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Chapter 4: Problem 11

Consider the integral $$ I_{R}=\frac{1}{2 \pi i} \oint_{C_{R}} \frac{d z}{\left(z^{2}-a^{2}\right)^{1 / 2}}, \quad a>0 $$ where \(C_{R}\) is a circle of radius R centered at the origin enclosing the points \(z=\pm a\). Take the principal value of the square root. (a) Evaluate the residue of the integrand at infinity and show that \(I_{R}=1\). (b) Evaluate the integral by defining the contour around the branch points and along the branch cuts between \(z=-a\) to \(z=a\), to find (see Section \(2.3\) ) that \(I_{R}=\frac{2}{\pi} \int_{0}^{a} \frac{d x}{\sqrt{a^{2}-x^{2}}}\). Use the well-known indefinite integral \(\int \frac{d x}{\sqrt{a^{2}-x^{2}}}=\sin ^{-1} x / a+\) const to obtain the same result as in part (a).

Short Answer

Expert verified
The residue at infinity shows \(I_R=1\). Evaluating the integral contour around the branch points confirms \(I_R=1\).

Step by step solution

01

Understanding the Problem

We begin with evaluating the integral \[ I_R = \frac{1}{2 \pi i} \oint_{C_{R}} \frac{d z}{(z^2 - a^2)^{1/2}}, \quad a>0 \]. where \(C_R\) is a circular path centered at the origin with radius \(R\). We are to compute the residue of the integrand at infinity and evaluate this integral.
02

Assessing the Integrand

The integrand is \frac{1}{(z^2 - a^2)^{1/2}}\ with branch points at \(z=\pm a\). The principal value of the square root function is assumed.
03

Finding Residue at Infinity

A useful approach is to consider the behavior at large \(|z|\). Transform the variable \(w = \frac{1}{z}\), such that \[ dz = -\frac{1}{w^2} dw, \quad z^2 = \frac{1}{w^2} \]. Substitute in the integral and find the behavior of the residue at infinity.
04

Contour Integral Simplification

On transforming to the variable \(w\), the integral changes form and we consider using the residue theorem for circular paths. Find the contribution of residues of the integral. This confirms that \(I_R=1\).
05

Branch Cut Considerations

Consider a contour integral around the branch cuts between the points \(z=-a\) and \(z=a\). Evaluate this by integrating from 0 to a.
06

Evaluate the New Integral

Express the integral in terms of the function \int\frac{dx}{\textrm{sqrt}(a^2 - x^2)}\, where the given integral \[ I_R = \frac{2}{\pi} \int_0^a \frac{dx}{\textrm{sqrt}(a^2 - x^2)} \]. Use \sin^{-1}(\frac{x}{a})\ to compute the value.
07

Final Expression

We conclude the integral calculation: the result \[ \frac{2}{\pi} \times \sin^{-1}(1) = 1 \] confirms that both parts (a) and (b) show that \ I_R = 1\.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Complex Analysis
Complex analysis is an area of mathematics that studies functions of complex numbers. These functions have both real and imaginary parts. In the context of contour integration, complex analysis provides various tools to evaluate integrals around closed curves in the complex plane.
One main tool is the Cauchy-Goursat theorem, which states that the integral of a function around a closed contour is zero if the function is analytic inside the contour. This simplifies evaluating complex integrals.
In problems like this, we often focus on the behavior of functions near specific points called singularities. These are points where the function is not defined or not analytic. Understanding how to approach these is crucial for solving complex integrals.
Residue Theorem
The residue theorem is a powerful tool in complex analysis, particularly for evaluating contour integrals. It relates the integral of a function around a closed contour to the sum of residues of the function's singularities enclosed by the contour.
The residue of a function at a given point is a coefficient that results from expanding the function into a Laurent series鈥攁 series that includes terms with negative powers. To find the residue, we often need to isolate the term with a coefficient of \frac{1}{z}\footnote{z}.
This problem involves computing the residue at infinity. By transforming the variable using \(w=\frac{1}{z}\), we see that the behavior of the integral at infinity can be derived. This structure allows us to confirm that the integral equals 1.
Branch Cuts
Branch cuts are lines or curves introduced in the complex plane to make multi-valued functions single-valued. For example, the complex square root function \((z^2 - a^2)^{1/2}\) has branch points at \(z = +/- a\).
To evaluate the integral, we must consider the branch cuts between these points. By properly choosing a branch cut, we ensure our function remains continuous and single-valued, which simplifies our calculations.
In this exercise, integrating around these branch cuts and considering the contributions from both sides allows us to resolve the integral. This step confirms the integral value obtained using residues.
Indefinite Integrals
Indefinite integrals, also known as antiderivatives, are integrals that represent a general form of the original function without specific boundaries. In this problem, the integral \(\int \frac{dx}{\textrm{sqrt}(a^2 - x^2)}\footnote{x} = \sin^{-1}(\frac{x}{a})\) is applied.
We use this antiderivative to evaluate the definite integral from 0 to a. This gives us the connection between the complex integral we started with and more familiar integrals over real numbers.
Evaluating this antiderivative provides a form that shows \(I_R=1\), consistent with the result obtained through the residue theorem.
Principal Value
The principal value of an integral is a method for assigning values to certain improper integrals which do not converge in the usual sense. By taking symmetric limits around singular points, we can define a meaningful result.
In this exercise, the principal value allows us to sensibly handle the square root function, ensuring the single-valued nature of the integrand. This is crucial for contour integrals because multi-valued functions can complicate the calculation.
In summary, the principal value provides a rigorous way to deal with otherwise problematic integrals, ensuring consistent and correct outcomes.

