For which values of $p,q$ does the integral $\int_0^1 x^p (\ln\frac{1}{x})^qdx$ converge? For which values of $p,q$ does the integral $\int_0^1 x^p (\ln\frac{1}{x})^qdx$ converge?
I use the substitution $t=1/x$ to obtain this better looking integral: $\int_1^\infty \frac{(\ln t)^q}{t^{p+2}}$. Integration by parts gives me a recurence formula for the integral, but $p,q$ are not necessarily integers so I don't think that's the right approach.
 A: $e^{1/x}>1+1/x \implies1/x>\ln(1+1/x)>\ln(1/x),\:\:x>0.$
$$\int_0^1 x^p\ln^q(1/x)dx\le\int_0^1x^{p-q}dx\\\int_c^dx^{p-q}\le\large\sigma_{\normalsize p,q}\normalsize=\frac{1}{1+p-q}\cdot (1-c^{1+p-q}),\:\:0<c<d<1;\:p\ne q-1\\\forall p>q-1,\:\:\:\lim_{c\to 0^+}\large\sigma_{\normalsize p,q}<\infty$$
A: By substituting $x=\frac{1}{t}$ so $dx=-\frac{1}{t^2}dt$ and the integral reads as follows
$$I_{p,q}=\int_1^\infty\frac{\left(\log{t}\right)^q}{t^{p+2}}dt$$
The interesting bound is $\infty$. If $p\gt -1$ the integral is convergent because the integrand is $o(t^{-p-2})$ and if $p\leq -1$ it is divergent because the integrand is larger than $t^{-p-2+\epsilon}$ for any $\epsilon\gt 0$ In the neighborhood of $\infty$
A: I understand this is a bit late.
But we can substitute $x=e^{-t}$ to make the integral nice looking: $$I=\int_{0}^{\infty}{e^{-(p+1)t}t^q}dt$$
We can split this into $I_1=\int_{0}^{1}{e^{-(p+1)t}t^q}dt$ and $I_2=\int_{1}^{\infty}{e^{-(p+1)t}t^q}dt$.
$I_1$ converges for $q>-1$ (LCT with $t^q$).
Now for $q>-1$, we consider $I$. The only point of concern is $\infty$.

*

*$p < -1: e^{-(p+1)t}t^q\rightarrow \infty$ as $x \rightarrow \infty \implies I$ diverges

*$p = -1: I=\int_{0}^{\infty}t^qdt$ diverges for $q>-1$

*$p > -1: \int_{0}^{\infty}{e^{-(p+1)t}t^q}dt=\frac{1}{(p+1)^q}\Gamma(q+1)$ converges.

Thus $I$ converges for $p>-1, q>-1$.
A: After substituting $x=\frac{1}{t}$, you can use two comparison test to check for divergence near 1, when $q<0$:


*

*Use the limit comparison test with $ln(t)^{q}$ - the limit is 1, so both of them will converge or diverge together
$$ \lim \limits_{t\to 1} \frac{\frac{ln(t)^q}{t^{p+2}}}{ln(t)^{q}} = \lim \limits_{t\to 1} \frac{1}{t^{p+2}} =1 $$

*Use the limit comparison test between $ln(t)^{q}$ and $(t-1)^q$:
$$ \lim \limits_{t\to 1} \frac{ln(t)^{q}}{(t-1)^q} = \lim \limits_{t\to 1} 
 (\frac{ln(t)}{(t-1)})^q $$
and since 
$$\lim \limits_{t\to 1} \frac{ln(t)}{(t-1)} = \lim \limits_{t\to 1} \frac{\frac{1}{t}}{1} = 1$$
you'll get that those two will also diverge/converge together. Since $\int_{1}^{a}\frac{1}{(t-1)^{-q}}dt$ will only converge if $-q<1 \Rightarrow q>-1$, the original integral will also converge.

