Integrating a power series (continued)

 In the previous post I showed how we can integrate a power series term by term to find the power representation of other functions. For example, we can integrate the power series of the function f(x) = ln(1+x) to find the power series representation of the f(x) = 1/1+x.

Example 

Solution

a) The derivative of f(x) = ln(1+x) is f'(x) = 1/1+x

Let's find a power series representation of f'(x) in order to integrate it:

The integral ∫f'(x)dx represents the antiderivatives of ln(1+x). We have:

ln(1+x) = C +  x  -x/2² +x³/3- x⁴/4 + ....

Since the power series representation of f'(x) is defined for ❙x❙<1 or -1<x<1, let's find the antiderivative for x = 0

ln( 1 + 0) = C + 0 + 0 + 0 + ....

ln1 = C lni1 = 0. Therefore C = 0.

Let's substitute C:

ln(1+x) =  x  -x²/2 +x³/3- x⁴/4 + 

 This represents an an alternating series that has the form:

The above expression is valid for ❘x❘<1. At x=1 we have by substituting x in the first expression of ln(x+1):

ln2 = 1-1/2 +1/3-1/4. We have an alternating harmonic series that converges.

At x =-1 we have ln0 = -1-1/2-1/3-1/4, which is an harmonic series that diverges. The interval of convergence is (-1,1)

Practice. Do the second exercise in the example