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Chapter 4: Problem 3

Show that the Fourier transform of the "Gaussian" \(f(x)=\exp \left(-\left(x-x_{0}\right)^{2} / a^{2}\right) x_{0}, a\) real, is also a Gaussian: $$ \hat{F}(k)=a \sqrt{\pi} e^{-(k a / 2)^{2}} e^{-i k x_{0}} $$

Short Answer

Expert verified
The Fourier transform of the Gaussian \(f(x)=\exp\left(-\frac{(x-x_{0})^2}{a^{2}}\right)\) is also a Gaussian: \[\hat{F}(k)=a \sqrt{\pi} e^{-(k a / 2)^2} e^{-i k x_{0}}.\]

Step by step solution

01

Write the Definition of Fourier Transform

Recall that the Fourier transform \(\hat{F}(k)\) of a function \(f(x)\) is given by \[\hat{F}(k) = \int_{-\infty}^{\infty} f(x) e^{-i k x} \, dx.\]
02

Substitute the Gaussian Function

Substitute the given Gaussian function \(f(x) = \exp\left(-\frac{(x-x_0)^2}{a^2}\right)\) into the integral: \[\hat{F}(k) = \int_{-\infty}^{\infty} \exp\left(-\frac{(x-x_0)^2}{a^2}\right) e^{-i k x} \, dx.\]
03

Complete the Square in the Exponent

Complete the square in the exponent to combine the Gaussian and exponential terms. Let \(u = x - x_0\), then the integral becomes: \[\hat{F}(k) = \int_{-\infty}^{\infty} \exp\left(-\frac{u^2}{a^2}\right) e^{-i k (u + x_0)} \, du.\]
04

Simplify the Exponent

Combine the exponents: \[\hat{F}(k) = e^{-i k x_0} \int_{-\infty}^{\infty} \exp\left(-\frac{u^2}{a^2} - i k u\right) \, du.\]
05

Use a Change of Variables

Use the substitution \(v = u - \frac{(i a^2 k)}{2}\) to shift and simplify the integral, so: \[\hat{F}(k) = e^{-i k x_0} \int_{-\infty}^{\infty} \exp\left(-\frac{v^2}{a^2}\right) \, dv \exp\left(-\frac{(i a k)^2}{4}\right).\]
06

Evaluate the Gaussian Integral

The remaining integral is a standard Gaussian integral which evaluates to \(a \sqrt{\pi}\): \[\int_{-\infty}^{\infty} \exp\left(-\frac{v^2}{a^2}\right) \, dv = a \sqrt{\pi}.\]
07

Combine the Results

Combine all the factors to obtain the final result: \[\hat{F}(k) = a \sqrt{\pi} \exp\left(-\frac{(k a / 2)^2}{1}\right) \exp\left(-i k x_{0}\right).\]

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Key Concepts

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

headline of the respective core concept
The Fourier transform is a key tool in many fields such as physics, engineering, and signal processing. It transforms a function of time (or space) into a function of frequency. Mathematically, given a function \( f(x) \), its Fourier transform \( \hat{F}(k) \) is defined as: \[ \hat{F}(k) = \int_{-\infty}^{\infty} f(x) e^{-i k x} \, dx. \] This formulation shows us how every point in the time domain can be transformed into the frequency domain by integrating over the entire range of the function multiplied by a complex exponential. The resulting function \( \hat{F}(k) \) describes the original function in terms of its frequency components.
headline of the respective core concept
The Gaussian integral is fundamental in probability, statistics, and quantum mechanics. The standard Gaussian function is of the form \( g(x) = \exp(-x^2) \). Its integral over all space is: \[ \int_{-\infty}^{\infty} \exp(-x^2) \ dx = \sqrt{\pi}. \] In the given exercise, we deal with a shifted and scaled Gaussian function \( f(x) = \exp\left(-\frac{(x - x_0)^2}{a^2}\right) \). Through the process of completing the square and clever substitutions, we transform and solve it using properties of Gaussian integrals, where the integral part evaluates to \( a \sqrt{\pi} \).
headline of the respective core concept
Complex exponentiation is a method to handle oscillatory behaviors in functions. It involves expressions like \( e^{-i k x} \), where \( i \) is the imaginary unit, and \( k x \) is a product of a frequency and a variable. It links directly to Euler's formula: \[ e^{i \theta} = \cos(\theta) + i\sin(\theta). \] This formula allows us to represent sinusoidal functions as exponentials, making them easier to manipulate analytically, especially in the Fourier transform. In the given problem, complex exponentials interplay with the Gaussian function to translate time-domain functions into the frequency domain.
headline of the respective core concept
The change of variables is a powerful technique in integral calculus to simplify the integration process. It involves substituting a new variable that makes the integral easier to solve. In the given problem, we used such a substitution: \( u = x - x_0 \) transforms the function to center around zero. Next, we utilize \( v = u - \frac{(i a^2 k)}{2} \), which shifts and simplifies the Gaussian in the exponential. This results in an integral of a standard Gaussian form, which greatly simplifies the evaluation and provides the necessary insight to solve the problem effectively.

