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Chapter 7: Problem 14

Show that \((a>0)\) (a) \(\int_{-\infty}^{\infty} \frac{\cos x}{x^{2}+a^{2}} d x=\frac{\pi}{a} e^{-a}\). How is the right-hand side modified if \(\cos x\) is replaced by \(\cos k x\) ? (b) \(\int_{-\infty}^{\infty} \frac{x \sin x}{x^{2}+a^{2}} d x=\pi e^{-a}\). How is the right-hand side modified if \(\sin x\) is replaced by \(\sin k x ?\) These integrals may also be interpreted as Fourier cosine and sine transforms see Chapter 15 .

Short Answer

Expert verified
(a) Integral value becomes \( \frac{\pi}{a} e^{-ak} \). (b) Integral value becomes \( \pi \sin(k) e^{-ak} \).

Step by step solution

01

Introduction to Integral Techniques

To tackle these integrals, we'll use contour integration, a technique in complex analysis. Because the integrands involve expressions like \(x^2 + a^2\), we can consider using the residue theorem to evaluate them. The integral limits from \(-\infty\) to \(\infty\) suggest considering the complex plane approach.
02

Analyze \\( \\int_{-\infty}^{\infty} \frac{\cos x}{x^{2}+a^{2}} dx=

For the integral \( \int_{-\infty}^{\infty} \frac{\cos x}{x^{2}+a^{2}} d x \), rewrite it in terms of complex exponentials: \( \cos x = \text{Re}(e^{ix}) \). The associated complex function is \( \frac{e^{ix}}{x^2 + a^2} \). Next, close the contour in the upper half-plane to avoid poles on the real axis.
03

Evaluate using the Residue Theorem

The pole at \( z = ai \) (where \( z = x+i*a \)) in the upper half-plane is simple. Compute the residue: \( \text{Res}(f, ai) = \lim_{z \to ai} (z-ai) \frac{e^{iz}}{z^2+a^2} = \frac{e^{-a}}{2ai} \). By the residue theorem, the integral is \( 2 \pi i \) times the sum of residues, leading to \( \pi e^{-a} / a \).
04

Modify RHS for \\cos kx\\

If \( \cos x \) is replaced by \( \cos kx \), replace \( e^{ix} \) by \( e^{ikx} \) in the integral. The same contour and residue analysis gives the result \( \frac{\pi}{a} e^{-ak} \), as \( k \) modifies the exponential decay.
05

Approach for \\( \\int_{-\infty}^{\infty} \frac{x \sin x}{x^{2}+a^{2}} dx \\)

For \( \int_{-\infty}^{\infty} \frac{x \sin x}{x^{2}+a^{2}} dx \), mimic the approach by considering \( \text{Im}(e^{ix}) \). However, this integral can be shown directly to equal to the same decay as the sine component amplifies the imaginary part linkage.
06

Modify RHS for \\sin kx\\

If \( \sin x \) is replaced by \( \sin kx \), analyze similarly. The integrand changes to \( \text{Im}(e^{ikx}) \), which adjusts the decay in \( e^{-ak} \), giving finally \( \pi \sin(k) e^{-ak} \) for the real-valued result.

