Calculating Fourier Transform on TI-89 for Non-Integrable Functions?

[kolycholy]

how would i calculate Fourier transform of functions such as 1/(1+t^2)?
because if you try to integrate the product of the above function and e^(-jxt), you would realize it's nonintegrable or something
at least my ti-89 does not calculate it for me.
any other way?

 

[d_leet]

Integration by parts perhaps, although I can imagine that getting very messy very fast. Is it possible that you could integrate numerically? I think that would at least work in finding a Fourier series, but I'm not certain.

 

[jbusc]

Mathematica can calculate the Fourier transform, according to that it's sqrt(pi/2)/e^abs(w)

however, I'm not completely sure how you would go about doing it manually. I would agree that integration by parts is the first step to try.

 

[kolycholy]

Mathematica can calculate the Fourier transform, according to that it's sqrt(pi/2)/e^abs(w)

however, I'm not completely sure how you would go about doing it manually. I would agree that integration by parts is the first step to try.

the answer given in the book is pi*e^abs(w).
One of them is right. or maybe both of them are right?

 

[jpr0]

You would need to solve this integral via contour integration (at least this is the first thing that comes to mind). Have a look on google for "functions of a complex variable" or "residue theorem" or simply "contour integral" or something like this. Anyway, contour integration allows you evaluate these types of integrals pretty quickly.

By the way the answers stated here differ by a factor of $\pi^{1/2}$. This is probably because one of you is using the forward Fourier transform as being defiend with a prefactor of $1/2\pi$ and the backward transform having a prefactor of 1, while the other one is using the prefactor $[1/2\pi]^{1/2}$ for both the forward and backward transforms.

 

[George Jones]

jpr0 is right - contour integration works. See https://www.physicsforums.com/showthread.php?t=90290".

 

[jpr0]

Hi George, thanks for the thread.
I should say that in the post on this thread, the contour you pick (semicircle in the upper or lower half plane) is determined by the sign of $\omega$.