Integration by Parts — Question 10

Question 10 – Solution

Because $\ln(x)\to-\infty$ as $x\to0^+$, write the integral as $$J_n=\lim_{\varepsilon\to0^+} \int_\varepsilon^1x^n\ln(x)\,dx.$$ For $\varepsilon>0$, use integration by parts with $$u=\ln(x),\quad dv=x^n\,dx, \qquad du=\frac1x\,dx,\quad v=\frac{x^{n+1}}{n+1}.$$ This choice is valid because $n>0$, so in particular $n+1\ne0$. Using $\int u\,dv=uv-\int v\,du$, we obtain $$\begin{align*} \int_\varepsilon^1x^n\ln(x)\,dx &=\left.\frac{x^{n+1}\ln(x)}{n+1}\right|_\varepsilon^1 -\frac1{n+1}\int_\varepsilon^1x^n\,dx \\ &=\left.\frac{x^{n+1}\ln(x)}{n+1}\right|_\varepsilon^1 -\left.\frac{x^{n+1}}{(n+1)^2}\right|_\varepsilon^1. \tag{1} \end{align*}$$

We must evaluate the boundary term at $x=0$. Since $n+1>0$, $$\lim_{x\to0^+}x^{n+1}\ln(x) =\lim_{x\to0^+}\frac{\ln(x)}{x^{-(n+1)}}.$$ This has the form $-\infty/\infty$. By l’Hôpital’s rule, $$\begin{align*} \lim_{x\to0^+}\frac{\ln(x)}{x^{-(n+1)}} &=\lim_{x\to0^+} \frac{1/x}{-(n+1)x^{-(n+2)}} \\ &=\lim_{x\to0^+}-\frac{x^{n+1}}{n+1}=0. \end{align*}$$ Also, $\lim_{x\to0^+}x^{n+1}=0$. At the upper endpoint, $\ln(1)=0$. Therefore, taking $\varepsilon\to0^+$ in equation (1) gives $$\begin{align*} J_n &=\frac{1^{n+1}\ln(1)}{n+1}-0 -\frac{1^{n+1}-0}{(n+1)^2} \\ &=-\frac1{(n+1)^2}. \end{align*}$$ Hence, $$\boxed{\displaystyle J_n=-\frac1{(n+1)^2}}.$$

For every $n>0$, $J_n<0$, which agrees with $\ln(x)<0$ on $0<x<1$. Moreover, as $n\to\infty$, $$\lim_{n\to\infty}J_n =\lim_{n\to\infty}-\frac1{(n+1)^2} =0.$$ The values increase toward zero through negative values. In short, $$\boxed{J_n\to0^-\quad\text{as}\quad n\to\infty}.$$