Suppose we are considering the following integral: $$ I(s) = \int_1^\infty t^{s-1}e^{-t\lambda}\;dt $$ where $s \in \mathbb{C}$ and $\lambda > 0$ is a fixed constant. I would like to know when this integral is holomorphic. So far I found out that, if instead $s$ were real and the integral proper, so we were to consider $$ \widetilde I(s) = \int_1^N t^{s-1} e^{-t\lambda} \;dt \quad (s \in \mathbb{R}) $$ (with $N > 1$ fixed) then the function would be differentiable in $s$ provided $$ \frac{\partial}{\partial s} \left(t^{s-1} e^{-t\lambda}\right) $$ is jointly continuous (which it is), and then $$ \widetilde I(s) = \int_1^N \frac{\partial}{\partial s} t^{s-1} e^{-t\lambda} \;dt \,. $$ What are the additional conditions that I need to account for the fact that the integral is improper, and for holomorphicity?
Many thanks!
P.S. If someone knows a good source to learn about the theoretical rules for manipulating integrals I'd be very grateful for literature references !!
Assuming that for some $s$ the integral defining $I(s)$ is convergent, then $I'(s)$ is well-defined, too, by: $$I'(s)=\int_{1}^{\infty}t^{s-1}\log(t) \,e^{-\lambda t}\,dt,$$ since the logarithmic factor that arises from differentiation has no way to affect convergence, due to $\log(t)\ll t^{\varepsilon}$. It follows that $I(s)$ is a holomorphic function over the whole complex plane, since: $$ |I(s)|\leq \int_{1}^{+\infty}t^{\text{Re}(s)-1}e^{-\lambda t}\,dt $$ is bounded by $\frac{1}{\lambda}$ if $\text{Re}(s)<1$ and by $\frac{1}{\lambda^{\text{Re}(s)}}\cdot\Gamma(\text{Re}(s))$ otherwise.