Key to "Evaluate the indefinite integral as a power series...."

Brennan Flores

Answered question

2021-10-09

Evaluate the indefinite integral as a power series. What is the radius of convergence?

\int \frac{t}{1} - t^{8} d t

Answer & Explanation

Velsenw

Skilled2021-10-10Added 91 answers

Let's transform the integrand into a more familiar form:
$\frac{t}{1 - t^{8}} = t \cdot \frac{1}{1 - t^{8}} = t \cdot \sum_{n = 0}^{\infty} {(t^{8})}^{n} = t \cdot \sum_{n = 0}^{\infty} t^{8 n} = \sum_{n = 0}^{\infty} t^{8 n + 1}$
Now we can easily integrate the function:
$\int \frac{t}{1 - t^{8}} d t = \int \sum_{n = 0}^{\infty} t^{8 n + 1} d t = \sum_{n = 0}^{\infty} \int t^{8 n + 1} d t = \sum_{n = 0}^{\infty} \frac{t^{8 n + 2}}{8 n + 2} + C$
This is possible on the interval of convergence, which is:
$| t^{8} | < 1 \Leftrightarrow | t | < 1 \Rightarrow R = 1$
Result: $\int \frac{t}{1 - t^{8}} d t = \int \sum_{n = 0}^{\infty} t^{8 n + 1} d t = \sum_{n = 0}^{\infty} \int t^{8 n + 1} d t = \sum_{n = 0}^{\infty} \frac{t^{8 n + 2}}{8 n + 2} + C$

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Integral Calculus

Didn't find what you were looking for?