# Verify convergence of improper integral

Let $$f: (0,\infty) \rightarrow \mathbb{R}$$, where $$f(x) = \int_{0}^{1} \frac{t^x-1}{\log{t}}dt$$. I want to check if this improper integral is convergent or not (when $$t>0$$), and have tried out different measures, but couldn't find any thing that works. For example, I've tried finding $$g: [0,1] \rightarrow \mathbb{R}$$ such that $$\lvert f(x,t) \leq g(t) \rvert, \forall x \in (0,\infty)$$ and $$\int_0^1 g(t)dt \lt \infty$$ using $$\lvert \log(t) \rvert \leq t$$, but this doesn't seem to work either. Any ideas?

• Your integrand has a removable discontinuity at $t=1,$ so you need only examine the behavior of this improper integral near $t=0$. To do this, you should first show that $\int_0^{\delta}-\frac{1}{\ln(t)}\mathrm{d}t$ converges for any $\delta\in (0,1)$. Nov 16, 2021 at 2:40

$$0< \frac{t^x-1}{\log(t)} < x$$
for $$x>0$$ and $$0. So that
$$f(x)\leq x\int_0^1 dt=x\ .$$
• @MarkViola Maybe I'm missing something. The denominator goes to $-\infty$ and the numerator goes to $-1$. Nov 16, 2021 at 3:49