What is the inverse Fourier transform of e^{-|x|}?

[MrSeaman]

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

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

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

$$\hat \hat f(-x) = e^{-|x|}$$

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

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

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.

 

[mathman]

I have never seen it derived directly. However, it is easily found in a table of definite integrals.

 

[George Jones]

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

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

One way to do this is by using complex contour integration and the residue theorem.

Integrate

$$\int \frac{1}{1+z^2} e^{i \omega z} \mathrm{d} t$$

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

Repeat for a clockwise semicircle in the lower halfplane.

Regards,
George

 

[MrSeaman]

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...