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Key Concepts

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

Residue Theorem
The Residue Theorem is a powerful tool in complex analysis used to evaluate complex integrals, especially those along closed curves or contours in the complex plane. It revolves around residues, which are a measure of the singularity or pole of a function within a given contour. When we have an analytic function with isolated singularities within a closed contour, the integral around that contour is equal to \( 2 \pi i \) times the sum of the residues of the singularities inside the contour.
  • A residue at a simple pole is essentially the coefficient of \( \frac{1}{z-a} \) in the Laurent series expansion of the function around \( z=a \).
  • For our exercise, the pole is located at \( z=ai \) in the upper half-plane.
  • Applying the residue theorem allows us to transform a complex two-dimensional problem into a simpler one-dimensional evaluation of residues, greatly simplifying the calculation of real integrals.
Contour Integration
Contour integration involves evaluating integrals of complex functions along specified paths in the complex plane, known as contours. This method is particularly useful for solving complex-valued integrals like the ones in the exercise.
  • In the problem, a key step is to close the contour in the upper half-plane when integrating from negative to positive infinity.
  • Closing the contour helps avoid singularities on the real axis and ensures the integral is properly evaluated over a boundary without change from infinity back to infinity.
  • The choice of contour can simplify the problem, such as eliminating contributions from large semi-circles if they tend to zero as their radius goes to infinity.
Fourier Transforms
Fourier transforms play a crucial role in translating complex functions from the time domain to the frequency domain, giving insight into the frequency components of a function. In the context of our integrals, they help express trigonometric functions as complex exponentials through relationships like \( \cos(x) = \text{Re}(e^{ix}) \) and \( \sin(x) = \text{Im}(e^{ix}) \).
  • These relationships allow us to convert real integrals involving sines and cosines into more manageable forms using complex exponentials.
  • In this exercise, modifying the functions to \( \cos(kx) \) and \( \sin(kx) \) impacts the integral's result, demonstrating the modulating effect of frequency \( k \).
  • The transformation particularly affects the exponential decay terms \( e^{-ak} \), showing how frequency components influence the decay rate in these transforms.
Integrals in the Complex Plane
Integrating over the complex plane can seem daunting, but it offers several advantages when dealing with functions that have symmetries or periodicities. The set-up of the integral reflects a mix of both real and imaginary axis considerations.
  • These integrals convert real plane behavior into complex analytics, allowing for holistic evaluation using the entire plane.
  • In these exercises, using the complex plane means turning trigonometric expressions into exponentials, thereby simplifying their multiplication and integration.
  • Ultimately, this approach emphasizes understanding the integral's path and singularities affecting the function's behavior across the entire complex domain.

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Most popular questions from this chapter

Determine the nature of the singularities of each of the following functions and evaluate the residues \((a>0)\). (a) \(\frac{1}{z^{2}+a^{2}}\). (b) \(\frac{1}{\left(z^{2}+a^{2}\right)^{2}}\). (c) \(\frac{z^{2}}{\left(z^{2}+a^{2}\right)^{2}}\). (d) \(\frac{\sin 1 / z}{z^{2}+a^{2}}\). (e) \(\frac{z e^{+i z}}{z^{2}+a^{2}}\) (f) \(\frac{z e^{+i z}}{z^{2}+a^{2}}\). (g) \(\frac{e^{+1 z}}{z^{2}-a^{2}}\) (h) \(\frac{z^{-k}}{z+1}, \quad 0

Show that $$ \int_{0}^{\pi} \frac{d \theta}{(a+\cos \theta)^{2}}=\frac{\pi a}{\left(a^{2}-1\right)^{3 / 2}}, \quad a>1 . $$

Most of the special functions of mathematical physics may be generated (defined) by a generating function of the form $$ g(t, x)=\sum_{n} f_{n}(x) t^{n} . $$ Given the following integral representations, derive the corresponding generating function: (a) Bessel $$ J_{n}(x)=\frac{1}{2 \pi i} \oint e^{(x / 2)(t-1 / t)} t^{-n-1} d t . $$ (b) Modified Bessel $$ I_{n}(x)=\frac{1}{2 \pi i} \oint e^{(x / 2)(t+1 / t)} t^{-n-1} d t . $$ (c) Legendre $$ P_{n}(x)=\frac{1}{2 \pi i} \oint\left(1-2 t x+t^{2}\right)^{-1 / 2} t^{-n-1} d t . $$ (d) Hermite $$ H_{n}(x)=\frac{n !}{2 \pi i} \oint e^{-t^{2}+2 x_{t}-n-1} d t $$ (c) Laguerre $$ L_{n}(x)=\frac{1}{2 \pi i} \oint \frac{e^{-x t /(1-t)}}{(1-t) t^{n+1}} d t $$ (f) Chebyshev $$ T_{n}(x)=\frac{1}{4 \pi i} \oint \frac{\left(1-t^{2}\right) t^{-n-1}}{1-2 t x+t^{2}} d t $$ Each of the contours encircles the origin and no other singular points.

As an example of an essential singularity consider \(e^{1 / z}\) as \(z\) approaches zero. For any complex number \(z_{0}, z_{0} \neq 0\), show that $$e^{1 / z}=z_{0}$$ has an infinite number of solutions.

Determine the asymptotic dependence of the modified Bessel functions \(I_{y}(x)\). given by $$ I_{v}(x)=\frac{1}{2 \pi i} \int_{C} e^{(x / 2)(t+1 / t)} \frac{d t}{t^{v+1}} $$ The contour starts and ends at \(t=-\infty\), encircling the origin in a positive sense. There are two saddle points. Only the one at \(z=+1\) contributes significantly to the asymptotic form.
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