When ones tries to show the convolution theorem:
$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(x') g(x-x')e^{-ikx}dx'dx = \int_{-\infty}^{\infty}dx'f(x') \int_{-\infty}^{\infty}g(x-x')e^{-ikx}dx $
then, it is made a change of variable, for instance, $y=x-x'$, such that
$\int_{-\infty}^{\infty}dx'f(x') \int_{-\infty}^{\infty}g(x-x')e^{-ikx}dx=\int_{-\infty}^{\infty}e^{-ikx'}dx'f(x') \int_{-\infty}^{\infty}g(y)e^{-iky}dx = \tilde{f}(k)\tilde{g}(k)$.
But why the limits on the integration of $g$ were not changes? Like to $\int_{-\infty+x'}^{\infty+x'}g(y)e^{-iky}dx$. First, I cannot see why this is irrelevant since $x'$ also goes to $\infty$.
This concern arouse to me when trying to investigate when the limits of the convolution are not infinte. The convolution theorem in finite domain would become
$\int_{-L}^{L}\int_{-L}^{L}f(x') g(x-x')e^{-ikx}dx'dx = \int_{-L}^{L}dx'f(x') \int_{-L}^{L}g(x-x')e^{-ikx}dx $
Second, but it is clear that now the intervals change
$\int_{-L}^{L}\int_{-L}^{L}f(x) g(x-x')e^{-ikx}dx'dx = \int_{-L}^{L}dx'f(x') e^{-ikx'}\int_{-L+x'}^{L+x'}g(y)e^{-iky}dx $
So, in this case is not possible to say that,
$\int_{-L}^{L}\int_{-L}^{L}f(x) g(x-x')e^{-ikx}dxdx'= \hat{f}(k)\hat{g}(k)$
where $\hat{f}=\int_{-L}^{L}f(x)e^{ikx}dx$ and same for $\hat{g}$?
Thank you in advance, this is my first post here.
tom