Show that $x^a \log{x}$ is integrable on $(0,1)$ for all $a>-1$

Question

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.

Answer 1

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!

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We can proceed further and evaluate the integral in closed form. Note that we can write

$$\begin{align} \int_0^1 \log(x)x^a\,dx&=\frac d{da}\int_0^1 x^a\,dx\\\\ &=\frac d{da}\left(\frac{1}{a+1}\right)\\\\ &=-\frac{1}{(a+1)^2} \end{align}$$

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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!

Answer 2

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.

Answer 3

Use that $|\log(x) | \leq c_\epsilon x^{-\epsilon}$ for $x\in (0,1)$ and a suitable (depending on $a$) $\epsilon >0$.