# Weighted integral of a convolution

Afer some experiments, it seems to me that:

the integral (in the entire domain) of a convolution $$(f*s)(x)$$, of a filter function $$f(x)$$ with a a signal function $$s(x)$$, multiplied by a weight function $$w(x)$$,

$$\int (f*s)(x) \cdot w(x) \, dx$$

, when all functions are integrable,

is equal to the integral of the signal function $$s(x)$$ weighted by the cross-correlation $$(f\star w)(x)$$ of the filter function $$f(x)$$ with the weight function $$w(x)$$,

$$\int (f\star w)(x) \cdot s(x) \, dx$$

Is that right?

If it is, what is the proof?

If it is not, is there something similar that let one merge the filter and the weight functions to create a new weight function?

If I am not mistaken, this is just changing the order of integration (which has to be justified by integrability conditions): \begin{align} \int (f*s)(x)w(x) dx & = \int\int f(y-x)s(y)dy\, w(x)dx\\ & = \int\int f(y-x)w(x)dx\, s(y)dy\\ & = \int (f\star w)(y)s(y)dy. \end{align}