Lebesgue Dominated Convergence Theorem example For $x>0$ we have defined $$\Gamma(x):= \int_0^\infty t^{x-1}e^{-t}dt$$ Im trying to use Lebesgue's Dominated Convergence theorem to show $$\Gamma'(x):=lim_{h\rightarrow 0}\frac{\Gamma(x+h)-\Gamma(x)}{h}$$ exists and equals $$\int_0^\infty(\frac{\partial}{\partial x}t^{x-1}e^{-t})dt$$ This is my first time dealing with L.D.C.T and I'm super lost. Ive been sifting through books and notes but I'm not the best at dealing with very theoretical stuff so I came across this example to help me maybe understand it more. Can any explain this?  
edit: I've also been going through the different feeds on here!
 A: 
The ensuing development relies on the elementary inequality for the logarithm function
$$\log(x)\le x-1 \tag 1$$
for all $x>0$.


Let $f(t,h)$ be the function given by
$$f(t,h)=\frac{t^h-1}{h}\tag 2$$
for $h\ne 0$.  Note that $f(t,h)>0$ for $t>1$ and $f(t,h)<0$ for $t<1$. 
We seek to find a function $g(t)$ such that (i) $|f(t,h)|\le g(t)$ and (ii) $g(t)t^{x-1}e^{-t}$ is integrable for $t\in (0,\infty)$.  
If successful, then the Dominated Convergence Theorem guarantees that the derivative of the Gamma function is given by
$$\begin{align}
\Gamma'(x)&=\lim_{h\to 0}\int_0^\infty \left(\frac{t^h-1}{h}\right)t^{x-1}e^{-t}\,dt\\\\
&=\int_0^\infty \lim_{h\to 0}\left(\frac{t^h-1}{h}\right)t^{x-1}e^{-t}\,dt\\\\
&=\int_0^\infty \log(t)t^{x-1}e^{-t}\,dt \tag 3
\end{align}$$

FINDING A BOUNDING FUNCTION
Proceeding to bound $f(t,h)$, we find immediately from $(1)$ to $(2)$ with $x=t^h$ that for $t<1$
$$|f(t,h)|\le |\log (t)| \tag 4$$
Next, note that for $h\ne 0$ and any fixed $t>0$, the partial derivative of $f(t,h)$ with respect to $h$ is given by
$$\frac{\partial f(t,h)}{\partial h}=\frac{\log(t^h)t^h-(t^h-1)}{h^2}\ge \frac{(t^h-1)^2}{h^2}\ge 0$$
Therefore $f(h)$ is an increasing function of $h$ and so for $h\le 1$, we have
$$\begin{align}
f(t,h)&\le f(t,1)\\\\
&=t-1
\end{align}$$
So, for $t\ge 1$ we find that
$$|f(t,h)|\le t-1$$
Putting $(4)$ and $(5)$ together reveals 
$$|f(t,h)|\le g(t)$$
where 
$$g(t)=\begin{cases}|\log(t)|&,0<t\le 1\\\\t-1&,1<t\end{cases}$$

Inasmuch as $g(t)t^{x-1}e^{-t}$ is integrable and an upper bound for $\left|f(h) t^{x-1}e^{-t}\right|$, then application of the Dominated Convergence Theorem yields the result in $(3)$
$$\Gamma'(x)=\int_0^\infty \log(t)t^{x-1}e^{-t}\,dt$$
as was to be shown!
A: here you won't need the dominated convergence theorem, since everything is absolutely convergent :
$$\frac{t^{x+h-1} e^{-t} -t^{x-1} e^{-t}}{h} = t^x e^{-t} \ln(t) + h \ r(t,h)$$
with $|r(t,h)| < C |t^x e^{-t} \ln(t)^2|$  for every $t > 0$ and $|h| < \delta$
hence 
$$\frac{\int_0^\infty t^{x+h} e^{-t} dt -\int_0^\infty t^{x} e^{-t} dt}{h} =  \int_0^\infty\frac{ t^{x+h}-t^x}{h} e^{-t} dt = \int_0^\infty t^x e^{-t} \ln( t) dt + h \int_0^\infty r(t,h) dt$$
and  since $|\int_0^\infty r(t,h) dt| < C \int_0^\infty |t^x e^{-t} \ln(t)^2| dt < A$ : $$\frac{\partial}{\partial x} \int_0^\infty t^{x-1} e^{-t} dt = \lim_{t \to 0} \int_0^\infty \frac{\int_0^\infty t^{x+h} e^{-t} dt -\int_0^\infty t^{x} e^{-t} dt}{h} = \int_0^\infty t^x e^{-t} \ln(t) dt$$
A: Fix $x>0.$  We want to consider, for $t>0,$
$$\tag 1 e^{-t}\frac{(t^{x-1 + h} - t^{x-1})}{h} = e^{-t}t^{x-1}\frac{t^h -1}{h}.$$
As $h\to 0,$ the last expression $\to e^{-t}t^{x-1}\ln t.$ If we can find an $L^1$ dominating function, we will have our answer:
$$\Gamma'(x) = \int_0^\infty e^{-t}t^{x-1}\ln t\, dt.$$
Suppose $|h| < x/2.$ Use the mean value theorem to see
$$\frac{t^h -1}{h} = (\ln t)t^c$$
for some $c \in (-x/2,x/2).$ Verify that $t^c \le t^{x/2}+t^{-x/2}.$ In absolute value then, $(1)$ is bounded above by
$$e^{-t}t^{x-1}|\ln t\,| (t^{x/2}+t^{-x/2}).$$
There's your fixed $L^1$ bounding function; you're ready for LDCT.
