Evaluating a Fourier Transform Integral

[wxstall]

Evaluating a "Fourier Transform" Integral

Homework Statement

Evaluate

I = ∫[0,∞] e^-ktw2 cos(wx) dw

in the following way: Determine ∂I/∂x, then integrate by parts.

Homework Equations

Possibly?

The Attempt at a Solution

Since integral limits do not depend on x, the partial with respect to x should simply be:

I = ∫[0,∞] e^-ktw2 cos(wx) (-w) dw

The integration by parts poses the main problem. I have done a change of variables allowing z = w², although it seems I will have a recursion issue with an extra integral that is unable to be evaluated after each integration by parts.

For integration by parts, I previously let u = sin(x√z) and dv = e^-ktzdz but this doesn't seem to lead anywhere good.

I considered using Euler's formula to replace the cosine but this seems to lead in the wrong direction also.

Any suggestions are appreciated. Thanks!

 

[X89codered89X]

I wish I had more time to help, but you can try doing the substitution of cosine as you suggested in 2). Maybe then try completing the square in the exponent?

Sorry if this isn't much help. t.

 

[jackmell]

2[/SUP], although it seems I will have a recursion issue with an extra integral that is unable to be evaluated after each integration by parts.
Thanks!

You should not get an endless recursion. If:

[tex]I(x)=\int_0^{\infty}e^{-ktw^2} \cos(wx)dw[/tex]
then:

[tex]\frac{dI}{dx}=-\int_0^{\infty} w e^{-ktw^2} sin(wx)dw[/tex]

now, letting [itex] u=\sin(wx)[/itex] and [itex]dv=we^{-ktw^2}[/itex] I get:

[tex]\frac{dI}{dx}=\frac{sin(wx)}{2kt}e^{-ktw^2}-\int_0^{\infty} \frac{1}{2kt} w \cos(wx) e^{-ktw^2}[/tex]
or:

[tex]\frac{dI}{dx}=\left(\frac{e^{ktw^2}}{2kt}\right) \sin(wx)-\frac{w}{2kt} I[/tex]

so that just looks like:

[tex]\frac{dI}{dx}=a \sin(wx)-b I[/tex]

Now, I did this quick and may have some arithemetic errors but I believe that's the general idea. Need to solve the DE for I(x) but I'm not too clear about what to do about the constant of integration or really sure about the method but need to try and see what happens ok?

 

[wxstall]

I have made it to:

∂I/∂x = -(e^-ktw2/2kt) sin(wx) + (x/2kt)I

Solving for I gives:

I = (2kt/x) ∂I/∂x + -(e^-ktw2/x) sin(wx)

Does this need to be solved further? I have done as the problem instructed, but is it acceptable to leave the expression in terms of ∂I/∂x?

 

[jackmell]

Dang it. I messed up in my first post. Didn't apply the limits to the partial integration and some other things.

Ok, no, that's not good enough what you're doing. Not good at all. First, we need to get the correct DE, then we have to ACTUALLY solve the DE for the funtion I(x).

I looked at it again and it seems we have to integrate by parts twice to cycle back to the cosine integral, THEN get the DE, then solve it.

I should have been more careful above. Sorry. Kinda' busy. Will try to look at it again. Really nice problem. :)

 

[wxstall]

That's what I was wondering :)

So from here:

∂I/∂x = -(e-ktw2/2kt) sin(wx) + (x/2kt)I

Evaluating the first term at the limits [0,∞] will give 0. For infinity, the negative exponential becomes 0, and at 0, sin(0)=0. So we should just be left with

∂I/∂x = (x/2kt)I

correct?

From there, simple separation of variables could be used.

Then again, maybe I'm going even further off the path...

 

[wxstall]

Thought I would also share that the Mathematica Online Integrator comes up with this disaster of a solution:

http://integrals.wolfram.com/index.jsp?expr=Exp[-kt(x^2)]*x*Sin[ax]&random=false

 

[jackmell]

That's what I was wondering :)

So from here:

∂I/∂x = -(e-ktw2/2kt) sin(wx) + (x/2kt)I

Evaluating the first term at the limits [0,∞] will give 0. For infinity, the negative exponential becomes 0, and at 0, sin(0)=0. So we should just be left with

∂I/∂x = (x/2kt)I

correct?

Yes, that's correct. Sorry if I caused problems. I went over it carefully and got the correct answer as well.