What is the derivative of $g_T(s) = \int_0^T f(t)e^{-st}dt$, where $f(t)$ be locally integrable and bounded?

Question

Let $f(t)$ be a locally integrable and bounded function defined for all $t \geq 0$. Suppose that $$ g_T(s) = \int_0^T f(t)e^{-st}dt, $$ where $\Re(s) > 1$.

In the paper I read the author says that $g_T(s)$ is clearly entire. However, I can not see this. For instance, what is the derivative of $g_T(s)$?

Answer 1

The derivative is $\int_0^{T} (-t) f(t) e^{-st} dt$. This is finite because $|\int_0^{T} (-t) f(t) e^{-st} dt| \leq \int_0^{T} T|f(t)|dt$. ($|e^{-st}|=e^{-t \Re s } \leq 1$).