# 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^9} dt {/eq}

## Indefinite Integrals:

The Taylor series expansion of the Maclaurin series expansion of the integrand expression is integrated and the result is the power series. Then the series' interval of convergence and the radius of convergence can be found with the ratio test or the root test.

Here we have to use power series to evaluate the indefinite integral.

Now, the indefinite integral given is:

{eq}\displaystyle \int \frac {t}{1-t^9} dt \\ {/eq}

Here at first, we use the series form by 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^9} dt\\ \displaystyle = \int t ((1-t^9))^{-1} dt\\ \displaystyle =\int t (\sum _{n=0}^{\infty }\:(-t^9)^n) dt\\ \displaystyle =\int (\sum _{n=0}^{\infty }\:(-1)^n(t^{9n+1})) dt\\ \displaystyle =\sum _{n=0}^{\infty }\:(-1)^n\frac{(t^{9n+1+1})}{9n+1+1}+c \\ \displaystyle =\sum _{n=0}^{\infty }\:(-1)^n\frac{(t^{9n+2})}{9n+2}+ c \\ {/eq} with the c as the integral constant.

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

{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^{9\left(n+1\right)+2}}{9\left(n+1\right)+2}}{\left(-1\right)^n\frac{t^{9n+2}}{9n+2}}\right|\right)\\ \displaystyle =\lim _{n\to \infty \:}\left(\left|-\frac{t^9\left(9n+2\right)}{9n+11}\right|\right)\\ =\left|t^9\right|~~~~~~~~~~~~~~~~~~~~\left [ \because \lim _{n\to \infty \:}\left(\left|\frac{9n+2}{9n+11}\right|\right)=1 \right ]\\ {/eq}

Then we apply the convergence condition:

{eq}\left|t^9\right|<1\\ \Rightarrow -1<t<1 {/eq} 