Show that $\Gamma(y) = \int_0^{\infty}{e^{-x}x^{y-1}\,dx}$ is finite for $y>0$ both as an improper Riemann integral and as a Lebesgue integral.
Show $\Gamma'(y) = \int_0^{\infty}{e^{-x}x^{y-1}\ln{x}\,dx}$ for $y>0$.
For one: I've tried simply integrating it as an improper Riemann integral, but you always end up with another integral of the "same type" (which is how you eventually show $\Gamma(y+1)=y\Gamma(y)$ ). How do I get around this? As for the Lebesgue integral, I think it'd be easiest to compare the integrand to a larger function whose integral converges, but I haven't come up with a good candidate.
For two: Fix $y_0>0$. Write $$\Gamma'(y) = \lim_{y\to y_0}{\int_0^{\infty}{\frac{e^{-x}x^{y-1}-e^{-x}x^{y_0-1}}{y-y_0}\,dx}}\,.$$ By the MVT, there exists $\eta$ between $y$ and $y_0$ such that the above limit is equal to $$\lim_{y\to y_0}{\int_0^{\infty}{e^{-x}x^{\eta-1}\ln{x}\,dx}}\,.$$ But now I'm not sure what to do. This is similar to a previous question I posted; for that problem we knew the derivative of the original integrand was bounded, so we applied the bounded convergence theorem. Would it be enough to prove that the derivative of my integrand is bounded, and apply BCT?