Inverse Fourier Transform: Impact on Integration Limits

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SUMMARY

The discussion centers on the impact of the inverse Fourier transform on integration limits. It is established that the integration limits can change after performing the inverse Fourier transform on a function. Specifically, the example provided illustrates that the integration of a function F(a) over the interval [0, a] can lead to different boundaries when transformed into the Fourier domain. The necessity of returning to the original function after transformation is emphasized as a fundamental property of Fourier transforms.

PREREQUISITES
  • Understanding of Fourier transforms, specifically inverse Fourier transforms.
  • Knowledge of integration techniques in calculus.
  • Familiarity with the notation and properties of integrals.
  • Basic grasp of complex exponential functions and their applications in transforms.
NEXT STEPS
  • Study the properties of inverse Fourier transforms in detail.
  • Explore integration techniques for transforming functions in the Fourier domain.
  • Learn about boundary conditions in integrals and their implications in transforms.
  • Investigate the relationship between Fourier transforms and original functions through examples.
USEFUL FOR

Mathematicians, physicists, and engineers involved in signal processing, anyone working with Fourier analysis and integration techniques.

hula
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an equation involves an integration. After an inverse Fourier transform of the equation, will the integration limits change? (maybe you can take a look at the attached file)

Thanks!
 

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I can't read your attachment, but I think in general the answer would be yes.
Consider the function
F(a) = \int_0^a f(x) \, dx
The Fourier-transform will be
\tilde F(k) = \frac{1}{\sqrt{2\pi}} \int e^{-ika} F(a) \, da<br /> = \frac{1}{\sqrt{2\pi}} \int \int_0^a f(x) e^{-ika} \, dx \, da<br />
If you are lucky you will be able to do the integral in a and are left with one other integral, whose boundaries are probably not the same. But of course, if you transform this back, you should get the original function again (otherwise it's not really a good Fourier transform, is it)
 

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