A question on Fourier Analysis

[tpm]

it's a curiosity more than a HOmework, given the integral:

[tex] \int_{-\infty}^{\infty}dx e^{if(x)+iwx}=g(w) [/tex]

Where g(w) can be viewed as the Fourier transform of exp(if(x)) then my question is if we can prove g(w) satisfies the ODE:

[tex] -if'(x)\frac{\partial g(w)}{\partial w}+wg(w)=0 [/tex]

for simplicity we can impose [tex] exp(if(\infty))=0 [/tex] and the same for

-oo

 

[AiRAVATA]

Why don't you derive and show us if it's true?

 

[tacman]

This is a very interesting question and it seems like you have a good understanding of Fourier analysis. To answer your question, yes, it is possible to prove that g(w) satisfies the ODE -if'(x)∂g(w)/∂w + wg(w) = 0. This can be done by using the properties of the Fourier transform and the inverse Fourier transform.

First, let's start with the Fourier transform of exp(if(x)), which is g(w). By definition, we have:

g(w) = ∫exp(if(x))e^{-iwx}dx

Now, taking the derivative of g(w) with respect to w, we get:

∂g(w)/∂w = ∂/∂w∫exp(if(x))e^{-iwx}dx

Using the Leibniz integral rule, we can rewrite this as:

∂g(w)/∂w = ∫∂/∂w(exp(if(x)))e^{-iwx}dx

Since exp(if(x)) is a function of x only, we can write this as:

∂g(w)/∂w = ∫if'(x)exp(if(x))e^{-iwx}dx

Now, using the fact that g(w) is the Fourier transform of exp(if(x)), we can rewrite this as:

∂g(w)/∂w = ∫if'(x)g(w)e^{-iwx}dx

Next, we use the property of the inverse Fourier transform, which states that:

∫g(w)e^{iwx}dw = 2πf(x)

Applying this property to the above equation, we get:

∂g(w)/∂w = 2πif'(x)f(x)

Finally, we substitute this back into the ODE we want to prove, and we get:

-if'(x)∂g(w)/∂w + wg(w) = -if'(x)(2πif'(x)f(x)) + w(2πf(x))

= -2πif'(x)^2f(x) + 2πwf(x)

= 2πf(x)(w - if'(x)^2)

Since we have imposed the condition that exp(if(∞)) = 0, we can conclude that f(∞) = 0 and therefore w - if'(x)^2 = 0. This means