Step in fourier transform derivation

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SUMMARY

The discussion focuses on the derivation of the Fourier transform from the Fourier series as the period approaches infinity. The key equations presented are: (x) = (2/π)∫(0 to ∞) (dk) [∫(0 to ∞) h(ξ) sin(kξ) sin(xξ) dξ] and (x) = (1/2π)∫(-∞ to ∞) (dk) [∫(-∞ to ∞) h(ξ) sin(kξ) sin(xξ) dξ]. The factor of 1/4 arises due to the change in limits of integration, where the integrals over symmetric functions yield a doubling effect. The discussion also references Morse and Feshbach's "Methods of Theoretical Physics" as a source for understanding this derivation.

PREREQUISITES
  • Understanding of Fourier series and Fourier transforms
  • Familiarity with integral calculus, particularly improper integrals
  • Knowledge of symmetric functions and their properties
  • Basic concepts from theoretical physics, especially related to wave functions
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  • Study the derivation of the Fourier transform from the Fourier series in detail
  • Explore the properties of symmetric functions in integral calculus
  • Read Morse and Feshbach's "Methods of Theoretical Physics" for deeper insights
  • Learn about the implications of changing limits of integration in Fourier analysis
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This discussion is beneficial for physicists, mathematicians, and engineering students who are delving into Fourier analysis and its applications in signal processing and theoretical physics.

Nick R
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Looking at how the Fourier transform comes about from the Fourier series when the period goes to infinity, they make the following step

h \left( x \right) = \frac{2}{\pi}\int^{\infty}_{0} \left( dk \right) \left[ \int^{\infty}_{0} h \left( \varsigma \right) sin \left( k \varsigma \right) sin \left( x \varsigma \right) d\varsigma \right]

and the limits of integration can be changed to

h \left( x \right) = \frac{1}{2 \pi}\int^{\infty}_{-\infty} \left( dk \right) \left[ \int^{\infty}_{-\infty} h \left( \varsigma \right) sin \left( k \varsigma \right) sin \left( x \varsigma \right) d\varsigma \right]

I'm trying to understand why this factor of 1/4 arises...

My understand is this
- in the first case, one integral gives us a Fourier coefficient A_{k} the other integration is effectively doing this; \sum^{\infty}_{k=0} A_{k} sin \left( kx \right)

- in the second case, one integral gives 2 A_{k}, because it integrates across 2 periods, but it can't matter which period you integrate across to find the Fourier coefficient. The other integral is effectively 2 \sum^{\infty}_{k=0} A_{k} sin \left( kx \right) since it is summing over 2 periods instead of 1.

Is this a correct way of looking at it? I'm more confident I know what's going on after writing this post and having to articulate it...
 
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Nick R said:
Looking at how the Fourier transform comes about from the Fourier series when the period goes to infinity, they make the following step

h \left( x \right) = \frac{2}{\pi}\int^{\infty}_{0} \left( dk \right) \left[ \int^{\infty}_{0} h \left( \varsigma \right) sin \left( k \varsigma \right) sin \left( x \varsigma \right) d\varsigma \right]

and the limits of integration can be changed to

h \left( x \right) = \frac{1}{2 \pi}\int^{\infty}_{-\infty} \left( dk \right) \left[ \int^{\infty}_{-\infty} h \left( \varsigma \right) sin \left( k \varsigma \right) sin \left( x \varsigma \right) d\varsigma \right]

I'm trying to understand why this factor of 1/4 arises...

My understand is this
- in the first case, one integral gives us a Fourier coefficient A_{k} the other integration is effectively doing this; \sum^{\infty}_{k=0} A_{k} sin \left( kx \right)

- in the second case, one integral gives 2 A_{k}, because it integrates across 2 periods, but it can't matter which period you integrate across to find the Fourier coefficient. The other integral is effectively 2 \sum^{\infty}_{k=0} A_{k} sin \left( kx \right) since it is summing over 2 periods instead of 1.

Is this a correct way of looking at it? I'm more confident I know what's going on after writing this post and having to articulate it...
The simplest way to see it is to note that in the first equation you cite, the integrals are both going form 0 to \infty while in the second equation, they are going from -\infty to \infty.
\int_{-\infty}^\infty} f(x)dx= \int_{-\infty}^0 f(x)dx= \int_0^\infty f(x)dx

and, since both integrands are symmetric,
\int_{-\infty}^0 f(x)dx= \int_0^\infty f(x)dx
which gives
\int_{-\infty}^\infty f(x)dx= 2\int_0^\infty f(x)dx
or
\int_0^\infty f(x)dx= \frac{1}{2}\int_{-\infty}^\infty f(x) dx
for each integral.

It has nothing to do with "periods" because, in the Fourier transform, the functions are not necessarily periodic.
 
The factor of 4 is easy enough to explain. I'm a little more concerned with where the :

\frac{2}{\pi} \int^{\infty}_{0} \left( dk \right)

bit comes from. That's bit is clearly infinite.

BTW. I'm reasonably familiar with the development of the Fourier transform from the Fourier series but I haven't seen that method before. Do you have a either a link or the equations/development that precedes the lines shown above?
 
uart said:
The factor of 4 is easy enough to explain. I'm a little more concerned with where the :

\frac{2}{\pi} \int^{\infty}_{0} \left( dk \right)

bit comes from. That's bit is clearly infinite.

BTW. I'm reasonably familiar with the development of the Fourier transform from the Fourier series but I haven't seen that method before. Do you have a either a link or the equations/development that precedes the lines shown above?

This all comes from Morse and Feshbach's "Methods of Theoretical Physics". They make a substitution, take the limit of the period going to infnity and the summation becomes an integral. Its the best description I've found but its taking me a while to assimulate it. Most explanations online are "heres the formula - magnificent!".

Also that whole thing could be rewritten \frac{2}{\pi} \int^{infty}_{0} \int^{infty}_{0} \left( \right) d\varsigma dk

So its not infinity. Its just a double integral.

edit: also I made a mistake in those formulas in the first post but it shouldn't matter now.
 

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