The second integral can be shown to converge in a variety of ways. In particular, it can be shown to converge by some very general approaches that have broad applicability. (This is also applicable to the first and third integrals).
Dirichlet test for convergence of an improper integral (under weak conditions)
Suppose $f:[a,\infty) \to \mathbb{R}$ is Riemann integrable on any bounded interval such that $\int_a^b f(x) \, dx $ is bounded for all $b > a$ and that the function $g:[a,\infty) \to \mathbb{R}$ is monotonically decreasing to zero as $x \to \infty$. Then the improper integral $\int_a^\infty f(x) g(x) \, dx$ is convergent.
In this case $f(x) = \sin x$ and $g(x) = [\log x]^{-1}$ are easily seen to satisfy the conditions for the Dirichlet test.
Cauchy criterion and second integral mean value theorem.
(These are more fundamental results that can be used to prove the Dirichlet test.)
The Cauchy criterion states that the improper integral $\int_a^b h(x) \, dx$ is convergent if and only if for any $\epsilon > 0$ there exists $K \geqslant a$ such that $\left| \int_\alpha ^\beta h(x) \, dx\right| < \epsilon$ for all $\beta > \alpha > K$.
A special case of the second integral mean value theorem for integrals is if $f:[a,b) \to \mathbb{R}$ is Riemann integrable and $g:[a,b] \to \mathbb{R}$ is non-negative and decreasing, then there exists $\xi \in [a,b]$ such that $\int_a^b f(x) g(x) \, dx = g(a)\int_a^\xi f(x) \, dx$.
For the problem at hand, we have by the second integral mean value theorem with $f(x) = \sin x$ and $g(x) = [\log x]^{-1}$,
$$\left|\int_\alpha^\beta \frac{\sin x}{\log x} \, dx \right| = \left|\int_\alpha^\beta \frac{\sin x}{\log x} \, dx \right| =\frac{1}{|\log \alpha|}\left|\int_\alpha^\xi \sin x \, dx \right|= \frac{|\cos \alpha - \cos \xi|}{|\log \alpha|}\leqslant \frac{2}{|\log \alpha|}$$
For any $\epsilon >0$ there exists $K$ such that if $\alpha > K$ then $|\log \alpha | = \log \alpha > \frac{2}{\epsilon}$. Hence, with $\beta > \alpha > K$ we have
$$\left|\int_\alpha^\beta \frac{\sin x}{\log x} \, dx \right| < \epsilon, $$
and the improper integral converges by the Cauchy criterion.
Best Answer
The second integral you wrote is $$\int_0^{+\infty}\dfrac{xe^{-x}}{\log^2(x)}dx.$$ We have to study the integral in a neighbourhood of the points $0,1$ and for $x\to +\infty$.
Let $x\in\mathcal U(0)$ we have that $\dfrac{xe^{-x}}{\log^2(x)}\sim\dfrac{x}{\log^2(x)}\underset{x\to 0^+}{\longrightarrow}0\implies$ the integrand is defined in a neighbourhood of zero and the integral is clearly convergent.
Let now $x\in \mathcal U(1)$, then the integrand is asymptotic to $\dfrac{c}{\log^2(x)}$ and since a function of the form $\dfrac{1}{|x|^p|\log x|^q}$ has improper integrand convergent iff $q>1$ we can state that the integral is divergent in a neighbourhood of $x=1$.
Finally if $x\in \mathcal U(+\infty)$, then $\dfrac{xe^{-x}}{\log^2(x)}\sim \tilde c/e^x\le\dfrac{1}{x^2}$, whose integral is convergent.
So the integral will be convergent only in the set of the form $[0,1-\delta]\cup[1+\delta,+\infty)$.
Let $f(x)=\dfrac{1}{|x|^p|\log x|^q}$ we have the following properties $$x\to 0\implies \begin{cases}\text{convergent integral}&&\text{for }p>1,\forall q\in\mathbb R\text{ or }p=1,q>1\\\text{divergent integral}&& \text{otherwise} \end{cases}$$
$$x\to 1\implies \begin{cases}\text{convergent integral}&&\text{for }p<1,\forall q\in\mathbb R\text{ or }p=1,q>1\\\text{divergent integral}&& \text{otherwise} \end{cases}$$
$$x\to \infty\implies \begin{cases}\text{convergent integral}&&\text{for }\forall p\in\mathbb R,q<1\\\text{divergent integral}&& \text{for }\forall p\in \mathbb R,q>1 \end{cases}$$