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Power Series Operations · Tutorial 762 of 998

The Termwise Integration Theorem

Learn the conditions and proof for integrating a power series term by term inside its radius of convergence, and verify that the resulting series is an antiderivative.

Intermediate 10 min read

What You'll Learn

  • State the termwise integration theorem for a power series centered at an arbitrary point
  • Explain why the integrated series converges at every point strictly inside the original radius
  • Use uniform convergence on closed subintervals to justify integrating partial sums and passing to a limit
  • Verify that the resulting series differentiates back to the original power series
  • Apply the theorem to ordinary, weighted, and sparse power series
  • Identify why omitting the factor of one over n plus one gives an incorrect result

Why Termwise Integration Needs Justification

A power series is an infinite sum of functions. Integrating its first few terms is straightforward, but integrating an infinite sum one term at a time requires a reason that the limiting process is valid. In particular, convergence at each individual point is not, by itself, enough to justify interchanging an infinite sum and an integral.

Inside its radius of convergence, a power series has stronger convergence properties: on every closed interval strictly inside that radius, its partial sums converge uniformly. That control over an entire interval allows us to justify the termwise integration rule and to show that the resulting series differentiates back to the original one.

Standing assumptions: Let \(c\in\mathbb R\), let \(a_n\in\mathbb R\), and suppose the power series \(\sum_{n=0}^{\infty}a_n(x-c)^n\) has radius of convergence \(R>0\). All claims below concern real \(x\) with \(|x-c|<R\). No conclusion about convergence at either endpoint is implied.

The Termwise Integration Theorem

Integrating a single term uses the power rule. For each nonnegative integer \(n\),

$$ \int a_n(t-c)^n\,dt = \frac{a_n}{n+1}(t-c)^{n+1}+C. $$

For a power series, this calculation suggests a new series whose terms have powers raised by one and coefficients divided by the new exponent. The theorem states precisely where this operation is valid.

Theorem: If \(\sum_{n=0}^{\infty}a_n(x-c)^n\) has radius of convergence \(R>0\), then the integrated power series $$ \sum_{n=0}^{\infty}\frac{a_n}{n+1}(x-c)^{n+1} $$ converges for every \(x\) satisfying \(|x-c|<R\). Define its sum by \(F(x)\), and define \(f(x)=\sum_{n=0}^{\infty}a_n(x-c)^n\) on that interval. Then \(F\) is differentiable there and \(F'(x)=f(x)\). In particular, \(F\) is an antiderivative of \(f\) throughout the open interval \((c-R,c+R)\).

The theorem is an interior statement. It does not say whether either series converges at \(x=c-R\) or \(x=c+R\), and endpoint behavior must not be inferred from it.

Why the Integrated Series Converges

Fix an \(x\) strictly inside the radius, and let \(r=|x-c|\). If \(r>0\), the original power series converges absolutely at \(x\), so \(\sum_{n=0}^{\infty}|a_n|r^n\) is finite. Since \(1/(n+1)\leq 1\),

$$ \sum_{n=0}^{\infty} \left|\frac{a_n}{n+1}(x-c)^{n+1}\right| = r\sum_{n=0}^{\infty}\frac{|a_n|r^n}{n+1} \leq r\sum_{n=0}^{\infty}|a_n|r^n <\infty. $$

Thus the integrated series converges absolutely at \(x\). If \(r=0\), every term of the integrated series is zero, so convergence holds there as well. This proves convergence at every interior point.

Pointwise convergence alone does not explain why integrating partial sums gives the integral of the limiting function. For that step, we use uniform convergence on closed intervals strictly inside the radius. The reason this uniform convergence holds can be seen directly: on any such interval, choose \(\rho<R\) large enough that \(|t-c|\leq\rho\) throughout the interval. The sum \(\sum |a_n|\rho^n\) converges, and it bounds the absolute values of the original series' terms uniformly there.

