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?
1
-
$\begingroup$ 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 $\delta\in (0,1)$. $\endgroup$– Matthew H.Nov 16, 2021 at 2:40
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
Note that
$$0< \frac{t^x-1}{\log(t)} < x$$
for $x>0$ and $0<t<1$. So that
$$f(x)\leq x\int_0^1 dt=x\ .$$
-
$\begingroup$ @MarkViola Maybe I'm missing something. The denominator goes to $-\infty$ and the numerator goes to $-1$. $\endgroup$– podikiNov 16, 2021 at 3:49
-