Let $n=1,2\dots$ and $\theta\in(0,\infty)$. I want to prove $$ \partial_\theta^n\int_0^1t^{\theta-1}\,\mathrm dt=\int_0^1\partial_x^nt^{\theta-1}\,\mathrm dt. $$ I was using this document as a reference for understanding the conditions under which integration and differentiation can be interchanged. Using the notation in the linked paper: $\Theta=(0,\infty)$, $X=[0,1]$, and $g:\Theta\times X\to \Bbb R$ with $g(\theta,x)=x^{\theta-1}$.
Clearly, $$ \int_0^1|g(\theta,x)|\,\mathrm dx<\infty,\quad\forall\theta\in\Theta. $$ Furthermore, its easy to show that $|\partial_\theta^ng(\theta,x)|=(-\log x)^{n}x^{a-1}$ and $$ \int_0^1|\partial_\theta^ng(\theta,x)|\,\mathrm dx<\infty,\quad\forall\theta\in\Theta. $$ If I am understanding correctly, all that is left to show is uniform local integrability. If so, I have to show for all $\theta_0\in\Theta$ that there is an $\epsilon>0$ such that $$ \sup_{\theta\in(\theta_0-\epsilon,\theta_0+\epsilon)}|\partial_\theta^ng(\theta,x)|\leq h(x), $$ where $\int_0^1h(x)\,\mathrm dx<\infty$. But how do I find the dominating function $h$? It is true that $$ \sup_{\theta\in(\theta_0-\epsilon,\theta_0+\epsilon)}|\partial_\theta^ng(\theta,x)|\leq (-\log x)^n x^{\theta_0-\epsilon-1}, $$ which is integrable if $\theta_0>\epsilon$. Since I could choose $\epsilon=\theta_0/2$ it would seem I can always find a dominating function that is integrbale, however, this dominating function is not independent of $\theta$.
How to I prove the uniform local uniform integrability condition here to finish the proof?
Notice that $$\partial^k_{\theta} (x^{\theta-1}) = x^{\theta-1} \log^k x.$$
Let $\theta_0 > 0.$ Then, for $\theta \in [\theta_{0},\infty)$ we have $$|x^{\theta-1}\log^k x| \leq x^{\theta_0-1} |\log^k x|$$ for all $x \in (0,1),$ and the right hand side is integrable, independent of $\theta$.
By the criterion above, the function $\theta \mapsto \int_0^1 x^{\theta-1} d x$ is $k$ times differentiable in $(\theta_0, \infty)$ with $k$-th derivative $\int_0^1 x^{\theta-1} \log^k x \,d x.$ Since $\theta_0 > 0$ was arbitrary, it follows that the function is $k$ times differentiable in $(0,\infty).$