Uniform convergence of $\int_{x=0}^{1}\frac{x^{-t}-1}{t}\mathrm{d}x$ I'm trying to show that this improper integral:
$$\int_{x=0}^{1}\frac{x^{-t}-1}{t}\mathrm{d}x$$
uniformly converge with respect to $t\in(0,1/2)$, for which pointwise convergence is obvious, because the explicit result is $\frac{1}{1-t}$.
I am using Cauchy criterion for uniform convergence of a limit but I can't reach nothing interesting (in the following $0<A_1<A_2<1$):
$$\left|\int_{A_1}^{A_2}\frac{x^{-t}-1}{t}\mathrm{d}x\right|=\frac{1}{t}\left|\frac{A_2^{-t+1}-A_1^{-t+1}}{-t+1}-A_2+A_1\right|$$
because I can't rearrange terms on the right in a useful way.
Could I have a little help please? Is Cauchy criterion test the right way?
 A: Defining $f(t) = x^{-t}= e^{-t\log x}$ with $t > 0$ and $0 < x \leqslant 1$, we have $f'(t) = -x^{-t}\log x\, $ and $f(0) = 1$. By the mean value theorem, there exists $\xi_t \in (0,t)$ such that
$$\tag{*}0 < \frac{x^{-t}-1}{t} = -x^{-\xi_t}\log x$$
The improper integral of $x \mapsto \frac{x^{-t} -1}{t}$ over $(0,1]$ is divergent for any $t \geqslant 1$ and convergent for any $0< t < 1$, since
$$\int_0^1 x^{-t} \, dx = \lim_{\delta \to 0}\int_\delta^1 x^{-t} \, dx = \lim_{\delta \to 0}\begin{cases}\frac{1- \delta^{1-t}}{1-t}, \quad t > 1 \\ -\log \delta, \quad t=1 \\\frac{1- \delta^{1-t}}{1-t}, 0 < t <1\end{cases}= \begin{cases}+\infty, \quad t >1\\+\infty, \quad t = 1\\\frac{1}{1-t}, \quad 0 < t < 1\end{cases}$$
For any $\beta \in (0,1)$ and all $0 < \xi_t < t \leqslant \beta$, we have $x^{-\xi_t} < x^{-\beta}$, and it follows from (*) that
$$\left|\frac{x^{-t}-1}{t} \right| < \frac{|\log x|}{x^\beta}$$
It is easy to show using the comparison test that the improper integral of $|\log x|/x^\beta$ over $(0,1]$ is convergent.
Hence, the improper integral  $$\int_0^1\frac{x^{-t} -1}{t}\, dx$$
is uniformly convergent for all $t \in (0,\beta]$ where $\beta < 1$ by the Weierstrass M-test.
