I want to determine when the integral $$\int_{0}^{1} \dfrac{\ln(x)}{x^p} dx$$ is convergent or divergent depending on $p$.
I tried applying comparison test:
$$\int_{0}^{1} \dfrac{\ln(x)}{x^p}dx < \int_{0}^{1} \dfrac{x}{x^p}dx = \int_{0}^{1} \dfrac{1}{x^{p-1}}dx$$
So: $$ \lim_{b\to0} \int_{b}^{1} \dfrac{1}{x^{p-1}}dx = \lim_{b\to0} \left(\dfrac{1}{2-p} - \frac{b^{2-p}}{2 - p}\right)$$
That means when $p < 2$ the integral converges.
If I assume $p < 0$ I seem to be able to calculate the integral's value:
$$\int_{0}^{1} \ln(x)\cdot x^{-p} dx = \ln(x)\cdot \left.\frac{x^{1-p}}{1-p} \right|_0^1 - \int_{0}^{1} \dfrac{1}{x}\cdot \dfrac{x^{1-p}}{1-p} dx$$
But what I got is not full range. How to get other values? Thanks in advance.
No, it is incorrect.
Note this upper bound is useless. Since $\int_{0}^{1} \frac{\ln(x)}{x^p}dx<0$, you need to determine in what case the integral will converge, or diverge to $-\infty$, hence your upper bound has no help on this.
Instead of considering this integral with a negative value, we can do the substitution $x=1/t$, then we get
$$\int_{0}^{1} \frac{\ln(x)}{x^p}dx =-\int_1^\infty \frac{\ln t}{t^{2-p}}~dt$$
It is easy to see the RHS integral is convergent when $2-p>1$, hence $p<1$