Find $\alpha(t,a)$, integrable w.r.t. to t, that bounds $\vert e^{-t}t^{a} \frac{t^h - 1}{h} \vert$ ; $a,t > 0$, for all $\vert h \vert < h_0$. 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 $Re(z) &gt; 1$ 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, i.e. $$ \lim_{h\to 0} \int^{\infty}_0 e^{-t}t^{z-1} \frac{t^h - 1}{h} dt =  \int^{\infty}_0 \lim_{h\to 0}  e^{-t}t^{z-1} \frac{t^h - 1}{h} dt $$
which is possible if the integrand is bounded by an $\mathbf{integrable}$ function $\alpha(t,z)$, i.e.
$$ \Bigg | e^{-t}t^{z-1} \frac{t^h - 1}{h}  \Bigg | &lt; \alpha(t,z)$$ for all $\vert h \vert &lt; h_0$ (for some real positive $h_0$) and $$ \int^{\infty}_0 \alpha dt &lt; \infty $$
This is equivalent to $$ \Bigg | e^{-t}t^{a} \frac{t^h - 1}{h}  \Bigg | &lt; \alpha(t,a)$$ for real $a&gt;0$.

So finding such a function $\alpha$ is the goal.

Edit: The answer that I had given before was wrong. I was treating h as a real no. when actually it can be complex.

Here is another solution that I found:

$$ \Bigg | \frac{t^h - 1}{h} \Bigg | = \Bigg |\frac{e^{h\log t} - 1}{h} \Bigg | $$
$$ = \Bigg | \frac{h\log t + \frac{(h\log t)^2}{2!} +  \frac{(h\log t)^3}{3!}+ \cdots}{h} \Bigg | $$
$$ \leq \vert \log t (1 + \frac{(h\log t)^1}{2!} +\frac{(h\log t)^2}{3!} + \cdots)\vert $$
 $$= \vert \log t \vert (1+  \frac{(\vert h\log t \vert)^1}{2!} + \frac{(\vert h\log t \vert)^2}{3!} + \cdots) $$
$$ \leq \vert \log t \vert (1+ \vert h \log t \vert + \frac{(\vert h\log t \vert)^2}{2!} + \cdots) $$
$$= \vert \log t \vert e^{\vert h\log t \vert } $$

Thus required function is
$$ \alpha(t) = e^{-t}t^{a} \vert \log t \vert e^{\vert h\log t \vert }$$
And it can be shown that it's integral exists and is finite.
There cannot be an intergable function $g$ on $(0,\infty)$ such that $|\frac {t^{h}-1} h | \leq g(t)$ for all $t&gt;0$ and all $h$ with $|h|&lt;h_0$. 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.
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.