Exponential Fourier transform

  • #1
agnimusayoti
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Homework Statement:
Find the exponential Fourier transform of
##f(x)=e^{-|x|}## and write the inverse transform. You should find:
$$\int_{0}^{\infty} \frac{\cos{ax}}{a^2+1} da = \frac {\pi}{2} e^{-|x|}$$
Relevant Equations:
Fourier transform:
$$g(a)=\frac{1}{2\pi} \int_{-\infty}^{\infty} f(x) e^{-iax} dx$$
Inverse Transform:
$$f(x)=\int_{-\infty}^{\infty} g(a) e^{iax} da$$
From the sketch, I know that this function is an even function. So, I simplify the Fourier transform in the limit of the integration (but still in exponential form). Then, I try to find the exponential FOurier transform. Here what I get:
$$g(a)=\frac{2}{2\pi} \int_{0}^{\infty} e^{-x} e^{-iax} dx$$,
$$g(a)=\frac{1}{\pi} \int_{0}^{\infty} e^{(-x)(1+a)} dx$$,
$$g(a)=\frac{1}{\pi} \left[\frac{e^{-ix(1+a)}}{-i(1+a)} \right]^{\infty}_{0}$$.
As x approaching infinite ##e^{-ix(1+a)}## approaching zero. So,
$$g(a)=\frac{1-ia}{\pi(1+a^2)}$$.

Knowing this transform, I did the inverse transformation.
$$f(x)=\int_{-\infty}^{\infty} \frac{1-ia}{\pi(1+a^2)} e^{iax} da$$, where ##e^{iax}=\cos {(ax)} + i \sin {(ax)}##
So,
$$f(x)=\int_{-\infty}^{\infty} \frac{(1-ia)\left(\cos{ax} + i \sin {ax}\right)}{\pi(1+a^2)} da$$.

I observe that ##\frac{\sin{ax}}{1+a^2}##; ##\frac{(-a)\cos{ax}}{1+a^2}## are odd functions. But, ##\frac{\cos{ax}}{1+a^2}##; ##\frac{(a)\sin{ax}}{1+a^2}## are even functions. So,
$$f(x)=\frac{2}{\pi}\int_{0}^{\infty} \frac{\cos {ax} + a \sin {ax}}{(1+a^2)} da$$.

The sin term of the answer shouldn't be there. I have double-checked my work and still haven't find the mistake. Could you please explain how I get the answer term, in the problem statement? Thanks.
 

Answers and Replies

  • #2
PeroK
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Then, I try to find the exponential FOurier transform. Here what I get:
$$g(a)=\frac{2}{2\pi} \int_{0}^{\infty} e^{-x} e^{-iax} dx$$,

How did you get that?
 
  • #3
agnimusayoti
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I change ##e^{-|x|}## from ##-\infty## to ##\infty## becomes ##e^{-x}## from 0 to ##\infty##.
 
  • #4
PeroK
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I change ##e^{-|x|}## from ##-\infty## to ##\infty## becomes ##e^{-x}## from 0 to ##\infty##.
What about the ##e^{-iax}##?
 
  • #5
agnimusayoti
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What about the ##e^{-iax}##?
It should be the same, isn't it? Because of the general formula?
 
  • #6
PeroK
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It should be the same, isn't it? Because of the general formula?
It's not an even function:
$$e^{-iax} = \cos(ax) - i\sin(ax)$$
And ##\sin(ax)## is an odd function.
 
  • #7
agnimusayoti
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Oh Gee. I forgot that ##f(x)## should be multiplied by ##e^{-iax}##. I will try to fix this.
 
  • #8
agnimusayoti
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I mean, an odd or even function is multiplied function; not f(x) itself.
 
  • #9
agnimusayoti
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Yeah, finally I can show the solution. Thanks, Pero K for the correction!
 

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