Solving Fourier Inverse: Integrals and Techniques for Exam Revision

[usn7564]

Homework Statement

Doing some exam revision and one of the questions from an old exam has me stuck at the last step, simply need to inverse the following

F( \omega ) = \frac{e^{i \omega}}{1+\omega ^2}


We're allowed to use a table on the exams but I cannot find anything quite resembling what I have (nor in any way that I can manipulate it). The inverse formula lead me to an integral that seems to be well over my head or the content of this course.

Is there some specific method to utilize when I have a product of two functions whose Fourier inverse I know? The exponential or denominator on its own wouldn't be a problem.

Thanks

 

[vela]

Use the convolution theorem.

 

[vanhees71]

For the inverse FT just use the formula. I don't know your convention. In theoretical high-energy physics it would read
f(t)=\int_{\mathbb{R}} \mathrm{d} \omega \frac{1}{2 \pi} F(\omega) \exp(-\mathrm{i} \omega t).
Now you can use the theorem of residues easily in your case. You just close the contour in the appropriate upper or lower half-plane for t<0 or t>0 respectively. Your function has only simple poles along the imaginary axis. So there is no big trouble getting the result.

 

[usn7564]

Use the convolution theorem.

Didn't think of going backwards with the convolution theorem, that should certainly work thank you.

For the inverse FT just use the formula. I don't know your convention. In theoretical high-energy physics it would read
f(t)=\int_{\mathbb{R}} \mathrm{d} \omega \frac{1}{2 \pi} F(\omega) \exp(-\mathrm{i} \omega t).
Now you can use the theorem of residues easily in your case. You just close the contour in the appropriate upper or lower half-plane for t<0 or t>0 respectively. Your function has only simple poles along the imaginary axis. So there is no big trouble getting the result.

Only half a year since I finished our course in complex analysis and the fact that I could use residues completely slipped my mind, thanks. Going to need to revise that a tiny bit, but was rather algorithmic if I recall correctly.