How do I convert f(x) into its Fourier Transform?

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SUMMARY

The discussion focuses on converting the function f(x) = δ(x + α) + δ(x - α) into its Fourier Transform. The user correctly identifies that multiplying f(x) by the exponential term (1/√(2π))e^(-ikx) and integrating leads to the Fourier Transform. The integration of the Dirac delta functions results in the expression e^(-ikα), confirming the properties of the delta function in Fourier analysis.

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  • Understanding of Dirac delta functions
  • Familiarity with Fourier Transform concepts
  • Knowledge of integration techniques, particularly involving delta functions
  • Basic principles of electrodynamics
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  • Study the properties of the Dirac delta function in detail
  • Learn the Fourier Transform of common functions
  • Explore integration techniques involving delta functions
  • Review applications of Fourier Transforms in electrodynamics
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Students in physics, particularly those studying electrodynamics, as well as anyone interested in understanding Fourier Transforms and their applications in signal processing and analysis.

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Homework Statement



I am really confused in my electrodynamics class. I have the following function.

f(x) = \delta (x + \alpha ) + \delta(x -\alpha)

How do i convert this into Fourier Tranform ?
Those are dirac delta functions on either sides of the origin.


Homework Equations



no clue

The Attempt at a Solution



I multiplied f(x) by 1/sqrt(2*pi)*e-ikt and integrated the whole thing.
 
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And what did you get?
 
1/\sqrt{2\pi}\int\delta(x+\alpha)e-ikx + \delta(x-\alpha)e-ikx dx

I tried to integrate those terms by parts (uv substitution).

\int\delta(x+\alpha) = 1. Isn't that right ?
Also I think f(x)\int\delta(x+\alpha) = f(\alpha). Am i right ?
 
All you need is the following property:

\int\limits_{-\infty}^{+\infty} f(x)\delta\left(x-a\right)=f(a)
 
so

\inte-ikx \delta (x - a) = e-ika ?
 
Exactly.
 

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