## Proving that we can differentiate the gamma function

Hi,

So we have $\Gamma(s)=\int_0^\infty t^{s-1} e^{-t} ~dt=\int_0^\infty f(s,t) ~dt$, $s \in \mathbb{R}^{>0}$
In order to say that $\Gamma'(s)=\int_0^\infty \frac{\partial f}{\partial s} (s,t) ~dt=\int_0^\infty t^{s-1} \ln(t) e^{-t} ~dt$, we have to prove, in particular, that there exists an integrable function g such that :
$\left|\frac{\partial f}{\partial s}(s,t)\right|\leqslant g(t)$

My idea is as following :
For any n, there exists a>1 such that $e^t>t^n \Rightarrow e^{-t}
For any t>a, $\ln(t)\leqslant t$
So we have :
$\left|\frac{\partial f}{\partial s}(s,t)\right|\leqslant \left|t^{s-1} \cdot t \cdot t^{-n}\right|$

By letting $n=s+2$, we have :
$\left|\frac{\partial f}{\partial s}(s,t)\right|\leqslant \frac{1}{t^2}$, which is integrable over $[a,+\infty)$

So I wanted to know if this part is correct... especially the red sentence :s

And for the integral over 0 and $\epsilon$, I don't know how to dominate