Question

Evaluate the indefinite integral $\int \frac{{\tan }^{-1}\left(x\right)}{x}dx$ as a power series. What is the radius of convergence $R$?

Answer 1

Step 1: Evaluate the indefinite in terms of power series:

Given,$\int \frac{{\tan }^{-1}\left(x\right)}{x}dx$

We know that the series of ${\tan }^{-1}\mathrm{x}$ can be written as:

$\begin{array}{rcl}{\tan }^{-1}\left(\mathrm{x}\right)& =& \mathrm{x}-\frac{{\mathrm{x}}^3}{3}+\frac{{\mathrm{x}}^5}{5}+....\\ \therefore {\tan }^{-1}\left(\mathrm{x}\right)& =& \sum _0^{\infty }{\left(-1\right)}^{\mathrm{n}}\frac{{\mathrm{x}}^{2\mathrm{n}+1}}{2\mathrm{n}+1}\end{array}$

So,

$$\begin{gathered} \Rightarrow \frac{{\tan }^{-1}\left(x\right)}{x}=\frac{\sum _0^{\infty }{\left(-1\right)}^n\frac{x^{2n+1}}{2n+1}}{x} \\ =\sum _0^{\infty }{\left(-1\right)}^n\frac{x^{2n}.\cancel{x}}{2n+1}\times \frac{1}{\cancel{x}}\left[\because x^{2n+1}=x^{2n}.x\right] \\ =\sum _0^{\infty }{\left(-1\right)}^n\frac{x^{2n}}{2n+1} \end{gathered}$$

Substitute this in given integral

$$\begin{gathered} \int \frac{{\tan }^{-1}\left(x\right)}{x}dx=\int \sum _0^{\infty }{\left(-1\right)}^n\frac{x^{2n}}{2n+1}dx \\ =\sum _0^{\infty }{\left(-1\right)}^n\times \int \frac{x^{2n}}{2n+1}dx \\ =\sum _0^{\infty }{\left(-1\right)}^n\times \frac{1}{2n+1}\times \int x^{2n}dx \\ =\sum _0^{\infty }{\left(-1\right)}^n\times \frac{1}{2n+1}\times \frac{x^{2n+1}}{2n+1}+C\left[\because \int x^{n}dx=\frac{x^{n+1}}{n+1}+C\right] \end{gathered}$$

Step 2: Evaluate $a_n$and $a_{n+1}$

The radius of the convergence $R=\underset{n\to \infty }{\lim }\frac{a_{n+1}}{a_n}$

Let

$a_n=\sum _0^{\infty }{\left(-1\right)}^n\times \frac{1}{2n+1}\times \frac{x^{2n+1}}{2n+1}$

So,

$$\begin{gathered} a_{n+1}=\sum _0^{\infty }{\left(-1\right)}^{n+1}\times \frac{1}{2\left(n+1\right)+1}\times \frac{x^{2\left(n+1\right)+1}}{2\left(n+1\right)+1} \\ =\sum _0^{\infty }{\left(-1\right)}^{n+1}\times \frac{1}{2n+2+1}\times \frac{x^{2n+2+1}}{2n+2+1} \\ =\sum _0^{\infty }{\left(-1\right)}^{n+1}\times \frac{1}{2n+3}\times \frac{x^{2n+3}}{2n+3} \end{gathered}$$

Step 3: Evaluate $\frac{a_{n+1}}{a_n}$:

$$\begin{gathered} \Rightarrow \frac{a_{n+1}}{a_n}=\frac{\sum _0^{\infty }{\left(-1\right)}^{n+1}\times \frac{1}{2n+3}\times \frac{x^{2n+3}}{2n+3}}{\sum _0^{\infty }{\left(-1\right)}^n\times \frac{1}{2n+1}\times \frac{x^{2n+1}}{2n+1}} \\ =\frac{\cancel{\sum _0^{\infty }{\left(-1\right)}^n}\times \left(-1\right)\times \frac{1}{2n+3}\times \frac{\cancel{x^{2n+1}}.x^2}{2n+3}}{\cancel{\sum _0^{\infty }{\left(-1\right)}^n}\times \frac{1}{2n+1}\times \frac{\cancel{x^{2n+1}}}{2n+1}} \\ =\frac{\left(-1\right)\times \frac{1}{2n+3}\times \frac{x^2}{2n+3}}{\frac{1}{2n+1}\times \frac{1}{2n+1}} \\ =\frac{\frac{-x^2}{{\left(2n+3\right)}^2}}{\frac{1}{{\left(2n+1\right)}^2}} \\ =\left(\frac{-x^2}{{\left(2n+3\right)}^2}\right)\times {\left(2n+1\right)}^2 \end{gathered}$$

Step 4: Finding the radius of the convergence:

We know that,

A power series about $x=a$ is $\sum _{n=0}^{\mathrm{\infty }}{a}_n{\left(x-a\right)}^n$, then $R=\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|$ if $R<1$ the series converges and After obtaining an expression |$\left|x-a\right|<r$, $r$ the radius of convergence.

The radius of the convergence $R=\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|$

$$\begin{gathered} \Rightarrow R=\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right| \\ =\underset{n\to \infty }{\lim }\left|\left(\frac{-x^2}{{\left(2n+3\right)}^2}\right)\times {\left(2n+1\right)}^2\right| \\ =\underset{n\to \infty }{\lim }\left|\left(\frac{-x^2}{{\left(n\left(2+\frac{3}{n}\right)\right)}^2}\right)\times {\left(n\left(2+\frac{1}{n}\right)\right)}^2\right| \\ =-x^2\underset{n\to \infty }{\lim }\left|\left(\frac{\cancel{n^2}{\left(2+\frac{1}{n}\right)}^2}{\cancel{n^2}{\left(2+\frac{3}{n}\right)}^2}\right)\right| \\ =\left|-x^2\times \frac{{\left(2\right)}^2}{{\left(2\right)}^2}\right| \\ =\left|-x^2\right| \\ =x^2 \end{gathered}$$

$$\begin{gathered} \Rightarrow x^2<1 \\ \Rightarrow x<\pm 1 \\ \Rightarrow -1<x<1 \\ \Rightarrow \left|x\right|<1 \end{gathered}$$

We know that,

$\left|x-a\right|<r$, then $r$ is the radius of convergence.

Here,

$$\begin{gathered} \Rightarrow \left|x\right|<1 \\ \Rightarrow |x-0|<1 \end{gathered}$$

Hence, the radius of the convergence is $1$.