Integrate[E^(I (x - a) t), {t, -Infinity, Infinity}]
simply returns the result $\int_{-\infty }^{\infty } e^{\,i\, t\,(x\,-a\,)} \, dt$
Why doesn't it yield something like 2 $\pi$ DiracDelta[x-a]?
I think you could if you define a new Integral function of your own and check if your integral is a fourier transform before passing it on to the regular internal integral
In[3]:= FourierTransform[Sqrt[2 Pi], t, (x - a)] Out[3]= 2 \[Pi] DiracDelta[-a + x]
works better
Thanks very much!
Is there a way to make Mathematica remember this rule easily, so that when it's asked to evaluate an integral like
$\int {\text d}^3x\ e^{\ \pm \ {\vec A}.{\vec x} }$
so the result is
(2 $\pi)^3$ DiracDelta[ $\vec A$] ?