# Use power series to evaluate the indefinite integral. What is the radius of convergence? \int...

## Question:

Use power series to evaluate the indefinite integral. What is the radius of convergence? {eq}\int \frac {t}{1-t^3} dt {/eq}

## Indefinite Integral:

The series representation of the indefinite integral will have the integration constant c along with the series form. then the interval of convergence and the radius of convergence of the series can also be found.

We have the indefinite integral:

{eq}\displaystyle \int \frac {t}{1-t^3} dt \\ {/eq}

So to integrate, we can use the series form by the help of the series expansion formula:

{eq}\displaystyle \left(1-x\right)^{-1} = \sum _{n=0}^{\infty }\:x^n\\ {/eq}

So the integral of the series form will be given by:

{eq}\int \frac {t}{1-t^3} dt\\ \displaystyle = \int t ((1-t^3))^{-1} dt\\ \displaystyle =\int t (\sum _{n=0}^{\infty }\:(-t^3)^n) dt\\ \displaystyle =\int (\sum _{n=0}^{\infty }\:(-1)^n(t^{3n+1})) dt\\ \displaystyle =\sum _{n=0}^{\infty }\:(-1)^n\frac{(t^{3n+1+1})}{3n+1+1}+c \\ \displaystyle =\sum _{n=0}^{\infty }\:(-1)^n\frac{(t^{3n+2})}{3n+2}+ c \\ {/eq} Here c is the integral constant.

So this is the series form of the given indefinite integral, whose radius of convergence is given by the ratio test as follows:

{eq}\displaystyle \lim _{n\to \infty }|\frac{a_{n+1}}{a_n}|\\ \displaystyle = \lim _{n\to \infty \:}\left(\left|\frac{\left(-1\right)^{\left(n+1\right)}\frac{t^{3\left(n+1\right)+2}}{3\left(n+1\right)+2}}{\left(-1\right)^n\frac{t^{3n+2}}{3n+2}}\right|\right)\\ \displaystyle =\lim _{n\to \infty \:}\left(\left|-\frac{t^3\left(3n+2\right)}{3n+5}\right|\right)\\ =\left|t^3\right|\\ {/eq}

Then we apply the convergence condition:

{eq}\left|t^3\right|<1\\ \Rightarrow -1<t<1 {/eq}

is the required convergence interval. 