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Chapter 4: Problem 7

Suppose we know that everywhere outside the circle \(C_{R}\), radius \(R\) centered at the origin, \(f(z)\) and \(g(z)\) are analytic with \(\lim _{z \rightarrow \infty} f(z)=C_{1}\) and \(\lim _{z \rightarrow \infty}(z g(z))=C_{2}\), where \(C_{1}\) and \(C_{2}\) are constant. Show $$ \frac{1}{2 \pi i} \oint_{C_{R}} g(z) e^{f(z)} d z=C_{2} e^{C_{1}} $$

Short Answer

Expert verified
Using the limits and properties of analytic functions, the integral \( \frac{1}{2\pi i} \oint_{C_R} g(z) e^{f(z)} \, dz \) simplifies to \( C_2 e^{C_1} \).

Step by step solution

01

State the given conditions and goal

We know that outside the circle \(C_R\) with radius \(R\) centered at the origin, the functions \(f(z)\) and \(g(z)\) are analytic. We are given the limits \( \lim_{z \to \infty} f(z) = C_1 \) and \( \lim_{z \to \infty} (zg(z)) = C_2 \). We need to show that \( \frac{1}{2\pi i} \oint_{C_R} g(z) e^{f(z)} \, dz = C_2 e^{C_1} \).
02

Behavior of \(f(z)\) and \(g(z)\) at infinity

Since \( \lim_{z \to \infty} f(z) = C_1 \), we know that for very large \(|z|\), \(f(z)\) approaches \(C_1\). Therefore, we can approximate \(f(z) \approx C_1\) for large \(|z|\). Similarly, since \( \lim_{z \to \infty} (zg(z)) = C_2 \), for very large \(|z|\), \(g(z)\) approaches \( \frac{C_2}{z} \).
03

Integral over the circle \(C_R\)

Recall that the integral we need to evaluate is \( \frac{1}{2\pi i} \oint_{C_R} g(z) e^{f(z)} \, dz \). Because both \(f(z)\) and \(g(z)\) are analytic outside \(C_R\), and by the limits given as \(z \to \infty\), we can substitute the approximations derived.
04

Substitution in the integral

Substitute \(g(z) \approx \frac{C_2}{z} \) and \( e^{f(z)} \approx e^{C_1} \) into the integral: \[ \oint_{C_R} g(z) e^{f(z)} \, dz \approx \oint_{C_R} \frac{C_2}{z} e^{C_1} \, dz \]. This simplifies to: \[ C_2 e^{C_1} \oint_{C_R} \frac{1}{z} \, dz \].
05

Evaluate the simplified integral

The integral \( \oint_{C_R} \frac{1}{z} \, dz \) around a circle centered at the origin is known to equal \(2\pi i\) because it's the integral of the function \( \frac{1}{z} \) which has a simple pole at the origin with residue 1. Thus, we have: \[ C_2 e^{C_1} \oint_{C_R} \frac{1}{z} \, dz = C_2 e^{C_1} \cdot 2\pi i \].
06

Final step

Substitute the result from the integral back into our original expression: \[ \frac{1}{2\pi i} \oint_{C_R} g(z) e^{f(z)} \, dz = \frac{1}{2\pi i} \left( C_2 e^{C_1} \cdot 2\pi i \right) = C_2 e^{C_1} \]. Therefore, the expression is proved as required.

