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Power Series Operations · Tutorial 763 of 998

Antiderivative Coefficients and Constants

Integrate power series term by term, track the shifted coefficients, and use an initial value to determine the unique constant.

Intermediate 9 min read

What You'll Learn

  • Convert each coefficient of a power series into the coefficient of its integrated term.
  • Write an antiderivative using either the original index or a shifted index.
  • Determine the integration constant from a value at the center or at another interior point.
  • Work with dense and sparse exponent patterns without inserting nonexistent terms.
  • Check an antiderivative by differentiating and checking its initial value.
  • Recognize common errors involving exponents, denominators, and the constant term.

From Power Series to Antiderivatives

A power series describes a function by its coefficients and powers of a centered variable. Integrating it term by term changes both the exponents and the coefficients: the exponent increases by one, and the coefficient is divided by the new exponent. A separate constant is needed because differentiation removes constant terms.

These changes are simple to state, but indexing matters. In particular, the constant of integration is the coefficient of the zeroth power, while the integrated terms begin with the first power. Once these roles are kept distinct, the same procedure works for series with consecutive exponents and for sparse series that use only selected powers.

The Coefficient Rule

Suppose a power series centered at \(c\) represents \(f\) on its interval of convergence, with radius \(R>0\). The termwise integration theorem gives an antiderivative on the interior \(|x-c|<R\). For each term, integration uses the power rule with the centered variable \(x-c\).

Coefficient rule: If \(f(x)=\sum_{n=0}^{\infty}a_n(x-c)^n\) has radius of convergence \(R>0\), then on \(|x-c|<R\) its antiderivatives have the form \(F(x)=C+\sum_{n=0}^{\infty}\frac{a_n}{n+1}(x-c)^{n+1}\), where \(C\) is an arbitrary constant.

To read the formula, start with the term \(a_n(x-c)^n\). Its integral is \(\frac{a_n}{n+1}(x-c)^{n+1}\). The denominator is \(n+1\), not \(n\), because it is the exponent after integration. The first integrated term, corresponding to \(n=0\), is \(a_0(x-c)\); it is not a constant term.

The same result can be written with the exponent of the integrated series as the index. Set \(m=n+1\), so \(n=m-1\). Then the formula becomes

$$ F(x)=C+\sum_{m=1}^{\infty}\frac{a_{m-1}}{m}(x-c)^m. $$

This shifted-index form makes the coefficient pattern especially clear: the coefficient of \((x-c)^m\) in the antiderivative is \(a_{m-1}/m\). The constant \(C\) sits outside the sum because the sum begins at \(m=1\).

Why the Rule Works

For a single term, the power rule gives the coefficient and exponent transformation directly. The termwise integration theorem allows this operation to be applied to the full power series at interior points of its interval of convergence.

1
Integrate one term.
For \(a_n(x-c)^n\), the power rule gives \(\frac{a_n}{n+1}(x-c)^{n+1}\).
2
Apply termwise integration.
On \(|x-c|<R\), the termwise integration theorem says the integrated series defines an antiderivative of the original series.
3
Include the full family of antiderivatives.
Any two antiderivatives on an interval differ by a constant, so add \(C\) to represent every possibility.

For the last step, if \(F\) and \(G\) are antiderivatives of the same function on an interval, then \((F-G)'=0\) throughout that interval. By the mean value theorem, \(F-G\) has the same value at any two points in the interval. Therefore \(F-G\) is constant there. This also explains why one initial value is sufficient to determine the constant uniquely.

Determining the Constant from an Initial Value

Let \(F(x_0)=K\) be given at a point \(x_0\) in the interior \(|x_0-c|<R\). Substituting \(x_0\) into the antiderivative formula gives the constant directly.

$$ C=K-\sum_{n=0}^{\infty}\frac{a_n}{n+1}(x_0-c)^{n+1}. $$

If the initial point is the center, then every integrated term vanishes there, so \(F(c)=C\). For a different interior point, the integrated series generally contributes a nonzero value, which must be subtracted from the prescribed value. The series in this expression is evaluated at an interior point; this formula does not assert convergence at an endpoint.

Worked Example: A Dense Coefficient Sequence

Worked Example: Integrating Consecutive Powers

Find the antiderivative \(F\) satisfying \(F(1)=-2\) for \(f(x)=\sum_{n=0}^{\infty}(n+1)(x-1)^n\), on its interval of convergence.

Here the center is \(c=1\), and the coefficient of \((x-1)^n\) is \(a_n=n+1\). Applying the coefficient rule gives the integrated coefficient \(\frac{a_n}{n+1}=\frac{n+1}{n+1}=1\). Thus each integrated term is \((x-1)^{n+1}\), and

$$ F(x)=C+\sum_{n=0}^{\infty}(x-1)^{n+1}. $$

At \(x=1\), every term in the sum is zero. Therefore \(F(1)=C\). The initial value \(F(1)=-2\) gives \(C=-2\), so

$$ F(x)=-2+\sum_{n=0}^{\infty}(x-1)^{n+1}. $$

Check the initial value: substituting \(x=1\) makes each positive power zero, leaving \(F(1)=-2\). Check the derivative by the termwise differentiation theorem in the interior of the convergence interval:

$$ F'(x)=\sum_{n=0}^{\infty}(n+1)(x-1)^n=f(x). $$

The recovered coefficient in each degree is exactly the original one.

