Hi EveryoneI asked Mathematica to evaluate the Fourier transform of $f(x) = e^{-x}$, and it returned $\hat f(t) = \delta (i+t)$, where $\delta$ is the Dirac delta function. I confirmed this calculation by taking the Fourier inversion of $\delta (i+t)$ (on Mathematica) and I did recover my original function. But here's my question, the Fourier inversion says$$f(x) = \int_{-\infty}^\infty \hat f(t) e^{i t x} dt. $$If you put $\hat f(t) = \delta (i+t)$ in that formula, shouldn't you get $f(x) = 0$? Because the integral is taken from t = minus infinity to infinity, and nowhere will $t$ ever equal to -i. Therefore, shouldn't the Dirac Delta kill the whole thing?

The problem here is out-of-control formal manipulations.Look at your original fourier trasnform integral if t is real as x -> -infinity. It is grossly divergent.Now, formally, int e^(i t x) dx = 2 pi delta(t), but this needs to be used carefully. Yes, this is also formal, but at least the improper behavior of the integrand is -bounded- here at +- infinity. You can get to this "less singular" integral from your original integral by generalizing t to complex t = t_r + i t_i and letting t_i = -i.Then, the inversion integral must -also- be along the contour t_i = -i. This justifies and explains the final result.If all this makes you unhappy, you need to look up the theory of distributions which is reallythe mathematically correct way to do delta functions.regards,

Thanks for the reply. I'll have a think about it. (4 hours later)Problem completely resolved. Thanks again.