\frac{1}{2} \sqrt{\frac{\pi}{2}}.\]
As a last example we propose to evaluate the following integral which plays a role in the study of the \(\Gamma\) -function: \[\int_{0}^{\infty} \frac{x^{\alpha-1}}{1+x} \,d x \quad (0<\alpha<1).\] To this end we consider the closed path \(C\) as in the adjoining figure and the integral \[\int_{C} \frac{z^{\alpha-1}}{1+z} \,d z=2 \pi i r_{1}\] where \(r_{1}\) is the residue of \(\frac{z^{\alpha-1}}{1+z}\) at \({z}=-1\) .
Clearly \(r_{1}=(-1)^{\alpha-1}=e^{i \pi(\alpha-1)}\) . Thus \begin{align} \int_{C} \frac{z^{\alpha-1}}{1+z} \,d z &=\int_{\rho}^{R} \frac{x^{\alpha-1}}{1+x} \,d x\\ & \quad +i \int_{0}^{2 \pi} \frac{R^{\alpha} e^{i \alpha \theta}}{1+R e^{i \theta}} \,d \theta \\ & \quad +\int_{R}^{\rho} \frac{x^{(\alpha-1)} e^{2 \pi i(\alpha-1)}}{1+x} \,d x\\ & \quad -i \int_{0}^{2 \pi} \frac{\rho^{\alpha} e^{i \alpha \theta}}{1+\rho e^{i \theta}} \,d \theta\\ &=2 \pi i e^{i \pi(\alpha-1)}, \end{align} or \begin{align} \left[1-e^{2 \pi i(\alpha-1)}\right] \int_{\rho}^{R} \frac{x^{(\alpha-1)}}{1+x} \,d x&=2 \pi i e^{\pi i(\alpha-1)}\\ & \quad -i \int_{0}^{2 \pi} \frac{R^{\alpha} e^{i\alpha \theta}}{1+R e^{i\theta}} \,d \theta \\ & \quad +i \int_{0}^{2 \pi} \frac{\rho^{\alpha} e^{i \alpha \theta}}{1+\rho e^{i \theta}} \,d \theta \end{align} But as \(R \rightarrow \infty\) , \(\int_{0}^{2 \pi} \frac{R^{\alpha} e^{i \alpha \theta}}{1+R e^{i \theta}} \,d \theta \rightarrow 0\) for \[\left|\int_{0}^{2 \pi} \frac{R^{\alpha} e^{i \alpha \theta}}{1+R e^{i\theta}} \,d \theta\right| \leq \frac{2 \pi R^{\alpha}}{R-1}.\] Similarly as \(\rho \to 0\) , \(\int_0^{2\pi} \frac{\rho^\alpha e^{i\alpha\theta}}{1 + \rho e^{i\theta}}\,d\theta \to 0\) for we have a singular inequality \[\left|\int_0^{2\pi} \frac{\rho^\alpha e^{i\alpha\theta}}{1 + \rho e^{i\theta}}\,d\theta\right| \leq \frac{2\pi\rho^\alpha}{1 - \rho}.\] Hence \begin{align} \int_{0}^{\infty} \frac{x^{\alpha-1}}{1+x} \,d x & =\lim _{\substack{R \rightarrow \infty \\ \rho \rightarrow 0}} \int_{\rho}^{R} \frac{x^{(\alpha-1)}}{1+x} \,d x\\ & =\frac{2 \pi i e^{i \pi(\alpha-1)}}{1-e^{2 \pi i(\alpha-1)}}\\ &=\frac{2 \pi i}{e^{-i \pi(\alpha-1)}-e^{i \pi(\alpha-1)}}\\ & =\frac{\pi}{\sin (\pi-\alpha \pi)}\\ &=\frac{\pi}{\sin \alpha \pi}. \end{align}
Exercises
Exercise 3.1 . Evaluate the following integrals:
- \(\displaystyle \int \cos z \,d z\) over the ellipse \(\frac{x^{2}}{2\pi^{2}}+\frac{y^{2}}{1}=1\) .
- \(\displaystyle \int \tan z \,d z\) over the circle \(x^{2}+y^{2}=9\) .
- \(\displaystyle \int \operatorname{cosec} z \,d z\) over the circle \(x^{2}+y^{2}=9\) .
Exercise 3.2 . Prove by using Cauchy’s Integral Formula that if \(f(z)\) is regular in a region \(R\) bounded by a simple closed curve \(C\) , and if \(f(z)\) is real on \(C\) , then \(f(z)= \textit{const.}\) in \(R\) and on \(C\) .
Exercise 3.3 . Evaluate the integral \(\displaystyle \frac{1}{2 \pi i}\int_{C} \frac{e^{s z}}{z}\,d z\) where \(s\) is real for each of the following curves:
- If \({C}\) is a closed path circling the origin.
- If \(C\) is a straight line parallel to the \(y\) -axis: \({x}={x}_{0}>0\) , and \({s}>0\) .
- If \(C\) is \(x=x_{0}>0\) and \(s<0\) .
Exercise 3.4 . Show that \(\displaystyle \int_{0}^{\infty} e^{-\alpha x^{2}} \cos\left(\sqrt{1-\alpha^{2}} x^{2}\right) d x=\frac{1}{2}\sqrt{\frac{\pi(1+\alpha)}{2}}\) .
Exercise 3.5 . By integrating \(\displaystyle \int_{C} e^{-z^{2}} \,d z\) over the following rectangle evaluate \(\int_{0}^{\infty} e^{-x^{2}} \cos 2ax \,d x\) .
Exercise 3.6 . Evaluate
- \(\displaystyle \int_{0}^{\infty} \frac{\cos x}{a^{2}+x^{2}} \,d x\) ;
- \(\displaystyle\int_{0}^{\infty} \frac{\sin ^{2} x}{x^{2}} \,d x\) .
Exercise 3.7 . Show that \(\displaystyle\frac{1-R^{2}}{2 \pi}\int_{0}^{2 \pi} \frac{d \theta}{1-2 R\cos \theta+R^{2}}=1\) , \((0 \leq R<1)\) , by considering the integral \(\displaystyle\int_{C} \frac{d z}{(z-R)(Rz-1)}\) over the unit circle.
Exercise 3.8 . Prove that if \(Q(z)\) is a polynomial of degree \(n\) having simple roots \(z_{1}, \ldots, z_{n}\) then for any function \(P(z)\) regular at these points the residue \(\dfrac{P(z)}{Q(z)}\) at \(z_i\) is given by \(\dfrac{P(z_{i})}{Q'(z_{i})}\) .
Exercise 3.9 . Let \(Q(z)\) be a polynomial of degree \(2 n\) with real coefficients and having simple non-real roots \(z_{i}\) and let \(P(z)\) be a polynomial of degree \(m\leq 2 n-2\) . Evaluate \(\displaystyle \int_{-\infty}^{\infty} \frac{P(x)}{Q(x)} \,d x\) .
Exercise 3.10 . Evaluate
- \(\displaystyle\int_{0}^{\infty} \frac{d x}{1+x^{4}}\) ;
- \(\displaystyle \int_{-\infty}^{\infty} \frac{d x}{\left(1+x^{2}\right)^{n}}\) .
Exercise 3.11 . Evaluate \(\displaystyle \int_{0}^{\infty} \frac{x^{\alpha-1} \log x}{1+x} \,d x\) ; \((0<\alpha<1)\) .