Show that integral is analytic Let $h:[0,\infty)$ be an integrable function. Prove that the function $$g(z)=\int_0^\infty h(t)e^{tz}\,dt$$ is analytic on $\{z=x+yi:x<0,y\in\mathbb{R}\}$.
How do I start for this question?
I know Cauchy's integral theorem, but that does not seem relevant here?
Also, I am not sure if this is related to my previous question (Lebesgue integral question (double integral)).
$$\int_{-\infty}^\infty g(x)^2h(x)\,dx=\int_0^\infty\int_{\{t\in\mathbb{R}:g(t)>x\}}2h(t)x\,dtdx.$$
Thanks for any help.
 A: To show this function is analytic, you need to show that the Cauchy Riemann equations are valid; equivalently, the function satisfies
$$ \frac{\partial g}{\partial \bar z} 
= \frac{1}{2}\bigg( \frac{\partial g}{\partial x} - \frac{1}{i}\frac{\partial g}{\partial y} \bigg) = 0.$$
If you are unfamiliar with this very useful formulation of the Cauchy Riemann equations, it's a nice, quick exercise to verify they are equivalent. We'll use it to finish the problem.
Now, we want to apply $\partial/ \partial \bar z$ to $g(z)$, and we want to show this is $0.$ To accomplish this, we want to switch the order of differentiation and integration for both $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial y}$. For example, the integrand satisfies the hypothesis of Theorem 2.27 in Folland's Real Analysis. To apply this, just differentiate the integrand and use the fact that $h(t)$ is integrable and $te^{tz}$ decays as $t$ goes to infinity because of the domain's restriction to $x<0$ to show the derivatives are integrable. The upshot is we can exchange the order of $\frac{\partial}{\partial \bar z}$ and integration. Now, since $e^{tz}$ is analytic, it satisfies the Cauchy Riemann equations, so 
$$\frac{\partial}{\partial \bar z}e^{tz} =0.$$
Therefore, the integral is 0, and $g$ is analytic, as desired.
