• Support PF! Buy your school textbooks, materials and every day products Here!

Fourier Transform / Integral

  • Thread starter MrSeaman
  • Start date
  • #1
6
0
Hi,

I got an exam in calculus in a few weeks, and lots of questions coming up. Here's one of them:

We learned that the Fourier Transform of

[tex] f(x) = e^{-|x|} [/tex]
is
[tex] \hat f(\omega) = \sqrt{\frac{2}{\pi}}\frac{1}{1+\omega^2}[/tex]

I've got no problem with this one. Now, since [tex] \hat f(\omega)[/tex] is Lebesgue - integrable, the inverse Fourier transform exists and should be

[tex] \hat \hat f(-x) = e^{-|x|} [/tex]

To show this the 'hard way', I want to calculate the integral

[tex] \sqrt{\frac{1}{2 \pi}} \int \limits_{-\infty}^\infty \frac{1}{1+t^2} e^{i \omega t} \mathrm{d} t [/tex]

Well, I just don't know how to do this one. Partial Integration doesn't seem to work, and I can't find neither a good substition nor a clever use of Fubini's Theorem.

Would be thankful for any help.
 

Answers and Replies

  • #2
mathman
Science Advisor
7,739
406
I have never seen it derived directly. However, it is easily found in a table of definite integrals.
 
  • #3
George Jones
Staff Emeritus
Science Advisor
Gold Member
7,259
790
MrSeaman said:
To show this the 'hard way', I want to calculate the integral

[tex] \sqrt{\frac{1}{2 \pi}} \int \limits_{-\infty}^\infty \frac{1}{1+t^2} e^{i \omega t} \mathrm{d} t [/tex]
One way to do this is by using complex contour integration and the residue theorem.

Integrate

[tex] \int \frac{1}{1+z^2} e^{i \omega z} \mathrm{d} t [/tex]

along the closed contour that consists of the portion of the real axis from [itex]-R[/itex] to [itex]R[/itex] together with a closing counterclockwise semicircle of radius [itex]R[/itex] in the upper halfplane. Let [itex]R\rightarrow \infty[/itex]. Then:the closed contour encloses a pole at [itex]z=i[/itex]; the contibution from the semicircle goes to zero.

Repeat for a clockwise semicircle in the lower halfplane.

Regards,
George
 
Last edited:
  • #4
6
0
As I was just told, it seems to be done by using the residue theorem of complex analysis.

edit: George was faster, thanks! Working it out...
 

Related Threads for: Fourier Transform / Integral

Replies
5
Views
3K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
1
Views
690
  • Last Post
Replies
0
Views
1K
Top