Let $f$ be a function on R and $\hat f$ its Fourier transform.
Consider a truncated version of $f$ called $\bar f$ whose value outside an interval is $0$. Formally, $\bar f(x) = f(x) * 1_{x \in I}$ where $I$ is a closed interval.
What can we say about $F{\bar f}$, the Fourier transform of $\bar f$, with respect to the original transform $\hat f$?
I guess $F{\bar f} = \hat f \times F{1_{x\in I}}$ where $\times$ is the convolution operator. Can I have a discrete representation of $F{\bar f}$ as a function of $\hat f$?
Hopefully this answers your question: Are you able to find/compute the inverse Fourier transform of the other function you are trying to take the $L^2$ inner product with? If your other function is $Fg \in L^2$, then by Parseval's Theorem you have that $\langle \bar{f} , g \rangle = \langle F\bar{f}, Fg \rangle$ (there may be a constant in front depending on which of the many versions of the Fourier transform you use). This avoids having to figure out what $Ff \star F1_I$ looks like as long as you can figure out what $g$ is and then you can use typical numerical integration techniques.