Suppose $g$ is a continuous function such that the integral $\int_0^\infty|g(t)|dt$ converges. Is $$F(z)=\int_0^\infty g(t)\sin(zt)dt$$ analytic? And if so, in what region?
My attempt: $$F(z)=\int_0^\infty g(t)\sin(zt)dt=\int_0^\infty g(t)\frac{e^{izt}-e^{-izt}}{2i}dt$$ $$=\int_0^\infty g(t)\frac{e^{izt}}{2i}dt-\int_0^\infty g(t)\frac{e^{-izt}}{2i}dt$$ This is equivalent to $$\frac{\mathcal{L}[g(t)](iz)}{2i}-\frac{\mathcal{L}[g(t)](-iz)}{2i},$$ where $\mathcal{L}$ is the Laplace transform. Since $g$ is continuous and convergent, the Laplace transform is analytic where $Re(iz)>0$ and $Re(-iz)>0$, which is impossible, so $F(z)$ is not analytic. Is this correct?