Question

The Calculus of Power Series: Let \(\sum_{k=0}^{\infty }a_{k}\left ( x-x_{0} \right )^{k}\) be a power series in \(x-x_{0}\) that converges to a function \(f(x)\) on an interval \(I\).

The indefinite integral of \(f\) is given by \(\int f\left ( x \right )dx =\) _____.

The interval of convergence for this new series is ______, with the possible exception that _____.

Short answer

The indefinite integral of \(f\) is given by \(\int f\left ( x \right )dx =\sum_{k=0}^{\infty }\frac{a_{k}}{k+1}\left ( x-x_{0} \right )^{k+1}+C\).

The interval of convergence for this new series is \(I\), with the possible exception that if \(I\) is a finite interval, the convergence at the endpoints of \(I\) may be different than it was for the initial series.

Step-by-step solution

Step 1. The indefinite integral of \(f\)

The given function is \(\sum_{k=0}^{\infty }a_{k}\left ( x-x_{0} \right )^{k}\).

So, the indefinite integral of the function is

\(\int f\left ( x \right )dx =\int \left ( \sum_{k=0}^{\infty }a_{k}\left ( x-x_{0} \right )^{k} \right )dx\)

\(\int f\left ( x \right )dx =\sum_{k=0}^{\infty }a_{k}\int \left ( x-x_{0} \right )^{k}dx\)

\(\int f\left ( x \right )dx =\sum_{k=0}^{\infty }\frac{a_{k}}{k+1}\left ( x-x_{0} \right )^{k+1}+C\)

Step 2. The indefinite integral of \(f\)

Therefore, the indefinite integral of \(f\) is given by \(\int f\left ( x \right )dx =\sum_{k=0}^{\infty }\frac{a_{k}}{k+1}\left ( x-x_{0} \right )^{k+1}+C\).

The interval of convergence for this new series is \(I\), with the possible exception that if \(I\) is a finite interval, the convergence at the endpoints of \(I\) may be different than it was for the initial series.