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!

  • $\begingroup$ search the web for "differentiation under the integral sign" $\endgroup$ – zhw. Apr 27 '16 at 17:36

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

$$\begin{align} \Gamma'(x)&=\lim_{h\to 0}\int_0^\infty \left(\frac{t^h-1}{h}\right)t^{x-1}e^{-t}\,dt\\\\ &=\int_0^\infty \lim_{h\to 0}\left(\frac{t^h-1}{h}\right)t^{x-1}e^{-t}\,dt\\\\ &=\int_0^\infty \log(t)t^{x-1}e^{-t}\,dt \tag 3 \end{align}$$


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

$$|f(t,h)|\le g(t)$$


$$g(t)=\begin{cases}|\log(t)|&,0<t\le 1\\\\t-1&,1<t\end{cases}$$

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!


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 $|\int_0^\infty r(t,h) dt| < C \int_0^\infty |t^x e^{-t} \ln(t)^2| dt < A$ : $$\frac{\partial}{\partial x} \int_0^\infty t^{x-1} e^{-t} dt = \lim_{t \to 0} \int_0^\infty \frac{\int_0^\infty t^{x+h} e^{-t} dt -\int_0^\infty t^{x} e^{-t} dt}{h} = \int_0^\infty t^x e^{-t} \ln(t) dt$$

  • $\begingroup$ This is very helpful. However, I was really hoping to use the LDCT to solve it. I knew this could be done without and I chose this just so I could check my answer to see if it was equivalent in any way @user1952009 $\endgroup$ – user316861 Apr 27 '16 at 17:41
  • $\begingroup$ @user316861 : what I wrote is enough for applying the dominated convergence theorem. can you see where is hidden the dominating function in what I wrote ? $\endgroup$ – reuns Apr 27 '16 at 17:43
  • $\begingroup$ I'm sorry, but I cannot $\endgroup$ – user316861 Apr 27 '16 at 17:45
  • $\begingroup$ @user316861 : what are $f_n$ and $f$ to which $f_n$ converges pointwise ? en.wikipedia.org/wiki/Dominated_convergence_theorem $\endgroup$ – reuns Apr 27 '16 at 17:46
  • $\begingroup$ so here my $f_n$ and $f$ are $t^{x-1}$ and $e^{-t}$ respectively? $\endgroup$ – user316861 Apr 27 '16 at 17:52

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.

  • $\begingroup$ this is starting to make sense. Now that I have the bounding function I apply the LDCT. Looking on wikipedia, this looks like you've given the answer. SO i guess my confusion is how to apply the LDCT when you already have? What am I missing (and please don't say it's obvious ha). $\endgroup$ – user316861 Apr 27 '16 at 20:08

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