Some questions about the gamma function 
*

*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? 
 A: A bit late to the game, but here is an answer:
1: To show it is finite:  Write $e^{-x}=e^{-x/2}\cdot e^{-x/2}$.  For every $y>0$ there exists $N$ such that $e^{-x/2}x^{y-1}<1$ when $x>N$ so that $$\int_N^\infty e^{-x} x^{y-1}dx\leq \int_N^\infty e^{-x/2} <\infty.$$  For $x$ between $0,1$ compare to $\int x^{y-1}dx$ which converges whenever $y-1>-1$, so for all $y$.  Lastly bounding $\int_1^N e^{-x}x^{y-1}dx$ I leave to you.  
2: For this part, what you have done so far is good.  All we need to do is switch the last integral with the limit.  To do this, notice that in some neighborhood of radius $\delta$ around $y_0$ we can bound $$\int_0^\infty e^{-x} x^{y-1}dx$$ by similar methods as in 1.  Then, applying the Dominated Convergence Theorem to that neighborhood, we can switch the limit and the integral, solving the problem.
Remark:  For that last part, notice we cannot bound $$\int_0^\infty e^{-x} x^{y-1}dx$$ uniformely for all $y\in(0,\infty)$ since the function blows up at $0$ and at $\infty$.  But fortunately, we can bound it uniformely for all $y$ in any compact subset of $(0,\infty)$, allowing us to use the DCT.
