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Keep the old version. Delete my work and update to the new version.We integrate power series term by term.
Suppose that the power series converges for all in some open interval . Then, on this interval, the power series has an anti-derivative which can be obtained by integrating term-by-term:
In other words, the indefinite integral of a power series is computed term by term, as we would anti-differentiate a polynomial.
example 1 The power series converges on the interval . On that interval, its anti-derivative is given by The interval of convergence of the anti-derivative is . In general, the anti-derivative might converge at an endpoint when the original series did not.
(problem 1) Find the anti-derivatives of the following power series. Also, compare the interval of convergence of the original series to its anti-derivative.
example 2 The power series expansion for the function is Integrate the function and the series to obtain a power series representation for a logaritmic function.
First, note that Integrating the power series: Equating these two antiderivatives, we have If we set , we can find : so . Hence, and we have a power series representiation for a logarithmic function.
Remark 1 Note that the above series converges when since it is the alternating harmonic series. Substituting gives its sum: Hence, the sum of the alternating harmonic series is .
Remark 2 Since we have . Thus, the absolute value bars in the logarithm can be removed to obtain:example 3 The power series representation for the function is Integrate the function and the series to obtain a power series representation for an inverse trigonometric function.
We have and integrating the power series gives: Equating these two antiderivatives, we have If we set , we can find : so and we have the following power series representation for the inverse tangent function:
Remark If we substitute into both sides of the power series representation, we obtain: Since , we can multiply by to obtain an infinite series whose sum is :