At what values of $\alpha$ and $\beta$ does $\int_0^1x^\alpha(1-x)^\beta \ln xdx$ converge? $$\int_0^1x^\alpha(1-x)^\beta \ln x dx$$
When $x\to 0$:
$(1-x)^\beta=1-\beta x+o(x)$
$$\int_0^1x^\alpha(1-\beta x)\ln xdx=\int_0^1x^\alpha\ln xdx-\int_0^1\beta x^{\alpha+1}\ln xdx$$
If we integrate by parts:
$$\int x^\alpha\ln xdx=\frac{x^{\alpha+1}}{\alpha+1}\ln x-\int\frac{x^\alpha}{\alpha+1}dx$$
$$\beta\int x^{\alpha+1}\ln xdx=\beta\frac{x^{\alpha+2}}{\alpha+2}\ln x-\beta\int\frac{x^{\alpha+1}}{\alpha+2}dx$$
So the integral converges when $\alpha>-1$, but in the answer it's written that $\beta>-2$. I don't see where they got that from. Are both conditions supposed to be met simultaneously or is one of them enough?
(By the way, I tried adding more members to the taylor expansion, but didn't really get to anything useful. Something tells me that that's where the answer lies but I can't how adding more members would affect the result.)

$$
\int_{0}^{1} x^{a}(1-x)^{b} \ln x \mathrm{~d} x=\int_{0}^{\frac{1}{2}} x^{a}(1-x)^{b} \ln x \mathrm{~d} x+\int_{\frac{1}{2}}^{1} x^{a}(1-x)^{b} \ln x \mathrm{~d} x :=I_{1}+I_{2}
$$
For $I_1$, Consider $x\to 0^+$.Since $x$ is between $0$ and $\displaystyle\frac12$, $(1-x)^b$ does not have to participate in the discussion of convergence.
When $a>0$, we have $\displaystyle\lim _{x \rightarrow 0^{+}} x^{a}(1-x)^{b} \ln x=0$,so $I_1$ is convergent.
When $a=0$, easy to know $\displaystyle\int_{0}^{\frac{1}{2}}(1-x)^{b} \ln x \mathrm{~d} x$ and
$$
\int_{0}^{\frac{1}{2}} \ln x \mathrm{~d} x=x \ln x\Bigg|_{0} ^{\frac{1}{2}}-\int_{0}^{\frac{1}{2}} x \cdot \frac{1}{x} \mathrm{~d} x=\frac{1}{2} \ln \frac{1}{2}-\frac{1}{2}
$$
have the same divergence.so $I_1$ is convergent.
When $-1<a<0$, take $\varepsilon$  satisfied
$0<-a<\varepsilon<1$. At this time $\displaystyle\int_{0}^{\frac{1}{2}} \frac{1}{x^{\varepsilon}} \mathrm{d} x$  converges, and
$$
\lim _{x \rightarrow 0^{+}} \frac{x^{a} \ln x}{\frac{1}{x^{\varepsilon}}} =\lim _{x \rightarrow 0^{+}} x^{a+\varepsilon} \ln x=0
$$
So $\displaystyle\int_{0}^{\frac{1}{2}} x^{a} \ln x \mathrm{~d} x$ converges,
$I_1$ converges.
When $a \leqslant-1$, we have
$$
\lim _{x \rightarrow 0^{+}} \frac{x^{a} \ln x}{\frac{1}{x}}=\lim _{x \rightarrow 0^{+}} x^{a+1} \ln x=\infty
$$
Because $\displaystyle\int_{0}^{\frac{1}{2}} \frac{1}{x} \mathrm{~d} x$ diverges,
so $\displaystyle\int_{0}^{\frac{1}{2}} x^{a} \ln x \mathrm{~d} x$ diverges, $I_1$
diverges.
For $I_2$, Consider $x\to 1^-$. $x^a$ does not have to participate in the discussion of convergence, and
$$
\ln x=\ln (1+x-1) \sim x-1\qquad x \rightarrow 1^{-}
$$
Therefore, it is equivalent to discussing the convergence of
$$
-\int_{\frac{1}{2}}^{1}(1-x)^{b+1} \mathrm{~d} x=-\int_{\frac{1}{2}}^{1} \frac{1}{(1-x)^{-(b+1)}} \mathrm{d} x
$$
When $-(b+1)<1$, that is $b>-2$, $I_2$ converges;
when $b \leqslant -2$, we have $I_2$ diverges.
