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.)

Remember that if $a>0$, then $x^a\log(x)\to0$ as $x\to0$ (you can check that easily). So if $a>0$ and $\beta\geq0$, the integral converges.
If $a\leq0$, then $\int_0^1x^a(1-x)^b\log(x)dx\geq\int_0^1(1-x)^b\log(x)dx\geq\int_0^{1/2}(1-x)^b\log(x)dx\geq c_b\int_0^{1/2}\log(x)dx=\infty$ for all $b\in\mathbb{R}$, where $c_b$ is a positive constant depending on $b$.
So what happens when $a>0$ and $b<0$? we have $$\int_0^1x^a(1-x)^b\log(x)dx=$$ $$=\int_0^{1/2}x^a(1-x)^b\log(x)dx+\int_{1/2}^1x^a(1-x)^bdx\sim \int_0^{1/2}x^a\log(x)dx+\int_{1/2}^1(1-x)^bdx$$
$\sim$ means that they behave in the same way with respect to convergence (I believe you can justify that yourself)
So if $b<-1$, the integral converges, but if $b\in(-1,0)$, the integral diverges.
To sum up: if $a>0$ and $b\geq0$, we have convergence. If $a\leq0$, we have divergence (no matter how $b$ behaves). If $a>0$ and $b<-1$, we have convergence. If $a>0$ and $b\in(-1,0)$, we have divergence.
