Fourier transform and inverse transform

The second thing he wants from you is to write f(x) = \int_{-\infty}^{\infty} f(x)\delta(t - a) dt and then switch the integrals.
  • #1
babyrudin
8
0

Homework Statement



Let f(x) be an integrable complex-valued function on [tex]\mathbb{R}[/tex]. We define the Fourier transform [tex]\phi=\mathcal{F}f[/tex] by
[tex]\[\phi(t)=\int_{\infty}^{\infty} e^{ixt} f(x) dx.\][/tex]

Show that if [tex]f[/tex] is continuous and if [tex]$\phi$[/tex] is integrable, then
[tex]\[f(x)=\frac{1}{2\pi} \int_{-\infty}^{\infty} e^{-ixt} \phi(t) dt.\][/tex]

The Attempt at a Solution



So far I have
[tex]\[\int_{-\infty}^{\infty} e^{-ixt} \phi(t) dt = \lim_{a \to \infty} \int_{-a}^{a} e^{-ixt} \phi(t) dt = \lim_{b \to \infty} \frac{1}{b} \int_0^b da \int_{-a}^a e^{-ixt} \phi(t) dt = \lim_{b \to \infty} \frac{1}{b} \int_0^b da \int_{-\infty}^\infty \frac{2 \sin((y-x)a)}{y-x} f(y) dy.\][/tex]

What to do next?
 
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  • #2
babyrudin said:

Homework Statement



Let f(x) be an integrable complex-valued function on [tex]\mathbb{R}[/tex]. We define the Fourier transform [tex]\phi=\mathcal{F}f[/tex] by
[tex]\[\phi(t)=\int_{\infty}^{\infty} e^{ixt} f(x) dx.\][/tex]

Show that if [tex]f[/tex] is continuous and if [tex]$\phi$[/tex] is integrable, then
[tex]\[f(x)=\frac{1}{2\pi} \int_{-\infty}^{\infty} e^{-ixt} \phi(t) dt.\][/tex]


The Attempt at a Solution



So far I have
[tex]\[\int_{-\infty}^{\infty} e^{-ixt} \phi(t) dt = \lim_{a \to \infty} \int_{-a}^{a} e^{-ixt} \phi(t) dt = \lim_{b \to \infty} \frac{1}{b} \int_0^b da \int_{-a}^a e^{-ixt} \phi(t) dt = \lim_{b \to \infty} \frac{1}{b} \int_0^b da \int_{-\infty}^\infty \frac{2 \sin((y-x)a)}{y-x} f(y) dy.\][/tex]

What to do next?
It's not a matter of "what to do next?" but "Why in the world are you doing that?". For one thing, there is no integral from 0 to [itex]\infty[/itex] in what you had before, where are you getting that? And even when you have two integrals, you still only have one variable: dt. You want to show that, for any continuous f (in which case [itex]\phi[/itex] must be integrable)
[tex]\frac{1}{2\pi}\int_{-\infty}^\infty e^{-ixt} \int_{-\infty}^\infty e^{iut}f(u)dudt= f(x)[/tex]
 
  • #3
What your coarse probably wants from you is to use the following,

[tex]\delta(t - a) = \frac{1}{2\pi}\int_{-\infty}^{\infty} e^{i \omega (t -a) } d\omega. [/tex]
If you want a full but rather simple proof, for a graduate student with a mathematical focus, of this identity I can send it to you.
 

1. What is the Fourier transform?

The Fourier transform is a mathematical operation that decomposes a function into its individual frequency components. In other words, it converts a function from its original time or space domain into a representation in the frequency domain.

2. What is the purpose of the Fourier transform?

The purpose of the Fourier transform is to analyze complex signals or functions and break them down into simpler components. This allows for easier analysis and understanding of the signal's frequency content.

3. What is the difference between Fourier transform and inverse transform?

The Fourier transform converts a function from the time or space domain into the frequency domain, while the inverse transform does the opposite - it converts a function from the frequency domain back into the time or space domain.

4. How is the Fourier transform used in science and engineering?

The Fourier transform is widely used in various fields of science and engineering, such as signal processing, image processing, communications, and quantum mechanics. It is used to analyze and manipulate signals and functions, and is especially useful for understanding the frequency content of complex signals.

5. Are there any limitations to the Fourier transform?

One limitation of the Fourier transform is that it assumes the signal being analyzed is periodic. If the signal is not periodic, then the Fourier transform may not accurately represent the frequency content of the signal. Additionally, the Fourier transform does not work well with signals that have sharp changes or discontinuities.

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