- #1
MrSeaman
- 9
- 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.
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.