## 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
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 Quote by 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
Hey michaelbarret and welcome to the forums.

Hint: Use the subsitution and try integration by parts twice and see what you get.
 Recognitions: Gold Member Science Advisor 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.