How to find If Mellin transform has no poles?

Question

I have the following integral: $$ F(s) = \int\limits_{\alpha}^{\delta} t^{s-3} f(t) dt. $$ The function $f(t)$ is real-valued, continuous, monotonically increasing on $[0,\delta]$, bounded, and is non-0 on the subinterval $[\alpha,\delta]$. $F(s)$ is a function with a complex number input which is analytic wherever it is defined. I would like to prove that $F(s)$ has no poles in its analytic continuation (that the function is entire). I have experimentally seen that this is the case, but I am struggling to actually prove it. $F(s)$ looks very much like a Mellin Transform, but I am unsure if that is actually helpful.

Answer 1

Premise. I answer to the question by showing that the integral $F(s)$ described in the OP is de facto an entire function: there are theorems due to Polya, Broggi and Pellegrino that give necessary and sufficient conditions for an analytic functions to have only pole-type singularities, but this does not seem what it is asked for.

Answer. The function $F(s)$ is entire holomorphic, and I think that the best way to proceed in order to prove this is is the following. Let's change the variable under the integral sign as it is done when passing from the Mellin to the Laplace transform and viceversa, i.e. $t=\ln z\iff z=e^t$. Then $$\DeclareMathOperator*{\d}{d\!} F(s) = \int\limits_{\alpha}^{\delta} t^{s-3} f(t) \d t = \int\limits_{\ln \alpha}^{\ln\delta} e^{z(s-2)}f(e^z)\d z $$ Now the integral at the right side can be shown to be holomorphic in every $z\in\Bbb C$ by proceeding along one of the two following ways:

1. by showing that it satisfies the Cauchy-Riemann equation i.e. $$ \frac{\partial F(s)}{\partial\bar{s}}=0\quad\forall s\in \Bbb C. $$ Explicitly $$ \begin{split} \frac{\partial F(s)}{\partial x} & =\frac{\partial F(x+iy)}{\partial x} =\int\limits_{\ln \alpha}^{\ln\delta} \frac{\partial e^{z(x+iy-2)}}{\partial x}f(e^z)\d z\\ & \qquad\qquad= \int\limits_{\ln \alpha}^{\ln\delta} ze^{z(s-2)}f(e^z)\d z\\ &= -i\int\limits_{\ln \alpha}^{\ln\delta} \frac{\partial e^{z(x+iy-2)}}{\partial y}f(e^z)\d z = -i\frac{\partial F(x+iy)}{\partial y} =-i\frac{\partial F(s)}{\partial y} \end{split} $$ or otherwise
2. by simply differentiating $F(s)$ under the integral sign, i.e. $$ \frac{\partial F(s)}{\partial{s}} = \int\limits_{\ln \alpha}^{\ln\delta} ze^{z(s-2)}f(e^z)\d z $$ and seeing that the integral exists and is finite for every $s\in\Bbb C$.

Final note. Each one of the two ways describes relies on the fact that $f(t)$ is bounded and thus the integral $$ F(s) = \int\limits_{\ln \alpha}^{\ln\delta} e^{z(s-2)}f(e^z)\d z $$ is bounded uniformly on every bounded neighborhood of the variable $s\in\Bbb C$: this implies that you can differentiate under the integral sign.