Integration by Parts — Question 6

Question 6

Use integration by parts to prove that $$\int_0^1\ln(x)\,dx=-1.$$ Because $\ln(x)$ is unbounded as $x\to0^+$, be sure to treat the lower endpoint correctly as an improper integral.

Question 6 – Solution

The function $\ln(x)$ is not defined at $x=0$ and satisfies $$\lim_{x\to0^+}\ln(x)=-\infty.$$ Therefore, we must interpret the integral as $$\int_0^1\ln(x)\,dx =\lim_{\varepsilon\to0^+}\int_\varepsilon^1\ln(x)\,dx.$$

For $\varepsilon>0$, evaluate the proper integral by integration by parts. Choose $$u=\ln(x),\quad dv=dx, \qquad du=\frac1x\,dx,\quad v=x.$$ Using $\int u\,dv=uv-\int v\,du$, we obtain $$\begin{align*} \int_\varepsilon^1\ln(x)\,dx &=\left.x\ln(x)\right|_\varepsilon^1 -\int_\varepsilon^1x\left(\frac1x\right)\,dx \\ &=\left.x\ln(x)\right|_\varepsilon^1 -\int_\varepsilon^1 1\,dx \\ &=\left[x\ln(x)-x\right]_\varepsilon^1. \end{align*}$$ Now substitute the bounds: $$\begin{align*} \int_\varepsilon^1\ln(x)\,dx &=\bigl(1\ln(1)-1\bigr) -\bigl(\varepsilon\ln(\varepsilon)-\varepsilon\bigr) \\ &=-1-\varepsilon\ln(\varepsilon)+\varepsilon. \end{align*}$$

It remains to evaluate the indeterminate endpoint term. Rewrite it as $$\varepsilon\ln(\varepsilon) =\frac{\ln(\varepsilon)}{1/\varepsilon}.$$ As $\varepsilon\to0^+$, this has the form $-\infty/\infty$. By l’Hôpital’s rule, $$\begin{align*} \lim_{\varepsilon\to0^+}\varepsilon\ln(\varepsilon) &=\lim_{\varepsilon\to0^+} \frac{1/\varepsilon}{-1/\varepsilon^2} \\ &=\lim_{\varepsilon\to0^+}(-\varepsilon)=0. \end{align*}$$ Also, $\lim_{\varepsilon\to0^+}\varepsilon=0$. Hence, $$\begin{align*} \int_0^1\ln(x)\,dx &=\lim_{\varepsilon\to0^+} \left(-1-\varepsilon\ln(\varepsilon)+\varepsilon\right) \\ &=-1. \end{align*}$$ Therefore, the improper integral converges and $$\boxed{\displaystyle \int_0^1\ln(x)\,dx=-1}.$$ This sign is reasonable because $\ln(x)<0$ for $0<x<1$.