Question is: For which values of $\alpha \in \mathbb R$ the following improper integrals converge:
a.$$\int_0^1\!\left|\ln(x)\right|^\alpha\,dx$$ b.$$\int_{-\pi/2}^{\pi/2}\!\left|\tan(x)\right|^\alpha\,dx$$ I don't have idea how to solve b but I thought maybe to write $$\int_0^1\!\left|\ln(x)\right|^\alpha\,dx \leq \int_0^1\!\left|x\right|^\alpha\,dx=\int_0^1\!x^\alpha\,dx$$ and then if $\alpha\ge 0$ the integral converges, but I don't have idea for $\alpha<0$.
Hint for part a:
If $\alpha > 0$, the integrand is unbounded near $0$. Use the substitution $u = 1/x$ to solve this case.
If $\alpha < 0$, the integrand is unbounded near $1$. Use the approximation $\ln x \approx x-1$ (valid near $x=1$) to solve this case.
Hint for part b:
If $\alpha > 0$, the integrand is unbounded near $\pm \pi/2$. Taking $x = +\pi/2$ for definiteness, use the approximation $\tan x \approx 1/(\pi/2-x)$ to solve this case.
If $\alpha < 0$, the integrand is unbounded near $0$. Use the approximation $\tan x \approx x$ to solve this case.