Derivative of a Gamma function

To prove $$\Gamma '(x) = \int_0^\infty e^{-t} t^{x-1} \ln t \> dt \quad \quad x>0$$

I.e. why can we put the derivative inside the integral? We have

$$\frac{\Gamma(x+h)-\Gamma(x)}{h}=\int_0^\infty e^{-t} t^{x-1} \left(\frac{t^h-1}{h}\right) dt$$

How to pass to the limit as $h \rightarrow 0$