Chapter 5, Problem 6P

To determine

To prove:

  ${\displaystyle {\int }_0^1{\left(\ln x\right)}^{n}dx={\left(-1\right)}^{n}n!}$

Explanation of Solution

Given information:

The positive integer is $n$

Formula used:

  ${\displaystyle \int udv=uv-{\displaystyle \int vdu}}$

Proof:

The given expression in the left hand side is,

  ${\displaystyle {\int }_0^1{\left(\ln x\right)}^{n}dx}$

Evaluate this integral by the use of integration by parts

  $\begin{array}{l}u=\ln x\\ du=\frac{1}{x}dx\\ dv=1\\ v=x\\ u\cdot v-{\displaystyle \int vdu=x\ln x-{\displaystyle \int x\frac{1}{x}dx}}\\ =x\ln x-x\end{array}$

Put back the limit, so

  ${x\ln x-x]}_0^1=1$

Repeat the process for $u={\left(\ln x\right)}^2$ , so

  $\begin{array}{l}u={\left(\ln x\right)}^2\\ du=2\ln x\cdot \frac{1}{x}dx\\ dv=1\\ v=x\\ u\cdot v-{\displaystyle \int vdu=x{\left(\ln x\right)}^2-2{\displaystyle \int x\ln x\cdot \frac{1}{x}dx}}\\ =x{\left(\ln x\right)}^2-2{\displaystyle \int \ln xdx}\\ =x{\left(\ln x\right)}^2-2\left(x\ln x-x\right)\end{array}$

Again put back the limit,

  ${\left[x{\left(\ln x\right)}^2-2\left(x\ln x-x\right)\right]}_0^1=2$

Now, repeat this process $\left(n-2\right)$ more times to obtain the value,

  $\left(-1\right)\cdot 2\cdot \left(-1\right)\cdot 3\cdot \left(-1\right)\cdots \cdots \left(n-1\right)\left(-1\right)\left(n\right)\left(-1\right)$

Hence, the $n$ element is obtained as $\left(-1\right)$

Therefore,

  ${\displaystyle {\int }_0^1{\left(\ln x\right)}^{n}dx={\left(-1\right)}^{n}n!}\left[\text{hence proved}\right]$