Chapter 8.6, Problem 23E

To determine

To find: the power series expansion of the given indefinite intrggral.

Answer to Problem 23E

The power series of ${\displaystyle \int \frac{t}{1-t^8}}dt$ is ${\displaystyle \sum _{n=0}^{\infty }\frac{t^{8n+2}}{8n+2}}+c$

Explanation of Solution

Given information: ${\displaystyle \int \frac{t}{1-t^8}}dt$

We know that ,

  $\frac{1}{t}={\displaystyle \sum _{n=0}^{\infty }{t}^n}\text{ , }\left|t\right|<1\mathrm{......}\text{(1)}$

Now, $\begin{array}{l}{\displaystyle \int \frac{t}{1-t^8}}dt={\displaystyle \int t\cdot \frac{1}{1-t^8}}dt\\ \text{ =}{\displaystyle \int t\cdot {\displaystyle \sum _{n=0}^{\infty }(t^8}}{)}^{n}dt[\text{help of (1)], }\left|t\right|<1\\ \text{ =}{\displaystyle \int {\displaystyle \sum _{n=0}^{\infty }{t}^{8n+1}}}dt\\ \text{ =}{\displaystyle \sum _{n=0}^{\infty }\frac{t^{8n+2}}{8n+2}}+c\end{array}$

Where c is integral constant.

Radius of convergence :

  $R=\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|$ where $a_n=\frac{1}{8n+2}$

And integral of convergence $(-1,1)$ .

Hence, the power series of ${\displaystyle \int \frac{t}{1-t^8}}dt$ is

  ${\displaystyle \sum _{n=0}^{\infty }\frac{t^{8n+2}}{8n+2}}+c$