Holomorphicity of an integral function

Question

I am looking for a way to prove that the the function \begin{equation} F(a) = \int_{-\infty}^{\infty} e^{-ax^2}dx \end{equation} is holomorphic for all $a \in \mathbb{C}$ such that $\Re(a) > 0$. I don't really know how to apply Morera's theorem on this example, or whether it is the right approach. All hints are welcome.

Answer 1

The statement is true when the interval of integration is finite, i.e. for $$F_R(z) = \int_{-R}^R e^{-zx^2}dx.$$ What you want to do is show that the integral converges uniformly on strips of the form $\textrm{Re}(z) > \delta$. Any triangle $T$ in the half-plane $\textrm{Re}(z) > 0$ is contained in such a strip. We want to show that $$\int_TF = \int_T\lim_{R \to \infty}F_R = 0,$$ as then $F$ is holomorphic by Morera's theorem. The uniform convergence allows us to switch the integral and the limit, thus $$\int_TF = \lim_{R \to \infty}\int_TF_R.$$ But since $F_R$ is holomorphic, the right-hand side integral is equal to zero by Goursat's theorem, thus the limit is also zero.

Answer 2

Hint: \begin{equation} \dfrac{\partial}{\partial\bar{a}}F(a) = \int_{-\infty}^{\infty} \dfrac{\partial}{\partial\bar{a}}e^{-ax^2}dx=0 \end{equation} iff ${\bf Re}\,a > 0$.