The problem is to find the derivative of Gamma function $\Gamma (y) = \int_0^{ + \infty } {{e^{ - x}}{x^{y - 1}}dx} $ using dominated convergence theorem. Although the following content is lengthy, most of them are background. You can scroll all the way down to my questions at the end in bold. My major problem is that I don't understand two inequities in the proof.
The dominated convergence theorem (DCT).
DCT states that for a sequence of function ${f_n}(x)$ defined on $E$ converging to $f$, if $|{f_n}(x)| \le g(x)$ where $g(x)$ is a Lebesgue integrable function on $E$, i.e. $\int_E {g(x)d\mu } < + \infty $, then $\lim_{n\to\infty} \int_E f_n\,d\mu = \int_E \lim_{n\to\infty}f_n\,d\mu= \int_E f\,d\mu$
Step 1.
$\begin{array}{l} \Gamma '(y) = \mathop {\lim }\limits_{h \to 0} \frac{{\Gamma (y + h) - \Gamma (y)}}{h}\\ = \mathop {\lim }\limits_{h \to 0} \frac{1}{h}[\int_0^\infty {{e^{ - x}}{x^{y + h - 1}}} dx - \int_0^\infty {{e^{ - x}}{x^{y - 1}}dx} ]\\ = \mathop {\lim }\limits_{h \to 0} \int_{[0,\infty )} {\frac{{{e^{ - x}}({x^{y + h - 1}} - {x^{y - 1}})}}{h}} dx \end{array}$
Step 2.
By mean value theorem, $\exists \xi\in(y,y+h)$ such that $(\frac{{d{x^{y - 1}}}}{{dy}})(\xi ) = {x^{\xi - 1}}\ln x = \frac{{{x^{y + h - 1}} - {x^{y - 1}}}}{h}$, thus $\Gamma '(y) = \mathop {\lim }\limits_{h \to 0} \int_{[0,\infty )} {{e^{ - x}}{x^{\xi - 1}}\ln xdx} $.
Note that $\xi\in(y,y+h)$, so $\xi$ is actually dependent on $h,y$,so we rewrite it as $\xi_{h,y}$. Thus $\Gamma '(y) = \mathop {\lim }\limits_{h \to 0} \int_{[0,\infty )} {{e^{ - x}}{x^{\xi_{h,y} - 1}}\ln xdx} $.
Denote ${f_h} = {e^{ - x}}{x^{{\xi _{h,y}} - 1}}\ln x$. Since $\mathop {\lim }\limits_{h \to 0} {\xi _{h,y}} = y$, we have $\mathop {\lim }\limits_{h \to 0} {f_h} = {e^{ - x}}{x^{y - 1}}\ln x$. Now if we can prove $|{f_h}(x)| \le g$ for some Lebesgue integrable function $g(x)$ on $[0,\infty)$, then
$\begin{array}{l} \Gamma '(y) = \mathop {\lim }\limits_{h \to 0} \int_{[0,\infty )} {{e^{ - x}}{x^{{\xi _{h,y}} - 1}}\ln xdx} \\ = \int_{[0,\infty )} {\mathop {\lim }\limits_{h \to 0} {e^{ - x}}{x^{{\xi _{h,y}} - 1}}\ln xdx} \\ = \int_{[0,\infty )} {{e^{ - x}}{x^{y - 1}}\ln xdx} \end{array}$
and we are done.
In step 2, I have one question. The index of $f_h(x)$ is continuous, so DCT holds when the index of function sequence is continuous? Can anyone provide a reference?
Step 3. is to prove $|{f_h}(x)| \le g(x)$ for some Lebesgue integrable function $g(x)$ on $[0,\infty)$, and
Since $\xi_{h,y} \in (y,y+h)$ and $h\rightarrow 0$, we can choose a fixed value $0<\delta<1$ such that $\xi_{h,y} \in (y-\delta,y+\delta)$ holds when $|h|$ is sufficiently small. Then
$\left| {{e^{ - x}}{x^{{\xi _{h,y}} - 1}}\ln x} \right| \le {e^{ - x}}\left| {\ln x} \right|{x^{y - \delta - 1}} \le C{e^{ - x}}|x{|^{y - 2\delta - 1}}$ when $0<x<1$
$\left| {{e^{ - x}}{x^{{\xi _{h,y}} - 1}}\ln x} \right| \le ({e^{ - x}}\ln x){x^{y + \delta - 1}} \le C{e^{ - \frac{x}{2}}}$ when $x\ge 1$
I have a major problem with this step. Why on earth the two inequities hold? I think $C$ represents a constant, and I under stand $\left| {{e^{ - x}}{x^{{\xi _{h,y}} - 1}}\ln x} \right| \le {e^{ - x}}\left| {\ln x} \right|{x^{y - \delta - 1}}$ and $\left| {{e^{ - x}}{x^{{\xi _{h,y}} - 1}}\ln x} \right| \le ({e^{ - x}}\ln x){x^{y + \delta - 1}}$, but I DON'T UNDERSTAND why inequity ${e^{ - x}}\left| {\ln x} \right|{x^{y - \delta - 1}} \le C{e^{ - x}}|x{|^{y - 2\delta - 1}}$ and $({e^{ - x}}\ln x){x^{y + \delta - 1}} \le C{e^{ - \frac{x}{2}}}$ hold.
This solution is copied from my professor's blackboard handwriting, so possibly there might be mistake in step 3. If the inequities in step 3 is indeed incorrect, can you help give the right inequities? Thank you so much!
well it's a general result that $$ \lim \limits_{y \to x} f(y) = l $$ if and only if $$ \lim \limits_{n \to \infty} f(y_n) = l, $$ for all sequences $y_n$ such that $y_n \to x, y_n \neq x. $ (This is easy to prove for further discussion see my book "Proof patterns")
So the DCT goes over.
For your second point. He is using basic properties of logs: $$ x^{\delta} \log x $$ is bounded as $x \to 0,$ for any $\delta >0.$
and $$ e^{-x/2} \log x $$ is bounded as $x \to \infty.$ this is obvious from the fact that $\log$ grows more slowly than any polynomial, and $e^{x}$ grows faster than any polynomial.