The integral is defined, for all $x\in\mathbb{R}$ as follows: $$I= \int_{0}^{1}\dfrac{\left(t-1\right)}{\ln t} t^x\mathrm{d}t.$$
When $I$ converges?
Let $t-1=u$, we have: $u\to 0$ when $t\to 1$.
$$\dfrac{\left(t-1\right)}{\ln t}=\dfrac{u}{\ln\left(u+1\right)}\sim_{0}1.$$
Then, $$\dfrac{\left(t-1\right)}{\ln t} t^x=\dfrac{u}{\ln\left(u+1\right)} \left(u+1\right)^x\sim_{0}\left(u+1\right)^x\sim_{0}1+ux.$$
And the integral converge when $t=1$. Is it correct and how to continue?
Use the inequality $1-t \leqslant - \ln t \leqslant (1-t)/t$ for $0 < t < 1$.
This inequality follows by bounding the integral representation of the logarithmic function.
For $0 < t < 1$ we have
$$-\ln t = \int_{t}^1 \frac{ds}{s},$$
and with $t \leqslant s \leqslant 1$ we have $1 \leqslant 1/s \leqslant 1/t$ and
$$ 1-t = \int_{t}^1 \frac{ds}{1} \leqslant -\ln t \leqslant \int_{t}^1 \frac{ds}{t} = \frac{1-t}{t}.$$
Then
$$\dfrac{\left(t-1\right)}{\ln t} t^x= \dfrac{\left(1-t\right)}{-\ln t} t^x \leqslant t^x.$$
Since the integral $\int_0^1 t^x \, dt$ converges for $x > -1$, the integral $I$ converges for $x > -1$ by comparison.
Let $u=-\ln t$, then the integral is \begin{equation} \int_0^\infty \frac{e^{-u}-e^{-2u}}{u}e^{-xu}du \end{equation} with some calculation. Then we can use Laplace transform to find when $I$ converges.