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Most popular questions from this chapter

Use Laplace transform methods to solve the ODE $$ L \frac{d y}{d t}+R y=f(t), \quad y(0)=y_{0}, \quad \text { constants } L, R>0 $$ (a) Let \(f(t)=\sin \omega_{0} t, \omega_{0}>0\), so that the Laplace transform of \(f(t)\) is \(\hat{F}(s)=\omega_{0} /\left(s^{2}+\omega_{0}^{2}\right) .\) Find $$ \begin{aligned} y(t)=& y_{0} e^{-\frac{\varepsilon}{L} t}+\frac{\omega_{0}}{L\left((R / L)^{2}+\omega_{0}^{2}\right)} e^{-\frac{k}{L} t}+\frac{R}{L^{2}} \frac{\sin \omega_{0} t}{\left((R / L)^{2}+\omega_{0}^{2}\right)} \\ &-\frac{\omega_{0}}{L} \frac{\cos \omega_{0} t}{\left((R / L)^{2}+\omega_{0}^{2}\right)} \end{aligned} $$ (b) Suppose \(f(t)\) is an arbitrary continuous function that possesses a Laplace transform. Use the convolution product for Laplace transforms (Section \(4.5\) ) to find $$ y(t)=y_{0} e^{-\frac{k}{L} t}+\frac{1}{L} \int_{0}^{t} f\left(t^{\prime}\right) e^{-\frac{k}{L}\left(t-t^{\prime}\right)} d t^{\prime} $$ (c) Let \(f(t)=\sin \omega_{0} t\) in (b) to obtain the result of part (a), and thereby verify your answer. This is an example of an "L,R circuit" with impressed voltage \(f(t)\) arising in basic electric circuit theory.

Establish the following results by formally inverting the Laplace transform. $$ \begin{aligned} &\hat{F}(s)=\frac{1}{s} \frac{1-e^{-\theta s}}{1+e^{-b^{-s}}}, \quad \ell>0, \\\ &f(x)=\sum_{n=1,3,5, \ldots}\left(\frac{4}{n \pi}\right) \sin \frac{n \pi x}{\ell} \end{aligned} $$ Note that there are an infinite number of poles present in \(\hat{F}(s)\); consequently, a straightforward continuous limit as \(R \rightarrow \infty\) on a large semicircle will pass through one of these poles. Consider a large semicircle \(C_{R_{N}}\), where \(R_{N}\) encloses \(N\) poles (e.g. \(\left.R_{N}=\frac{\pi i}{\ell}\left(N+\frac{1}{2}\right)\right)\) and show that as \(N \rightarrow \infty, R_{N} \rightarrow \infty\), and the integral along \(C_{R_{N}}\) will vanish. Choosing appropriate sequences such as in this example, the inverse Laplace transform containing an infinite number of poles can be calculated.

Let \(f(z)\) be analytic outside a circle \(C_{R}\) enclosing the origin. (a) Show that $$ \frac{1}{2 \pi i} \oint_{C_{R}} f(z) d z=\frac{1}{2 \pi i} \oint_{C_{p}} f\left(\frac{1}{t}\right) \frac{d t}{t^{2}} $$ where \(C_{\rho}\) is a circle of radius \(1 / R\) enclosing the origin. For \(R \rightarrow \infty\) conclude that the integral can be computed to be Res \(\left(f(1 / t) / t^{2} ; 0\right)\). (b) Suppose \(f(z)\) has the convergent Laurent expansion $$ f(z)=\sum_{j=-\infty}^{-1} A_{j} z^{j} $$ Show that the integral above equals \(A_{-1 .}\) (See also Eq. (4.1.11).)

(a) Assume that \(u(\infty)=0\) to establish that $$ \int_{0}^{\infty} \frac{d^{2} u}{d x^{2}} \cos k x d x=-\frac{d u}{d x}(0)-k^{2} \hat{U}_{c}(k) $$ where \(\hat{U}_{c}(k)=\int_{0}^{\infty} u(x) \cos k x d x\) is the cosine transform of \(u(x)\). (b) Use this result to show that taking the Fourier cosine transform of $$ \frac{d^{2} u}{d x^{2}}-\omega^{2} u=-f(x), \quad \text { with } \frac{d u}{d x}(0)=u_{0}^{\prime}, \quad u(\infty)=0, \quad \omega>0 $$ yields, for the Fourier cosine transform of \(u(x)\) $$ \hat{U}_{c}(k)=\frac{\hat{F}_{c}(k)-u_{0}^{\prime}}{k^{2}+\omega^{2}} $$ where \(\hat{F}_{c}(k)\) is the Fourier cosine transform of \(f(x)\) (c) Use the analog of the convolution product of the Fourier cosine transform $$ \begin{aligned} &\frac{1}{2} \int_{0}^{\infty}(f(|x-\zeta|)+f(x+\zeta)) g(\zeta) d \zeta \\ &=\frac{2}{\pi} \int_{0}^{\infty} \cos k x \hat{F}_{c}(k) \hat{G}_{c}(k) d k \end{aligned} $$ to show that the solution of the differential equation is given by $$ u(x)=-\frac{u_{0}^{\prime}}{\omega} e^{-\omega x}+\frac{1}{2 \omega} \int_{0}^{\infty}\left(e^{-\omega|x-\zeta|}+e^{-\mathfrak{\omega}|x+\zeta|}\right) f(\zeta) d \zeta $$ (The convolution product for the cosine transform can be deduced from the usual convolution product (4.5.17) by assuming in the latter formula that \(f(x)\) and \(g(x)\) are even functions of \(x\).)

In this problem we obtain the Green's function of Laplace's equation in the upper half plane, \(-\infty
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