New Reply

Fourier Transform Tricky Integral

 
Share Thread
May25-12, 04:05 AM   #1
 

Fourier Transform Tricky Integral


Hi I am trying to analytically calculate the Fourier transform attached.

I am getting really stuck with the integral, can anyone help?

I've attached how far I've got with it, any help much appreciated!

Kind Regards,

Mike
Attached Thumbnails
maths question.png  
PhysOrg.com science news on PhysOrg.com

>> New language discovery reveals linguistic insights
>> US official: Solar plane to help ground energy use (Update)
>> Four microphones, computer algorithm enough to produce 3-D model of simple, convex room
May25-12, 08:41 AM   #2
 
Quote by michaelbarret View Post
Hi I am trying to analytically calculate the Fourier transform attached.

I am getting really stuck with the integral, can anyone help?

I've attached how far I've got with it, any help much appreciated!

Kind Regards,

Mike
Hey michaelbarret and welcome to the forums.

Hint: Use the subsitution and try integration by parts twice and see what you get.
May25-12, 08:46 AM   #3
 
Recognitions:
Gold Membership Gold Member
Science Advisor Science Advisor
Retired Staff Staff Emeritus
The only thing complicated about that integral is your way of doing it. Instead, do it by parts:
[tex]\int cos(\omega x)e^{ax}dx[/tex]
Let [itex]u= cos(\omega x[/itex], [itex]dv= e^{ax}dx[/itex] so that [itex]du= -\omega sin(\omega x)dx[/itex] and [itex]v= (1/a)e^{ax}[/itex] so we have
[tex]\int udv= uv- \int vdu= \frac{1}{a}e^{ax}cos(\omega x)+ \frac{\omega}{a}\int e^{ax}sin(\omega x)dx[/tex]

Now, do it again. Let [itex]u= sin(\omega x)[/itex] and [itex]dv= e^{ax}[/itex] so that [itex]du= \omega cos(\omega x)dx[/itex] and [itex]v= (1/a) e^{ax}[/itex].

Now you have
[tex]\int e^{ax}cos(\omega x) dx= \frac{1}{a}e^{ax}cos(\omega x)+\frac{\omega}{a^2}e^{ax}sin(\omega x)- \frac{\omega^2}{a^2}\int e^{ax}cos(\omega x)dx[/tex]

Add [itex]\int e^{ax}cos(\omega x) dx[/itex] to both sides and divide by 2.
May25-12, 01:13 PM   #4
 

Fourier Transform Tricky Integral


Thanks HallsofIvy for your reply.

I think the biggest thing I'm unsure on is how my original signal can be generalised to ∫cos(wx) e^ax dx. In my case does w=w, x=t(1-qt) and dx=dt?

Also sorry If it's obvious but I dont really understand why I should add ∫e^ax cos(ωx)dx to both sides and divide by 2.

I can't figure out how to integrate the complex exponential associated with the FT either, ∫e^-jwt dt
May26-12, 10:56 AM   #5
 
Yeah last helper might not have noticed the squaring of t in the original problem, it also looks like you didn't write it after the first line. But it looks like you were close to done, I like what you have worked out on the PDF. I think the next goal might be to complete the square and use some substitutions and move things around so you almost have that thing that's equal to root pi over a.
New Reply

Similar Threads for: Fourier Transform Tricky Integral
Thread Forum Replies
Tricky Fourier Transform problem for an exponential function Calculus & Beyond Homework 1
Fourier transform of integral Calculus 8
Tricky Integral (fourier transforms) Calculus & Beyond Homework 5
Fourier Transform / Integral Introductory Physics Homework 3
Fourier integral / transform ? What is it really? Introductory Physics Homework 5