Proof that a Function is Entire For every $z \in \mathbb{C}$, let us define
\begin{equation}
H(z) = \int_{0}^{\infty} t^{-t} e^{tz} dt.
\end{equation}
I have tried to prove that $H$ is holomorphic in the whole plane and that
\begin{equation}
H(z) = \sum_{n=0}^{\infty} \frac{z^n}{n!} \int_{0}^{\infty} t^{n-t} dt,
\end{equation}
but I am stuck. One way would be to prove directly that $H$ is complex-differentiable in the whole plane and 
\begin{equation}
H'(z) = \int_{0}^{\infty} t^{1-t} e^{tz} dt,
\end{equation}
and then iterate the argument to compute the higher order derivatives. Another way would be to prove that the series
\begin{equation}
\sum_{n=0}^{\infty} \frac{|z|^n}{n!} \int_{0}^{\infty} t^{n-t} dt,
\end{equation}
converges for every $z$, since this would justify the interchange of integral and limit, so that 
\begin{equation}
H(z) = \sum_{n=0}^{\infty} \frac{z^n}{n!} \int_{0}^{\infty} t^{n-t} dt,
\end{equation}
and clearly $H$ would turn to be analytic in the whole plane. Unfortunately, I couldn't fill the details of the two proofs I have in mind, and I am stuck now. Any help is welcome.
 A: Yet another approach is using complex contour integrals.
By Morera's theorem, a continuous complex function is entire if its integral over every triangle $\Delta \subset \mathbb C$ equals zero.
Continuity of $H(z)$ follows from the Dominated convergence theorem. To show that $\oint_\Delta H(z)dz = 0$, you may apply the Fubini-Tonelli theorem to each side of $\Delta$ to change the order of integration.
You may also use this interchanging trick to compute the derivatives of $H$ with Cauchy's formula: for point $z$ insinde $\Delta$
$$
H^{(n)}(z) = \frac{n!}{2\pi i} \oint\limits_{\;\Delta} \frac{H(\zeta)\,d\zeta}{(\zeta-z)^{n+1}}
$$
A: Now I see how to complete the two sketches of proof I quoted in the post. Let $(w_n)$ a sequence of non-zero complex numbers converging to 0, and such that $|w_n| < M$ for all $n$. Then we have (use the power series expansion of $e^z$ and the convexity of $f(x)=e^{x}$) 
\begin{equation}
\left| \frac{e^{tw_n} - 1}{tw_n} \right| \leq \frac{e^{t |w_n|} - 1}{t|w_n|} \leq e^{t M},
\end{equation}
and so we can use the Dominated Converge Theorem to show that $H$ is complex-differentiable, with
\begin{equation}
H'(z) = \int_{0}^{\infty} t^{1-t} e^{tz} dt.
\end{equation}
Now, we can repeat the argument to show that
\begin{equation}
H^{(n)}(z) = \int_{0}^{\infty} t^{n-t} e^{tz} dt.
\end{equation}
The other proof I sketched is based on founding a bound for the integral
\begin{equation}
\int_{0}^{\infty} t^{n-t} dt.
\end{equation}
Define $f_n(t) = t^{n - t/2}$. For each n, $f_n$ is bounded, with maximum $M_n$. If $s_n$ is the point in which $f_n$ reaches the maximum, we have, by using the first order condition $n - s_n/2 = (s_n/2) \log s_n$, $s_n \log(s_n) \leq 2n$, and $M_n =  e^{(s_n/2)\log^2 (s_n)}$, so that $\sqrt[n]{M_n/(n!)} \leq s_n /\sqrt[n]{n!} \leq 2n/(\log(s_n)\sqrt[n]{n!})$, which tends to 0 as $n \rightarrow \infty$, since $s_n \rightarrow \infty$, and $n/ \sqrt[n]{n!} \rightarrow e$ (use Stirling's formula). So by Cauchy-Hadamard formula, the power series
\begin{equation}
\sum_{n=0}^{\infty} \frac{z^n}{n!} \int_{0}^{\infty} t^{n-t} dt,
\end{equation}
has radius of convergence $R=\infty$, and we have done.