Worked Example: A Sparse Sequence

Worked Example: Integrating Even Powers

Find the antiderivative \(F\) satisfying \(F(-2)=3\) for \(f(x)=\sum_{k=0}^{\infty}(2k+1)(x+2)^{2k}\), on its interval of convergence.

The center is \(-2\), and only even powers occur in \(f\). For the term with index \(k\), the exponent is \(2k\), so integration gives

$$ \int (2k+1)(x+2)^{2k}\,dx = \frac{2k+1}{2k+1}(x+2)^{2k+1} = (x+2)^{2k+1}. $$

Consequently, the antiderivative series is \(F(x)=C+\sum_{k=0}^{\infty}(x+2)^{2k+1}\). At the center \(x=-2\), all these terms vanish. The initial value therefore gives \(C=3\), and

$$ F(x)=3+\sum_{k=0}^{\infty}(x+2)^{2k+1}. $$

For the initial-value check, \(F(-2)=3+\sum_{k=0}^{\infty}0^{2k+1}=3\). Differentiating term by term at interior points gives

$$ F'(x)=\sum_{k=0}^{\infty}(2k+1)(x+2)^{2k}=f(x). $$

The integrated series contains only odd powers. No even-power terms need to be inserted with zero coefficients; the sparse indexing already records which powers occur.

Worked Example: An Initial Value Away from the Center

Worked Example: Finding the Constant at a Different Point

Find the antiderivative \(F\) satisfying \(F(2)=10\) when \(f(x)=2+6(x-1)+4(x-1)^2\).

The center is \(c=1\). Integrate the constant term, the first-power term, and the second-power term separately:

$$ 2 \longmapsto 2(x-1),\qquad 6(x-1) \longmapsto 3(x-1)^2,\qquad 4(x-1)^2 \longmapsto \frac{4}{3}(x-1)^3. $$

Thus \(F(x)=C+2(x-1)+3(x-1)^2+\frac{4}{3}(x-1)^3\). At \(x=2\), \(x-1=1\), so the nonconstant terms have total value

$$ 2(1)+3(1)^2+\frac{4}{3}(1)^3 =2+3+\frac{4}{3} =\frac{6}{3}+\frac{9}{3}+\frac{4}{3} =\frac{19}{3}. $$

The condition \(F(2)=10\) therefore requires \(C=10-\frac{19}{3}=\frac{30}{3}-\frac{19}{3}=\frac{11}{3}\). The answer is

$$ F(x)=\frac{11}{3}+2(x-1)+3(x-1)^2+\frac{4}{3}(x-1)^3. $$

Check the initial value: \(F(2)=\frac{11}{3}+2+3+\frac{4}{3} =\frac{11+6+9+4}{3} =\frac{30}{3}=10\). Check by differentiation: \(F'(x)=2+6(x-1)+4(x-1)^2=f(x)\).

A Common Coefficient Error

A frequent mistake is to divide by the original exponent instead of by the exponent after integration. For instance, consider the term \(4(x-c)^2\). The incorrect calculation

$$ \int 4(x-c)^2\,dx \ \mathrel{\color{red}{\not=}}\ \frac{4}{2}(x-c)^3 $$

uses \(2\) in the denominator. Differentiating the proposed result gives \(6(x-c)^2\), not \(4(x-c)^2\), so it fails the check. The correct calculation uses the new exponent \(3\):

$$ \int 4(x-c)^2\,dx = \frac{4}{3}(x-c)^3, \qquad \frac{d}{dx}\left(\frac{4}{3}(x-c)^3\right) = 4(x-c)^2. $$

Another indexing error is to treat the integration constant as the \(n=0\) term of the original series. That term integrates to \(a_0(x-c)\), not to a constant. The arbitrary constant is added separately and is determined by the initial value.

Using the Rule Reliably

For a dense series, apply the transformation to every index in turn. For a sparse series, apply it to each term that actually occurs: an original exponent \(q\) becomes \(q+1\), and its coefficient is divided by \(q+1\). In either case, keep the arbitrary constant outside the integrated sum until the initial value is imposed.

The derivative check is especially useful because it reverses the coefficient transformation: differentiating \(\frac{a_n}{n+1}(x-c)^{n+1}\) multiplies by \(n+1\) and lowers the exponent to \(n\), recovering \(a_n(x-c)^n\). The initial-value check then verifies the one remaining choice, the constant.

The series formulas here apply on the interior of the original interval of convergence. Endpoint behavior requires separate analysis; an interior antiderivative formula alone does not settle whether the integrated series converges at an endpoint.

Check Your Understanding

For each question, attend to both the coefficient transformation and the role of the constant.

  1. If \(f(x)=\sum_{n=0}^{\infty}a_n(x-c)^n\), what is the coefficient of \((x-c)^m\) in the nonconstant part of an antiderivative?
  2. Why is the constant of integration not the integrated term corresponding to \(n=0\)?
  3. Find the antiderivative series for \(\sum_{k=0}^{\infty}5(x-3)^{4k+2}\), leaving the arbitrary constant as \(C\).
  4. For an initial value \(F(c)=K\), what must the integration constant be?
  5. How can differentiating a proposed antiderivative detect an incorrect denominator?

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