Chapter 8.4, Problem 29E

(a)

To determine

To state: To state the reason for the given integral to be improper.

Answer to Problem 29E

The given integral is an improper integral because it has an infinite discontinuity at the point $x=0$ .

Explanation of Solution

Given information: The integral ${\displaystyle \underset{0}{\overset{1}{\int }}x\ln \left(x\right)}dx$ is given.

Any definite integral is said to be an improper integral if the integrand approaches infinity (infinite discontinuity) at one or more points that lies in the interval of integration.

Since the integrand in the integral ${\displaystyle \underset{0}{\overset{1}{\int }}x\ln \left(x\right)}dx$ has an infinite discontinuity at the point $x=0$ , the given integral is an improper integral.

(b)

To determine

To evaluate: The given improper integral or state that it diverges.

Answer to Problem 29E

The given improper integral is ${\displaystyle \underset{0}{\overset{1}{\int }}x\ln \left(x\right)}dx=-\frac{1}{4}$ .

Explanation of Solution

Given information: The integral ${\displaystyle \underset{0}{\overset{1}{\int }}x\ln \left(x\right)}dx$is given.

Concept Used: If $f\left(x\right)$ is continuous on the interval $\left(a,b\right]$ , then ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx=\underset{c\to a^{+}}{\lim }{\displaystyle \underset{c}{\overset{b}{\int }}f\left(x\right)}dx$ .

Calculation:

Simplify the integral ${\displaystyle \underset{0}{\overset{1}{\int }}x\ln \left(x\right)}dx$ using the fact that the integrand in the given integral has an infinite discontinuity at the point $x=0$ and continuous on $\left(0,1\right]$ .

Therefore,

  ${\displaystyle \underset{0}{\overset{1}{\int }}x\ln \left(x\right)}dx=\underset{c\to 0^{+}}{\lim }{\displaystyle \underset{c}{\overset{1}{\int }}x\ln \left(x\right)}dx$

Consider $u=\ln \left(x\right)$ and $dv=xdx$ such that $du=\frac{dx}{x}$ and $v={\displaystyle \int x}dx=\frac{x^2}{2}$ . Simplify the integral ${\displaystyle \underset{c}{\overset{1}{\int }}x\ln \left(x\right)}dx$ using integration by parts, that is, ${\displaystyle \int udv=uv-{\displaystyle \int vdu}}$ by substituting $u=\ln \left(x\right)$ , $dv=xdx$ , and $v=\frac{x^2}{2}$ into ${\displaystyle \int udv=uv-{\displaystyle \int vdu}}$ .

  $\begin{array}{c}{\displaystyle \underset{0}{\overset{1}{\int }}x\ln \left(x\right)}dx=\underset{c\to 0^{+}}{\lim }{\left[\ln \left(x\right)\frac{x^2}{2}\right]}_c^1-{\displaystyle \underset{c}{\overset{1}{\int }}\frac{x^2}{2}\frac{1}{x}}dx\\ =\underset{c\to 0^{+}}{\lim }{\left[\frac{1}{2}{x}^2\ln \left(x\right)\right]}_c^1-\underset{c\to 0^{+}}{\lim }{\displaystyle \underset{c}{\overset{1}{\int }}\frac{1}{2}x}dx\\ =\underset{c\to 0^{+}}{\lim }{\left[\frac{1}{2}{x}^2\ln \left(x\right)\right]}_c^1-\underset{c\to 0^{+}}{\lim }{\left[\frac{1}{4}{x}^2\right]}_c^1\\ =\underset{c\to 0^{+}}{\lim }\frac{1}{2}\left[1\ln \left(1\right)-c^2\ln \left(c\right)\right]-\underset{c\to 0^{+}}{\lim }\frac{1}{4}\left[1-c^2\right]\\ =\underset{c\to 0^{+}}{\lim }\frac{1}{2}\left[0-c^2\ln \left(c\right)\right]-\underset{c\to 0^{+}}{\lim }\frac{1}{4}\left[1-c^2\right]\\ =\frac{1}{2}\left[0-0\right]-\frac{1}{4}\left[1-0\right]\\ =-\frac{1}{4}\end{array}$

As a result, ${\displaystyle \underset{0}{\overset{1}{\int }}x\ln \left(x\right)}dx=-\frac{1}{4}$ .