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Most popular questions from this chapter

(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 $$ \int_{0}^{\infty} \frac{\sin ^{2} x}{x^{2}} d x=\frac{\pi}{2} $$

Obtain the inverse Laplace transforms of the following functions, assuming \(\omega, \omega_{1}, \omega_{2}>0\) (a) \(\frac{s}{s^{2}+\omega^{2}}\) (b) \(\frac{1}{(s+\omega)^{2}}\) (c) \(\frac{1}{(s+\omega)^{n}}\) (d) \(\frac{s}{(s+\omega)^{n}}\) (e) \(\frac{1}{\left(s+\omega_{1}\right)\left(s+\omega_{2}\right)}\) (f) \(\frac{1}{s^{2}\left(s^{2}+\omega^{2}\right)}\) (g) \(\frac{1}{\left(s+\omega_{1}\right)^{2}+\omega_{2}^{2}}\) (h) \(\frac{1}{\left(s^{2}-\omega^{2}\right)^{2}}\)

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

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).

Projection operators can be defined as follows. Consider a function \(F(z)\) $$ F(z)=\frac{1}{2 \pi i} \int_{C} \frac{f(\zeta)}{\zeta-z} d \zeta $$ where \(C\) is a contour, typically infinite (e.g. the real axis) or closed (e.g. a circle) and \(z\) lies off the contour. Then the "plus" and "minus" projections of \(F(z)\) at \(z=\zeta_{0}\) are defined by the following limit: $$ F^{\pm}\left(\zeta_{0}\right)=\lim _{z \rightarrow \zeta_{0}}\left[\frac{1}{2 \pi i} \int_{C} \frac{f(\zeta)}{\zeta-z} d \zeta\right] $$ where \(\zeta_{0}{ }^{\pm}\)are points just inside \((+)\)or outside (-) of a closed contour (i.e., \(\lim _{z \rightarrow 5_{0}+\text { denotes the limit from points } z \text { inside the contour } C \text { ) or to }}\) the left (t) or right (-) of an infinite contour. Note: the "+" region lies to the left of the contour; where we take the standard orientation for a contour, that is, the contour is taken with counterclockwise orientation. To simplify the analysis, we will assume that \(f(x)\) can be analytically extended in the neighborhood of the curve \(C\). (a) Show that $$ F^{\pm}\left(\zeta_{0}\right)=\frac{1}{2 \pi i} \int_{C} \frac{f(\zeta)}{\zeta-\zeta_{0}} d \zeta \pm \frac{1}{2} f\left(\zeta_{0}\right) $$ where \(f_{C}\) is the principal value integral that omits the point \(\zeta=\zeta 0\). (b) Suppose that \(f(\zeta)=1 /\left(\zeta^{2}+1\right)\), and the contour \(C\) is the real axis (infinite), with orientation take from \(-\infty\) to \(\infty\); find \(F^{\pm}\left(\zeta_{0}\right)\). (c) Suppose that \(f(\zeta)=1 /\left(\zeta^{2}+a^{2}\right), a^{2}>1\), and the contour \(C\) is the unit circle centered at the origin with counterclockwise orientation; find \(F^{\pm}\left(\zeta_{0}\right)\)
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