Convolution Theorem involving a constant.

Question

Should one have f(x) and g(x), and wants $f(x) \ast g(x) $ from what i understand this can be quite difficult, however should $f(x)=\alpha$, a constant, what is $f(x) \ast g(x) $?

Answer 1

Then we would have $$ (f\ast g)(x) = \int f(x-y)g(y) \mathrm{d}y= \alpha \int g(y) \mathrm{d}y $$ so just a multiple of the integral over $g$. Especially the convolution itself would be constant then.