Proof of the Antiderivative Claim

Let \(P_N(t)=\sum_{n=0}^{N}a_n(t-c)^n\) be the \(N\)th partial sum of the original series. Fix any \(x\) with \(|x-c|<R\). The whole line segment from \(c\) to \(x\) lies in the open interval of convergence. Choose \(\rho\) so that \(|x-c|\leq\rho<R\). On that segment, \(|t-c|\leq\rho\), and the tail of the series satisfies

$$ |f(t)-P_N(t)| \leq \sum_{n=N+1}^{\infty}|a_n|\rho^n. $$

The right side tends to zero as \(N\) tends to infinity and does not depend on \(t\) on the segment. Therefore the partial sums converge uniformly there. In particular, \(f\) is continuous on every closed interval strictly inside the radius, since it is a uniform limit of the continuous polynomials \(P_N\).

Uniform convergence also justifies passing from the integrals of the partial sums to the integral of \(f\). Indeed, the absolute difference between those integrals is bounded by the length of the segment times the uniform error. Integrating each finite partial sum term by term gives

$$ \int_c^x P_N(t)\,dt = \sum_{n=0}^{N}\frac{a_n}{n+1}(x-c)^{n+1}. $$

Taking the limit as \(N\) tends to infinity yields

$$ \int_c^x f(t)\,dt = \sum_{n=0}^{\infty}\frac{a_n}{n+1}(x-c)^{n+1} = F(x). $$

Since \(f\) is continuous in the open interval, the Fundamental Theorem of Calculus applies to \(F(x)=\int_c^x f(t)\,dt\). It follows that \(F'(x)=f(x)\) at every point strictly inside the radius. This proves both the termwise integration formula and the antiderivative claim.

1
Restrict to an interior segment.
Choose a closed segment from \(c\) to \(x\), with \(|x-c|<R\), and bound \(|t-c|\) there by some \(\rho<R\).
2
Control the series tail.
The convergent numerical series \(\sum |a_n|\rho^n\) bounds the power-series terms uniformly on the segment.
3
Integrate finite sums and pass to the limit.
Uniform convergence makes the integral error tend to zero, giving the integrated power series.
4
Differentiate the integral representation.
Continuity of \(f\) and the Fundamental Theorem of Calculus give \(F'=f\).

Worked Examples

Worked Example: Integrating a Geometric-Type Power Series

Consider \(\sum_{n=0}^{\infty}(x-1)^n/2^n\). Its terms have ratio \((x-1)/2\), so the series converges when \(|x-1|<2\) and has radius \(R=2\). Apply the theorem term by term:

$$ \begin{aligned} \sum_{n=0}^{\infty}\frac{(x-1)^n}{2^n} &\longmapsto \sum_{n=0}^{\infty}\frac{1}{n+1}\frac{(x-1)^{n+1}}{2^n}\\ &=(x-1)+\frac{(x-1)^2}{2\cdot2} +\frac{(x-1)^3}{3\cdot4} +\frac{(x-1)^4}{4\cdot8}+\cdots. \end{aligned} $$

Thus \(F(x)=\sum_{n=0}^{\infty}(x-1)^{n+1}/(2^n(n+1))\) is defined for \(|x-1|<2\). Check by differentiating each displayed term: the first derivative is \(1\), the second is \((x-1)/2\), the third is \((x-1)^2/4\), and the fourth is \((x-1)^3/8\). In general, the derivative of the \(n\)th term is \((x-1)^n/2^n\), which is exactly the original \(n\)th term. The theorem justifies this derivative identity throughout the open interval.

