Is the Lebesgue Integral of the Convolution of the Sinc function and a $L^1[R]$ Function Equal to the Product of the Integrals?

Question

There is a well known theorem which states that if $f, g\in L^1[R]$, then $f*g$, the convolution of $f$ and $g$, is also in $L^1[R]$, and in fact \begin{equation} \int_{R}(f*g)(x)\ dx=\int_Rf(x)\ dx\int_R g(x)\ dx. (1) \end{equation} Is this equality between these integrals in (1) still true if $f(x)=\frac{sin(\pi x)}{\pi x}$ and $g\in L^1[R]$ (where the integral of the Sinc function is an improper Riemann integral)? If this is true, I am interested in obtaining a reference, including any reference on this website (since this question seems a bit standard).

Answer 1

Let $F$ be the antiderivative of $f$ which vanishes at $0$. Since $F$ is bounded with bounded derivative, and since $g$ is integrable, an application of Lebesgue theorem of derivation of integrals with parameter show that function $F*g$ is differentiable with derivative $f*g$. And application of Lebesgue dominated convergence theorem shows also that $$F*g(x) = \int F(x-y)g(y)dy\to \int F(\pm\infty)g(y)dy \text{ as } x \to \pm\infty.$$ Hence the generalised integral of $f*g$ exists and equals $[F(+\infty)-F(-\infty)] \times \int g(y)dy$.