# Fourier Transform Tricky Integral

1. May 25, 2012

### michaelbarret

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 Files:

• ###### maths question.png
File size:
29.1 KB
Views:
103
2. May 25, 2012

### chiro

Hey michaelbarret and welcome to the forums.

Hint: Use the subsitution and try integration by parts twice and see what you get.

3. May 25, 2012

### HallsofIvy

Staff Emeritus
The only thing complicated about that integral is your way of doing it. Instead, do it by parts:
$$\int cos(\omega x)e^{ax}dx$$
Let $u= cos(\omega x$, $dv= e^{ax}dx$ so that $du= -\omega sin(\omega x)dx$ and $v= (1/a)e^{ax}$ so we have
$$\int udv= uv- \int vdu= \frac{1}{a}e^{ax}cos(\omega x)+ \frac{\omega}{a}\int e^{ax}sin(\omega x)dx$$

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

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

Add $\int e^{ax}cos(\omega x) dx$ to both sides and divide by 2.

4. May 25, 2012

### michaelbarret

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

5. May 26, 2012

### algebrat

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.

Last edited: May 26, 2012