Standard deviation is diverging... I've made some mistake. I'm supposed to find the standard deviation of the Fourier transformed function $f(t) = e^{-|t|/a}$.  Just a note: to make it easier on you guys I'll be leaving off the normalization constant.  Let me know for some reason you think I should add it back in.
The Fourier transform of this is $$\hat f(\omega) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} e^{-|t|/a - i\omega t}\ dt = \frac{1}{\sqrt{2\pi}}\left[\int_0^{\infty} e^{-t((1/a)-ik)}\ dt + \int_0^{\infty}e^{-t((1/a)+ik)}\ dt\right] = \cdots = \frac{1}{\sqrt{2\pi}(a^2\omega^2+1)}$$
I don't think that I've done that part wrong as I used this result from an integral table:
$$\int_0^{\infty} e^{-ax}\ dx = \frac{1}{a},\qquad \text{for }\operatorname{Re}(a)>0$$
But then to find the standard deviation I do the following integral
$$\sigma_{\omega} = \sqrt{\langle(\omega - \langle \omega\rangle )^2\rangle} = \sqrt{\langle \omega^2\rangle} = \int_{-\infty}^{\infty} \frac{\omega^2}{a^2\omega^2 + 1}\ d\omega$$  where $\langle(\omega - \langle \omega\rangle )^2\rangle = \langle \omega^2\rangle$ because $\langle \omega \rangle = 0$.  But this last integral clearly diverges (it's even and approaches $1$ as $\omega \to \pm\infty$).
So I must have made a mistake somewhere.  Can anyone explain to me where I've erred?  Thanks.
 A: Comment: This is a Laplace distribution with median 0 and scale parameter $\alpha.$
The Wikipedia article on this distribution states that its variance is $2\alpha^2.$ It also givee the correct characteristic function ('CF').
I don't think it is difficult to get the variance directly
from the density function. 
If your homework is to get the variance via
the Fourier transform (known as a 'characteristic function' in probability),
it may be helpful to know the answer. If your assignment is just to find the variance,
then I think you may find a more direct approach easier.
This is sometimes called the double exponential
distribution. To see why, look at the density function. Also,
consider that if $X_1, X_2$ are iid $Exp(rate = 1/\alpha),$ then $Y = X_1 - X_2$ is 
$Laplace(0, \alpha).$ 

Here is a brief simulation in R statistical software, based on a million
realizations of $Y$ with $\alpha = 4,$
which gives numerical results accurate to about two or three places.
It is followed by approximate values of $E(Y)$ and $SD(Y) = \sqrt{Var(Y)},$
and a histogram of the simulated distribution of $Laplace(0, 4)$
along with the actual density function.
m = 10^6;  alp = 4
x1 = rexp(m, 1/alp);  x2 = rexp(m, 1/alp)
y = x1 - x2 
mean(y);  sd(y);  sqrt(2*alp^2)
## -0.001015087  # aprx E(Y) = 0
## 5.659927      # aprx SD(Y)
## 5.656854      # exact SD(Y)
hist(y, br=50, prob=T, col="skyblue")
curve((.5/alp)*exp(-abs(x)/alp), lwd=2, col="blue", add=T)


