Evaluate indefinite integral as power series

Consider a power series [latex]\displaystyle\sum _^<\infty >_^=_+_x+_^+\cdots [/latex] that converges on some interval I, and let [latex]f[/latex] be the function defined by this series. Here we address two questions about [latex]f[/latex].

We know that, for a polynomial with a finite number of terms, we can evaluate the derivative by differentiating each term separately. Similarly, we can evaluate the indefinite integral by integrating each term separately. Recall the general expression of the power rules for derivatives and integrals of polynomials.

Recall: Power rule for Derivatives and Integrals

  1. [latex] \frac\left (x^n \right) = nx^ [/latex]
  2. [latex] \int = \fracx^ + C \:(\text\: n \ne -1 [/latex])

In this section we show that we can take advantage of the simplicity of integrating and differentiating polynomials to do the same thing for convergent power series. That is, if

[latex]f\left(x\right)=_^=_+_x+_^+\cdots [/latex]

converges on some interval I, then

[latex]^<\prime >\left(x\right)=_+2_x+3_^+\cdots [/latex] [latex]\displaystyle\int f\left(x\right)dx=C+_x+_\frac^>+_\frac^>+\cdots [/latex].

Evaluating the derivative and indefinite integral in this way is called term-by-term differentiation of a power series and term-by-term integration of a power series , respectively. The ability to differentiate and integrate power series term-by-term also allows us to use known power series representations to find power series representations for other functions. For example, given the power series for [latex]f\left(x\right)=\frac[/latex], we can differentiate term-by-term to find the power series for [latex]^<\prime >\left(x\right)=\frac<<\left(1-x\right)>^>[/latex]. Similarly, using the power series for [latex]g\left(x\right)=\frac[/latex], we can integrate term-by-term to find the power series for [latex]G\left(x\right)=\text\left(1+x\right)[/latex], an antiderivative of g. We show how to do this in the next two examples. First, we state Term-by-Term Differentiation and Integration for Power Series, which provides the main result regarding differentiation and integration of power series.

theorem: Term-by-Term Differentiation and Integration for Power Series

Suppose that the power series [latex]\displaystyle\sum _^<\infty >_<\left(x-a\right)>^[/latex] converges on the interval [latex]\left(a-R,a+R\right)[/latex] for some [latex]R>0[/latex]. Let f be the function defined by the series

for [latex]|x-a|f is differentiable on the interval [latex]\left(a-R,a+R\right)[/latex] and we can find [latex]^<\prime >[/latex] by differentiating the series term-by-term:

The proof of this result is beyond the scope of the text and is omitted. Note that although Term-by-Term Differentiation and Integration for Power Series guarantees the same radius of convergence when a power series is differentiated or integrated term-by-term, it says nothing about what happens at the endpoints. It is possible that the differentiated and integrated power series have different behavior at the endpoints than does the original series. We see this behavior in the next examples.

Example: Differentiating Power Series

      Use the power series representation
    [latex]g\left(x\right)=\frac<1><<\left(1-x\right)>^>[/latex]



    on the interval [latex]\left(-1,1\right)[/latex]. Determine whether the resulting series converges at the endpoints.

    Use the result of part a. to evaluate the sum of the series [latex]\displaystyle\sum _^<\infty >\frac^>[/latex].

    Show Solution
    1. Since [latex]g\left(x\right)=\frac<<\left(1-x\right)>^>[/latex] is the derivative of [latex]f\left(x\right)=\frac[/latex], we can find a power series representation for g by differentiating the power series for f term-by-term. The result is