Convolution

It follows directly from convolution theorem and Formula 15 that

$$aR_{q}(t)\ast bR_{p}(t)=abR_{q+p+1}(t).$$

Moreover, if $f_{1}(t)\stackrel {q}{\sim }aR_{q}(t)$ and $f_{2}(t) \stackrel{p}{\sim } bR_{p}(t)$, then  

$$f_{1}(t) \ast f_{2}(t) \stackrel{q+p+1}{\sim } a \cdot b R_{q+p+1}(t).$$ (16)

Proof: Let us express Fourier-transformations of f₁(t) and f₂(t) according to equation (2), taking N=r=q (for f₁(t)) and N=r=p (for f₂(t)). Performing multiplication, we get:

$$F_{1}(\omega)F_{2}(\omega)={ab \over {(i\omega)}^{p+q+2}}+0 \left( {1 \over \omega^{p+q+2}} \right).$$

That is exactly the same that equation (16) expresses in temporal domain. The convolution

$$f(t) \ast R_{q}(t)=f_{q+1}(t)$$

is the q^th integration of the f(t).