Find an integrable function in $t$ that bounds $\vert e^{-t}t^{a} \frac{t^h - 1}{h} \vert$ for all $a,t > 0$ and $\vert h \vert < h_0$.

Question

I think I have found an answer to this which I have given at the end but is there a better answer?

I am trying to prove differentiability of gamma function $\Gamma(z)$ for $Re(z) > 1$ using dominated convergence theorem. So for differentiability, $$ \lim_{h\to 0} \frac{\Gamma(z+h) - \Gamma(z)}{h} $$ must converge. To prove this, I want to be able to take the limit inside the integral as shown below: $$ \lim_{h\to 0} \int^{\infty}_0 e^{-t}t^{z-1} \frac{t^h - 1}{h} dt = \int^{\infty}_0 \lim_{h\to 0} e^{-t}t^{z-1} \frac{t^h - 1}{h} dt $$ which is possible, according to Dominated Convergence Theorem, if the integrand is bounded by an integrable function $\alpha(t,z)$, i.e. $$ \Bigg | e^{-t}t^{z-1} \frac{t^h - 1}{h} \Bigg | < \alpha(t,z)$$ for all $\vert h \vert < h_0$ (for some real positive $h_0$) and $$ \int^{\infty}_0 \alpha(t,z) dt < \infty $$ This is equivalent to finding an integrable function $\alpha(t,a)$ such that $$ \Bigg | e^{-t}t^{a} \frac{t^h - 1}{h} \Bigg | < \alpha(t,a)$$ for real $a>0$. So the goal is finding such a function $\alpha$.

Edit: The answer that I had given initially was wrong. I was treating $h$ as a real number instead of a complex.

Here is another solution (edited) that I found:

$$ \Bigg | \frac{t^h - 1}{h} \Bigg | = \Bigg |\frac{e^{h\log t} - 1}{h} \Bigg | $$ $$ = \Bigg | \frac{h\log t + \frac{(h\log t)^2}{2!} + \frac{(h\log t)^3}{3!}+ \cdots}{h} \Bigg | $$ $$ \leq \vert \log t (1 + \frac{(h\log t)^1}{2!} +\frac{(h\log t)^2}{3!} + \cdots)\vert $$ $$= \vert \log t \vert (1+ \frac{(\vert h\log t \vert)^1}{2!} + \frac{(\vert h\log t \vert)^2}{3!} + \cdots) $$ $$ \leq \vert \log t \vert (1+ \vert h \log t \vert + \frac{(\vert h\log t \vert)^2}{2!} + \cdots) $$ $$= \vert \log t \vert e^{\vert h\log t \vert } \leq \vert \log t \vert e^{h_0\vert\log t \vert } $$ for all $ \lvert h \rvert \leq h_0 $. Hence, the required function is $$ \alpha(t,a) = e^{-t}t^{a} \vert \log t \vert e^{h_0 \vert \log t \vert }$$ It can be shown that its integral w.r.t. $t$ exists and is finite.

Answer 1

Hint : Usually to show the derivability of the gamma function, you use theorems of dominated convergence. You first derive the function under the integral and show that it can be dominated.

Answer 2

There cannot be an intergable function $g$ on $(0,\infty)$ such that $|\frac {t^{h}-1} h | \leq g(t)$ for all $t>0$ and all $h$ with $|h|<h_0$. For, it you take limit as $h \to 0$ you will get $|\log (t)| \leq g(t)$ for all $t$ and $|\log (t)|$ is not integrable.