Find the Fourier transform of the following function: $$I(x) = \int_0^{1/2} e^{-(x-t)^2}dt$$
I tried using the convolution product, which is defined as: $$(f \ast g)(t) = \int_{-\infty}^{+\infty} dt' f(t') g(t-t')$$
So that I can use the property $\mathcal{F} \{ f \ast g \} = \mathcal{F} \{f \}\mathcal{F}\{g\}$. But it looks like I can't find the functions $f$ and $g$.
Thanks in advance!
Best Answer
Reverse the order of integration $$ \begin{split} \mathcal{F}\{I\}(s) &= \int_{x=-\infty}^{x=\infty} \int_{t=0}^{t=1/2} \exp\left(-(x-t)^2-2\pi i s x\right)\ dt\ dx \\ &= \int_{t=0}^{t=1/2} e^{-t^2}\int_{x=-\infty}^{x=\infty} \exp\left(-x^2-2x (t-i\pi s)\right)\ dx\ dt \end{split} $$ now complete the square in the exponent in the inner integral and you will get a term that will cancel out $e^{-t^2}$ leaving you with an $e^{-t}$ like integral to take wrt $t$.