Determine all values of $p$ and $q$ that the integral convergens of $$\int_0^1 x^p \ln^q \left(\frac{1}{x}\right)dx$$ without using Gamma function.
First attempt: Since $$\int_0^1 x^p \ln^q \left(\frac{1}{x}\right)dx \le \int_0^1x^{p-q}dx$$ the integral diverges when $p-q>-1$
Second attempt: Substituting $x=1/t$ the integral will look like $$\int_1^{\infty}\frac{\ln^qt}{t^{p+2}}dt \le \int_1^\infty\frac{1}{t^{p-q+2}}dt$$ which converges when $p-q+2>1$ or $p-q>-1$
But the answer in my book says the integral converges when $p>-1$ and $q>-1$. Why is this?
Hint: The substitution $y=-\log x$ changes the integral to $\int_0^{\infty} e^{-(1+p)y} y^{q}dy$. $\int_0^{1} e^{-(1+p)y} y^{q}dy$ converges iff $q >-1$ and $\int_1^{\infty} e^{-(1+p)y} y^{q}dy$ converges iff $1+p >0$ or $p >-1$.