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Key Concepts

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

Cauchy's Integral Theorem
Cauchy's Integral Theorem is a fundamental result in complex analysis. It states that if a function is analytic and the region it is defined on is simply connected, then the integral of that function over any closed curve within that region is zero. This theorem is vital because it helps simplify complex integrals by showing that the value of these integrals depends only on the properties of the function inside the curve, not on the specific path taken.
Analytic Functions
Analytic functions, also known as holomorphic functions, are those complex functions that are differentiable at every point within some region of the complex plane. Differentiability here means that the derivative exists and is also a complex function. What makes these functions special is they can be expanded into a power series around any point in their domain. This property makes them very predictable and easy to work with. In our exercise, both \( f(z) \) and \( g(z) \) are analytic outside the circle \( C_R \), allowing us to use various integral theorems to evaluate expressions involving them.
  • Key Properties: They are infinitely differentiable.
  • Power Series: They can be expressed as a convergent power series.
  • Implications: If they are analytic in a region, many powerful theorems (like Cauchy's Integral Theorem) can be applied.
Residue Theorem
The Residue Theorem is another powerful tool in complex analysis. It allows us to evaluate complex integrals by summing up residues (coefficients of the \( \frac{1}{z} \) term in the Laurent series expansion) at the poles within the integration path. Specifically, if \( f(z) \) has isolated singularities at \( z = z_1, z_2, ..., z_n \) inside a closed contour \( C \), then the integral of \( f(z) \) around \( C \) is \( 2 \pi i \) times the sum of residues at these singularities. In our exercise, when we simplify the integral to \( C_2 e^{C_1} \oint_{C_R} \frac{1}{z} dz \), we apply this theorem to find that the result is \( 2 \pi i \).
  • Application: Simplifies evaluation of integrals around singularities.
  • Steps: Identify residues and apply \( 2\pi i \) factor.
Asymptotic Behavior
Asymptotic behavior refers to the properties and tendencies of functions as the input grows very large. For complex functions, understanding asymptotic behavior helps in approximating the function when \( |z| \) becomes very large. In our problem, given \( \lim_{z \rightarrow \infty} f(z) = C_1 \) and \( \lim_{z \rightarrow \infty} (zg(z)) = C_2 \), the asymptotic behavior allows us to approximate \( f(z) \approx C_1 \) and \( g(z) \approx \frac{C_2}{z} \) for large \( |z| \). These approximations are critical in simplifying the integral and eventually proving the required result. Understanding the asymptotic behavior helps in predicting and simplifying complex functions, which is essential for solving integrals and other problems in complex analysis.
  • Usage: Simplifies functions for large inputs.
  • Application: Used in approximations and limit evaluation.

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Most popular questions from this chapter

Use multiple Fourier transforms to solve $$ \frac{\partial \phi}{\partial t}-\left(\frac{\partial^{2} \phi}{\partial x^{2}}+\frac{\partial^{2} \phi}{\partial y^{2}}\right)=0 $$ on the infinite domain \(-\infty0\), with \(\phi(x, y) \rightarrow 0\) as \(x^{2}+y^{2} \rightarrow \infty\), and \(\phi(x, y, 0)=f(x, y) .\) How does the solution simplify if \(f(x, y)\) is a function of \(x^{2}+y^{2} ?\) What is the Green's function in this case?

Show explicitly that the Laplace transform of the second derivative of a function of \(x\) satisfies $$ \int_{0}^{\infty} f^{\prime \prime}(x) e^{-s x} d x=s^{2} \hat{F}(s)-s f(0)-f^{\prime}(0) $$

Obtain the inverse Fourier transform of the following functions: (a) \(\frac{1}{k^{2}+w^{2}}, \quad w^{2}>0\) (b) \(\frac{1}{\left(k^{2}+w^{2}\right)^{2}}, \quad w^{2}>0\)

(a) Use principal value integrals to show that $$ \int_{0}^{\infty} \frac{\cos k x-\cos m x}{x^{2}} d x=\frac{-\pi}{2}(|k|-|m|), \quad k, m \text { real. } $$ Hint: note that the function \(f(z)=\left(e^{i k z}-e^{i m z}\right) / z^{2}\) has a simple pole at the origin. (b) Let \(k=2, m=0\) to deduce that $$ \int_{0}^{\infty} \frac{\sin ^{2} x}{x^{2}} d x=\frac{\pi}{2} $$

Given the linear "free" Schr枚dinger equation (without a potential) $$ i \frac{\partial u}{\partial t}+\frac{\partial^{2} u}{\partial x^{2}}=0, \quad \text { with } u(x, 0)=f(x) $$ (a) solve this problem by Fourier transforms, by obtaining the Green's function in closed form, and using superposition. Recall that \(\int_{-\infty}^{\infty} e^{i u^{2}} d u=e^{i \pi / 4} \sqrt{\pi}\) (b) Obtain the above solution by Laplace transforms.
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