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

Question

Two Notes :

1. 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.

2. 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.

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

My questions are :

1. Why $I_2$ is an entire function?

2. Is there a self-study book on be or not be analytic of integrals, esp improper integrals?

Answer 1

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.