Inverse Fourier Transform of $\frac{b}{\pi(x^2+b^2)}$: Solving the Problem

In summary, the conversation discusses taking the inverse Fourier transform of the function \frac{b}{\pi(x^2+b^2)}. The resulting inverse Fourier transform is e^{(-b|t|)}, which can be obtained using a standard contour integral and closing the contour with a semicircle above the real axis. The conversation also mentions that this type of problem typically involves integrals with absolute values.
  • #1
buttersrocks
29
0

Homework Statement


I need to take the inverse Fourier transform of

[tex]\frac{b}{\pi(x^2+b^2)}[/tex]

Homework Equations



f(t)=[tex]\int_{-\infty}^{\infty}e^{itx}\frac{b}{\pi(x^2+b^2)}dx[/tex]

It might be useful that [tex]\frac{2b}{\pi(x^2+b^2)}=\frac{1}{b+ix}+\frac{1}{b-ix}[/tex]

The Attempt at a Solution


I know the result is [tex]e^{(-b|t|)}[/tex], and I can get from [tex]e^{(-b|t|)}[/tex] to
[tex]\frac{b}{\pi(x^2+b^2)}[/tex], but how do I do it in reverse if I didn't already know the pair existed? This doesn't require complex integration does it?
 
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  • #2
I have to admit my first thought was a contour integral...

In my experience, things involving |x| tend to require them.
 
  • #3
It is a standard contour integral. Close the contour with a semicircle above the real axis.
 
  • #4
Okay guys, thanks, that is what I was thinking, but the book I'm in doesn't have anything else involving complex integration, so I assumed that I was just missing a trick.
 

Q1: What is the inverse Fourier transform of $\frac{b}{\pi(x^2+b^2)}$?

The inverse Fourier transform of $\frac{b}{\pi(x^2+b^2)}$ is given by the function $f(x) = \frac{1}{\sqrt{2\pi}}e^{-bx}$.

Q2: What is the purpose of solving the inverse Fourier transform of $\frac{b}{\pi(x^2+b^2)}$?

The purpose of solving the inverse Fourier transform of $\frac{b}{\pi(x^2+b^2)}$ is to find the original function in the time domain that corresponds to the given function in the frequency domain.

Q3: How is the inverse Fourier transform of $\frac{b}{\pi(x^2+b^2)}$ calculated?

The inverse Fourier transform can be calculated using the formula $\mathcal{F}^{-1}\{\frac{b}{\pi(x^2+b^2)}\} = \frac{1}{\sqrt{2\pi}}e^{-bx}$. Alternatively, it can also be calculated using integration techniques such as contour integration or residue theorem.

Q4: What are the properties of the inverse Fourier transform of $\frac{b}{\pi(x^2+b^2)}$?

The inverse Fourier transform of $\frac{b}{\pi(x^2+b^2)}$ is a real-valued function. It is also symmetric about the y-axis and decays exponentially as x approaches infinity.

Q5: What are some applications of solving the inverse Fourier transform of $\frac{b}{\pi(x^2+b^2)}$?

The inverse Fourier transform of $\frac{b}{\pi(x^2+b^2)}$ is commonly used in signal processing, image reconstruction, and solving differential equations in physics and engineering. It is also used in probability and statistics to calculate the probability distribution function of certain random variables.

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