Improper integrals involving logarithm in denominator Discuss/comment on the convergence/divergence of the following improper integrals:
(A)
$$\int_0^{+\infty}\dfrac{e^{-x}}{x^{\alpha}(|\log x|)^{\beta}},\text{ with }\alpha,\beta>0$$
(B)
$$\int_0^{+\infty}\dfrac{xe^{-x}}{\log^2(x)}$$
I find it difficult to deal with improper integrals when logarithms are in the denominator. Could you suggest some method to tackle them?
I understand that we'll first divide the domain of integrals so that we have only one type of improper integral to deal with in that domain. Then we try to come up with some function using which we apply the limit comparison test and proceed further.
I face difficulties in finding that function.
 A: 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
$$x\to 0\implies \begin{cases}\text{convergent integral}&&\text{for }p>1,\forall q\in\mathbb R\text{ or }p=1,q>1\\\text{divergent integral}&& \text{otherwise} \end{cases}$$
$$x\to 1\implies \begin{cases}\text{convergent integral}&&\text{for }p<1,\forall q\in\mathbb R\text{ or }p=1,q>1\\\text{divergent integral}&& \text{otherwise} \end{cases}$$
$$x\to \infty\implies \begin{cases}\text{convergent integral}&&\text{for }\forall p\in\mathbb R,q<1\\\text{divergent integral}&& \text{for }\forall p\in \mathbb R,q>1 \end{cases}$$
