Is the function $F(z)=\int_0^\infty e^{-zt}\,\Bbb dt$ for $z$ such that $\operatorname{Re}z>0$ holomorphic?
I wanted to prove that it is analytic and find its power series:
$$F(z)={\int_0^\infty e^{-zt}\,\Bbb dt}={\int_0^\infty \sum_{k=0}^\infty \frac{(-zt)^k}{k!}\,\Bbb dt}$$
And if I could change the order of summation and integration it would be over, but it is a complex-valued function under the integral and sum, so Fubini’s theorem does not apply.
I tried this also:
$${\sum_{k=0}^{\infty} \int_0^\infty \frac{(-zt)^k}{k!}\,\Bbb dt}={\sum_{k=0}^{\infty} \int_0^\infty \frac{t^k}{k!}(-z)^k\,\Bbb dt}={\sum_{k=0}^\infty \left(\int_0^\infty \frac{t^k}{k!}\,\Bbb dt\right)(-z)^k}={\int_0^\infty \sum_{k=0}^\infty \frac{t^k}{k!}\,\Bbb dt (-z)^k}={\int_0^\infty \sum_{k=0}^\infty \frac{t^k}{k!}(-z)^k\,\Bbb dt}$$
But the two last equations are suspicious (I changed the order in the second-to-last equation because function is non negative).
Can I do this? Is this function even holomorphic. How can it be proved?
They're power series about $z=0$ for obvious reasons. But things are well behaved for $z\in H=(0,\infty)\times\mathbb{R}$.
Here is a prove that shows directly that the function $$F(z)=\int^\infty_0 e^{-zt}\,dt, \qquad z\in H=(0,\infty)\times\mathbb{R}$$ is holomorphic at any $z\in H$. This relies on basic integration theory.
Clearly $F$ is well defined in $H$ since $f_{z}(t)=e^{-zt}\in L_1(0,\infty)$ for all $z\in H$.
Fix $z_0=a+ib\in H= (0,\infty)\times\mathbb{R}$. Clearly $f_{z_0}(t)=e^{-z_0t}\in L_1(0,\infty)$. For let $\delta>0$ so that $0<a-\delta$. For $h\in\mathbb{C}$ with $|h|<\delta$, $z_0+h\in H$.
Notice that
\begin{align} \Big|\frac{e^{-th}-1}{h}\Big|\leq\frac{e^{|h||t|}-1}{|h|}\leq \frac{e^{\delta|t|}-1}{\delta} \leq \frac{e^{\delta t}+e^{-\delta t}}{\delta}. \end{align}
Then
$$\frac{F(z_0+h)-F(z_0)}{h}=\int^\infty_0 e^{-z_0t} \frac{e^{-th}-1}{h}\,dt$$
Observe that $$\Big|e^{-z_0t} \frac{e^{-ith}-1}{h}\Big|\leq \frac{e^{-(a+\delta)t}+e^{-a(t- \delta)}}{\delta}\in L_1(0,\infty)$$
The conclusion follows from dominated convergence; furthermore,
$$F'(z_0)=-\int^\infty_0 te^{-z_0 t}\,dt$$
Applying a similar argument one obtains that
$$F^{(n)}(z_0)=(-1)^n\int^\infty_0 t^n e^{-z_0 t}\,dt$$
Which yields a power series about $z_0$ as
$$F(z)=\sum_{n\geq0} a_n(z-z_0)^n$$ where $$a_n=(-1)^n\int^\infty_0 t^n e^{-z_0t}\,dt$$