I have trouble determining for which $a\in \mathbb{R}\;$ the following improper integral converges:
$$\int_0^1\frac{\ln(x)}{x^a}dx$$ I have tried the following:
$\left|\frac{\ln(x)}{x^a}\right|=\frac{-\ln(x)}{x^a}$ for $0<x\leq 1$
and: $\frac{1}{x}>-\ln(x)$ on this interval.
So $\frac{\frac{1}{x}}{x^a}>\frac{-\ln(x)}{x^a}$ (on this interval)
$\frac{\frac{1}{x}}{x^a}$ can be written as a p-integral: $\int_0^1 x^{-p} dx$ with $p = a+1.$ Thus, I concluded, the improper integral converges for $a<0$ since the p-integral does too.

However, I'm asking myself whether this is the right way to tackle these kinds of problems, especially since I have no idea how to prove/disprove convergence for $a\geq 0.$ Is what I have done above right? And how should I go about testing convergence for $a\geq 0$?


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