Show that $x^a \log{x}$ is integrable on $(0,1)$ for all $a>-1$ In my lecture notes it is stated that $x^a \log{x}$ is integrable on $(0,1)$ for all $a>-1$, but a proof is not given. It doesn't seem to satisfy any of the comparison tests we have learned. What is the best way to go about showing this? Thanks in advance.
 A: In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's inequality that the logarithm function satisfies the inequalities
$$\frac{x-1}{x}\le \log(x)\le x-1\tag 1$$
For $x\in (0,1]$, $\log(x)<0$ and we see from $(1)$ that 
$$|\log(x)|\le \frac{1}{x}-1$$
Then, using $\log(x^b)=b\log(x)$ along with $(1)$, we find that for $b>0$
$$\left| x^a\log(x)\right|\le \frac{x^{a-b}-x^a}{b}\tag 2$$
Note that $(2)$ holds for all $b>0$. In particular, $(1)$ holds for $0<b<1+a$.  
Inasmuch as $\frac{x^{a-b}-x^a}{b}$ is integrable for $0<b<1+a$, $x^a\log(x)$ is integrable for $a>-1$ as was to be shown!

We can proceed further and evaluate the integral in closed form.  Note that we can write
$$\begin{align}
\int_0^1 \log(x)x^a\,dx&=\frac d{da}\int_0^1 x^a\,dx\\\\
&=\frac d{da}\left(\frac{1}{a+1}\right)\\\\
&=-\frac{1}{(a+1)^2}
\end{align}$$

In fact, we can prove convergence and evaluate the integral in one fell swoop.  Note that integrating by parts with $u=\log(x)$ and $v=\frac{x^{a+1}}{a+1}$ reveals 
$$\begin{align}
\lim_{\epsilon\to0}\int_0^1 \log(x)x^a\,dx&=\lim_{\epsilon\to0}\left(\left.\left(\log(x)\frac{x^{a+1}}{a+1}\right)\right|_\epsilon^1-\int_\epsilon^1 \frac{x^a}{a+1}\right)\\\\
&=-\frac{1}{(a+1)^2}
\end{align}$$
as was to be shown!
A: Let $0<\varepsilon<1$ and consider the integral 
$$I(\varepsilon):=\int_{\varepsilon}^1x^a\log x\,dx$$
A change of variable $\log x=u$ yields
$$I(\varepsilon)=\int_{\log\varepsilon}^0e^{u(a+1)}u\,du$$
Using integration by parts 
\begin{align}
I(\varepsilon)& =\frac{1}{a+1}e^{u(a+1)}u\Big|^0_{\log\varepsilon}-\frac{1}{a+1}\int^0_{\log\varepsilon}e^{u(a+1)}\,du\\ &=\frac{1}{a+1}e^{u(a+1)}u\Big|^0_{\log\varepsilon}-\frac{1}{(a+1)^2}e^{u(a+1)}\Big|^0_{\log\varepsilon}\\&=-\frac{\varepsilon^{a+1}}{a+1}\log\varepsilon-\Big[\frac{1}{(a+1)^2}-\frac{\varepsilon^{a+1}}{(a+1)^2}\Big]\\&=-\frac{\varepsilon^{a+1}}{a+1}\log\varepsilon+\frac{\varepsilon^{a+1}}{(a+1)^2}-\frac{1}{(a+1)^2}
\end{align}
The last expression is well defined whenever $a\neq -1$. In particular when $a>-1$ we take the limit $$\lim_{\varepsilon\to 0}I(\varepsilon)=\lim_{\varepsilon\to 0}\Big[-\frac{\varepsilon^{a+1}}{a+1}\log\varepsilon+\frac{\varepsilon^{a+1}}{(a+1)^2}-\frac{1}{(a+1)^2}\Big]=-\frac{1}{(a+1)^2}<\infty$$
The second term vanishes, while the first term by L'Hospital's rule vanishes as well. Note also that for $x\in(0,1)$ it holds $|\log x|=-\log x$ so the above estimate suffices for the claim. 
A: Use that $|\log(x) | \leq c_\epsilon x^{-\epsilon}$ for $x\in (0,1)$ and a suitable (depending on $a$) $\epsilon >0$.
