$\displaystyle \int_0^1 x^p \left ( \ln \dfrac 1 x \right)^q \, dx $
given $f(x) = (-1)^qx^p(\log(x))^q$
if we are able to show that integral f(x) absolutely converges then integral converges
using limit comparison test with $g(x) = 1/x^m$ where $m<1$ since limit $g(x)$ converges for $m<1$.
therefore we have $\lim_{x\to0} f(x)/g(x)$
which is giving me $\lim_{x\to 0} x^{p+m}(\log(x))^q$
for this limit to exist $p+m > q$
and hence $p > q-1$
but I am not sure of my method.
There are several cases to consider.
A. If $q<0$, then as $\log(x^{-1})\sim(x-1)$ as $x\rightarrow1$ the integral (denoted as $J_{p,q}$ from now on) would diverge if $q\leq -1$ since $$ J_{p,q}:=\int^1_0 x^p\log^q(x^{-1})\,dx \geq C_p\int^1_{1/2}\log^q(1/x)\,dx \geq C'_p\int^1_0(1-x)^q\,dx $$ This shows that $J_{p,q}$ diverges for $q\leq -1$ (regardless of $p$).
B. From (A) it follows that if $p>0$ and $-1<q<0$, $J_{p,q}$ converges since $J_{p,q}=\int^{1/2}_0+\int^1_{1/2} x^p\log^q(x^{-1})\,dx$. The second integral is bounded by $A_p\int^1_{1/2}(1-x)^p\,dx$ for some constant $A_p$. The first integral has not problem as the integrand can be defined as a continuous function by setting its value at $x=0$ as $0$.
C. If $p,q>0$, then $J_{p,q}$ converges since $\lim_{x\rightarrow0}x^\alpha\log(x)=0$ for all $\alpha>0$, and so $g(x)= x^{p/q}\log(1/x))^q$ for $0<x\leq1$, $g(0)=0$ ddefines a nice continuous function.
The case $p<0$ and $-1<q<0$ may be analyzed along similar lines.
$$J_{p,q}=\int^{1/2}_0 +\int^1_{1/2} x^p\log^q(x^{-1})\,dx = I^1_{p,q} + I^2_{p,q}$$
The second integral converges since $I^2_{p,q}\leq B_p\int^1_{1/2}(1-x)^q\,dx$. Thus, it all boils down to see what happens to $I^1_{p,q}$.
D. If $-1<p<0$ (and $-1<q<0$) then $J_{p,q}$ converges since $I^1_{p,q}\leq \log^q(1/2)\int^{1/2}_1 x^p\,dx<\infty$.
E. If $p<-1$ (and $-1<q<0$) then $J_{p,q}$ diverges, for $\lim_{x\rightarrow0}x^\alpha\log x=0$ for all $\alpha>0$ and so, $$ I^1_{p,q}=\int^{1/2}_0 x^{-1}\big(x^{(p+1)/q}\log(x^{-1})\big)^q\,dx\geq D\int^{1/2}_1x^{-1}\,dx =\infty$$
F. For the case $p=1$ (and $-1<q<0$) $J_{p,q}$ diverges. One can use substitution to directly evaluate the integral. $$\int^{1/2}_0x^{-1}(-\log(x))^p\,dx=\int^\infty_{\log2} u^q\,du=\infty$$
You can complete the remaining case ($p=0$, $-1<q\leq 0$)
\begin{equation} J=(-1)^{q}\int\limits_{0}^{1} x^{p}\ln(x)^{q} \,dx \end{equation}
Let $x=e^{-u}$, then this implies that: $dx=-e^{-u}du$. In terms of the limits, given that $u=-\ln(x)$, when $u(x=0)\rightarrow +\infty$ and also, when $u(x=1)=0$. If we use the negative sign in $dx$ to inverse the limits of integration, then we have that:
\begin{equation} J=(-1)^{q}\int\limits_{0}^{+\infty} e^{-up}(-u)^{q} \,e^{-u} du =(-1)^{2q}\int\limits_{0}^{+\infty} e^{-u(p+1)}u^{q} \,du \end{equation}
This almost has the form of the gamma function, now consider the susbtitution: $z=u(p+1) \Leftrightarrow \frac{dz}{p+1}=du$. Also, the limits of integration remain the same. Then:
\begin{equation} J=\frac{(-1)^{2q}}{p+1}\int\limits_{0}^{+\infty} e^{-z}\left(\frac{z}{p+1}\right)^{q} \,dz = \frac{(-1)^{2q}}{(p+1)^{1+q}}\underbrace{\int\limits_{0}^{+\infty} e^{-z}z^{q} \,dz}_{\Gamma(q+1)} \end{equation}
Then, we can conclude that:
\begin{equation} J=\frac{(-1)^{2q}}{(p+1)^{1+q}}\Gamma(q+1)=\frac{q!}{(p+1)^{1+q}} \end{equation}
From the result, we can see that the integral will converge as long as $q,p>-1$. In these conditions, the denominator is never zero and therefore, the integral never diverges.
