Chapter 8.6, Problem 25E

To determine

To find: To determine the power series expansion of indefinite integral and it’s Radius of convergence.

Answer to Problem 25E

  ${\displaystyle \int \frac{x-{\tan }^{-1}\left(x\right)}{x^3}dx={\displaystyle \sum _{n=1}^{\infty }\left(-1\right)\frac{x^{2n-1}}{\left(2n+1\right)\left(2n-1\right)}}}$

And Radius of Convergence is $1$ .

Explanation of Solution

Given information: ${\displaystyle \int \frac{x-{\tan }^{-1}\left(x\right)}{x^3}dx}$

Calculation:

We know that, $\begin{array}{l}{\tan }^{-1}x={\displaystyle \sum _{n=0}^{\infty }{\left(-1\right)}^n}\frac{x^{2n+1}}{2n+1}\left|x\right|<1\\ \Rightarrow \frac{{\tan }^{-1}x}{x^3}={\displaystyle \sum _{n=0}^{\infty }{\left(-1\right)}^n}\frac{x^{2n-2}}{2n+1}\left|x\right|<1\end{array}$

Now,

  $\begin{array}{l}{\displaystyle \int \frac{x-{\tan }^{-1}\left(x\right)}{x^3}dx}\\ ={\displaystyle \int \frac{1}{x^2}dx}-{\displaystyle \int \frac{{\tan }^{-1}\left(x\right)}{x^3}dx}\\ ={\displaystyle \int \frac{1}{x^2}dx}-{\displaystyle \int {\displaystyle \sum _{n=0}^{\infty }{\left(-1\right)}^n\frac{x^{2n-2}}{2n+1}}dx}\\ ={\displaystyle \int \frac{1}{x^2}dx}-{\displaystyle \int \frac{1}{x^2}dx}-{\displaystyle \int {\displaystyle \sum _{n=1}^{\infty }{\left(-1\right)}^n\frac{x^{2n-2}}{2n+1}}dx}\\ =-{\displaystyle \int {\displaystyle \sum _{n=1}^{\infty }{\left(-1\right)}^n\frac{x^{2n-2}}{2n+1}}dx}\\ =-{\displaystyle \sum _{n=1}^{\infty }{\left(-1\right)}^n\frac{x^{2n-1}}{\left(2n+1\right)\left(2n-1\right)}}+c\end{array}$

Put, $x=0$

Then, $c=0$

Hence, ${\displaystyle \int \frac{x-{\tan }^{-1}\left(x\right)}{x^3}dx={\displaystyle \sum _{n=1}^{\infty }\left(-1\right)\frac{x^{2n-1}}{\left(2n+1\right)\left(2n-1\right)}}}$

And Radius of Convergence is $1$ .