Show that $\int_1^{\infty} t^{x-1} e^{-t} dt$ is entire. Two Notes :

*

*The following is a half-line claim in a several pages of proof for some theorem in a book that I am studying. It is not an exercise.


*I can't include any "my attempt" because I don't have more background than Churchill's Complex Analysis book which doesn't include integral representation of analytic functions and this book is not about complex analysis so left out details.

$$\Gamma(x) = \int_0^{\infty} t^{x-1} e^{-t} dt = \int_0^1 t^{x-1} e^{-t} dt + \int_1^{\infty} t^{x-1} e^{-t} dt := I_1 + I_2.$$
My questions are :

*

*Why $I_2$ is an entire function?


*Is there a self-study book on be or not be analytic of integrals, esp improper integrals?
 A: Morera proof. details spelled out.
Let $I_2(x) := \int_1^{\infty} t^{x-1} e^{-t} dt$.  I claim
$I_2$ is an entire function
First, for a fixed real $t > 1$, the function
$$
F_t(x) := t^{x-1}e^{-t} = e^{(x-1)\log(t)-t}
$$
is entire.  (Choose the positive logarithm of the number $t > 1$.)
Let $\gamma$ be a rectifiable closed curve in $\mathbb C$.  For every $t>0$ we have
$$
\int_\gamma F_t(x)\;dx = 0 .
$$
Next consider
$$
\int_\gamma I_2(x)\;dx = \int_\gamma \left[\int_1^\infty F_t(x)\;dt\right]\;dx
\tag1$$
We want to interchange the integrals.  Note that $\gamma$ is a bounded set in $\mathbb C$.  There s a constant $K>0$ so that $|x-1| < K$ for all $x \in \gamma$.
Compute
$$
\int_\gamma \left[\int_1^\infty |F_t(x)|\;dt\right]\;|dx|
=\int_\gamma \left[\int_1^\infty |e^{(x-1)\log(t)-t}|\;dt\right]\;|dx|
=\int_\gamma \left[\int_1^\infty e^{\operatorname{Re}(x-1)\log(t)-t}\;dt\right]\;|dx|
\le \int_\gamma \left[\int_0^\infty e^{K\log(t)-t}\;dt\right]\;|dx|
=\int_\gamma \Gamma(K+1)\;|dx| < +\infty .
$$
(I wrote $|dx|$ for the arc-length measure on $\gamma$.  This
is a finite measure since $\gamma$ is rectifiable.)
(We used: for $t>1$, $\log t > 0$.)
Therefore, by Fubini's theorem, from $(1)$ we get
$$
\int_\gamma I_2(x)\;dx
= \int_1^\infty \left[\int_\gamma F_t(x)\;dx\right]\;dt
= \int_1^\infty 0\;dt = 0.
\tag2$$
But $(2)$ holds for all rectifiable contours $\gamma$.  So from Morera's theorem we
conclude that $I_2(x)$ is entire.