Worked Example: Coefficients That Cancel the Integration Factor

Consider \(\sum_{n=0}^{\infty}(n+1)(x+2)^n/4^n\). For \(x\neq-2\), the ratio of successive term magnitudes tends to \(|x+2|/4\), so the radius is \(R=4\). Integrating term by term divides the coefficient by \(n+1\):

$$ \begin{aligned} F(x) &=\sum_{n=0}^{\infty} \frac{n+1}{n+1}\frac{(x+2)^{n+1}}{4^n}\\ &=(x+2)+\frac{(x+2)^2}{4} +\frac{(x+2)^3}{16} +\frac{(x+2)^4}{64}+\cdots,\qquad |x+2|<4. \end{aligned} $$

To check, differentiating the \(n\)th general term gives

$$ \frac{d}{dx}\left(\frac{(x+2)^{n+1}}{4^n}\right) = \frac{n+1}{4^n}(x+2)^n. $$

This is precisely the \(n\)th term of the original series. For instance, the first four derivatives are \(1\), \(2(x+2)/4\), \(3(x+2)^2/16\), and \(4(x+2)^3/64\). These equal \(1\), \(2(x+2)/4\), \(3(x+2)^2/4^2\), and \(4(x+2)^3/4^3\), respectively, as required.

Worked Example: Integrating a Sparse Power Series

Consider \(\sum_{n=0}^{\infty}(-1)^n(x-3)^{2n}/5^n\). The absolute values form a geometric series with ratio \(|x-3|^2/5\), so the original series converges when \(|x-3|<\sqrt{5}\). Its radius is \(\sqrt{5}\). The exponent of each term is \(2n\), so integration raises it to \(2n+1\):

$$ \begin{aligned} F(x) &=\sum_{n=0}^{\infty} \frac{(-1)^n(x-3)^{2n+1}}{5^n(2n+1)}\\ &=(x-3)-\frac{(x-3)^3}{5\cdot3} +\frac{(x-3)^5}{25\cdot5} -\frac{(x-3)^7}{125\cdot7}+\cdots, \qquad |x-3|<\sqrt{5}. \end{aligned} $$

For a general term, differentiation gives

$$ \frac{d}{dx}\left( \frac{(-1)^n(x-3)^{2n+1}}{5^n(2n+1)} \right) = \frac{(-1)^n(x-3)^{2n}}{5^n}. $$

The factor \(2n+1\) from differentiating the power cancels the denominator \(2n+1\). The result is exactly the original term, including its sign and coefficient, so the theorem confirms that \(F'(x)\) is the original power series on the stated open interval.

A Common Mistake

A frequent error is to increase the exponent by one but leave the coefficient unchanged. For example, an incorrect attempt to integrate \(3(x-c)^2\) term by term might write \(3(x-c)^3\). Differentiating exposes the error:

$$ \frac{d}{dx}\bigl(3(x-c)^3\bigr)=9(x-c)^2\neq 3(x-c)^2. $$

The correction is to divide by the new exponent, \(3\):

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

For a power series, that same factor must be included in every term. More broadly, the theorem cannot be used to assert endpoint convergence merely because integration is valid at every point strictly inside the radius. Its proof relies on choosing a bound \(\rho<R\), so an endpoint is outside the scope of the result.

Why the Result Matters

The theorem turns an infinite power series into a practical representation for an antiderivative: the operation can be performed directly on its coefficients and powers, and the proof guarantees that the resulting series represents an actual antiderivative throughout the interior interval. The radius condition is essential to the justification: it supplies absolute convergence and uniform control on each closed segment where integration is carried out.

For the centered power \((x-c)^n\), the integrated term is \((x-c)^{n+1}/(n+1)\). Remembering both changes makes the calculation easy to check: differentiate the proposed integrated series term by term, and each term should return to its original value.

Check Your Understanding

Use the theorem and its hypotheses to answer these questions.

  1. What interval is covered by the termwise integration theorem if the original power series has center \(c\) and radius \(R\)?
  2. Why does the integrated series converge absolutely at an interior point \(x\)?
  3. What role does uniform convergence on the segment from \(c\) to \(x\) play in the proof?
  4. Integrate \(\sum_{n=0}^{\infty}2^n(x+1)^n\) term by term, and state the interval on which the theorem applies.
  5. Why does the theorem not determine convergence of the integrated series at the endpoints of the original interval of convergence?

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