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Integral of a Power Series

Integral of a Power Series

The indefinite integral of a function defined by a power series can be found by integrating the series term-by-term.

 

Power series integral formula: ∫f(x)dx = C + sum of c_n/(n+1)(x-a)^(n+1), integrating term-by-term over interval (c,d).

 

 

See also

Derivative of a power series, convergent series, divergent series

Key Formula

∫∑n=0∞an(x−c)n dx=C+∑n=0∞ann+1(x−c)n+1\int \sum_{n=0}^{\infty} a_n (x - c)^n \, dx = C + \sum_{n=0}^{\infty} \frac{a_n}{n+1}(x - c)^{n+1}
Where:
  • ana_n = The coefficient of the nth term in the original power series
  • xx = The variable of integration
  • cc = The center of the power series
  • nn = The index of summation, starting at 0
  • CC = The constant of integration

Worked Example

Problem: Find the integral of the power series f(x)=∑n=0∞xn=1+x+x2+x3+⋯f(x) = \sum_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + \cdots (the geometric series, valid for ∣x∣<1|x| < 1).
Step 1: Write out the term-by-term integration rule. Each term xnx^n integrates to xn+1n+1\frac{x^{n+1}}{n+1}.
∫∑n=0∞xn dx=C+∑n=0∞xn+1n+1\int \sum_{n=0}^{\infty} x^n \, dx = C + \sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1}
Step 2: Expand the first few terms to see the pattern clearly.
=C+x+x22+x33+x44+⋯= C + x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \cdots
Step 3: Re-index by letting m=n+1m = n + 1, so the sum runs from m=1m = 1 to ∞\infty.
=C+∑m=1∞xmm= C + \sum_{m=1}^{\infty} \frac{x^m}{m}
Step 4: Recognize this as a known series. Since f(x)=11−xf(x) = \frac{1}{1-x}, its integral is −ln⁡∣1−x∣+C-\ln|1 - x| + C. Indeed, ∑m=1∞xmm=−ln⁡(1−x)\sum_{m=1}^{\infty} \frac{x^m}{m} = -\ln(1-x) for ∣x∣<1|x| < 1.
∫11−x dx=−ln⁡(1−x)+C\int \frac{1}{1-x}\,dx = -\ln(1-x) + C
Answer: ∫∑n=0∞xn dx=C−ln⁡(1−x)=C+∑n=1∞xnn\displaystyle\int \sum_{n=0}^{\infty} x^n \, dx = C - \ln(1 - x) = C + \sum_{n=1}^{\infty} \frac{x^n}{n}, valid for ∣x∣<1|x| < 1.

Another Example

This example shows how to use the initial condition (evaluating at x = 0) to determine the constant of integration, and it produces a well-known series that students are often asked to derive.

Problem: Find a power series representation for ln⁡(1+x)\ln(1 + x) by integrating the power series for 11+x\frac{1}{1+x}.
Step 1: Start with the geometric series with −x-x substituted for xx.
11+x=∑n=0∞(−1)nxn=1−x+x2−x3+⋯ ,∣x∣<1\frac{1}{1+x} = \sum_{n=0}^{\infty} (-1)^n x^n = 1 - x + x^2 - x^3 + \cdots, \quad |x| < 1
Step 2: Integrate both sides term by term.
∫11+x dx=C+∑n=0∞(−1)nxn+1n+1\int \frac{1}{1+x}\,dx = C + \sum_{n=0}^{\infty} \frac{(-1)^n x^{n+1}}{n+1}
Step 3: The left side integrates to ln⁡(1+x)+C\ln(1+x) + C. Find CC by setting x=0x = 0: ln⁡(1)=0\ln(1) = 0 and every term of the series equals 0, so C=0C = 0.
ln⁡(1+0)=0  ⟹  C=0\ln(1 + 0) = 0 \implies C = 0
Step 4: Write the final series.
ln⁡(1+x)=∑n=0∞(−1)nxn+1n+1=x−x22+x33−x44+⋯\ln(1+x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{n+1}}{n+1} = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots
Answer: ln⁡(1+x)=∑n=0∞(−1)nxn+1n+1\ln(1+x) = \displaystyle\sum_{n=0}^{\infty} \frac{(-1)^n x^{n+1}}{n+1}, valid for ∣x∣<1|x| < 1 (and also at x=1x = 1).

Frequently Asked Questions

Does the radius of convergence change when you integrate a power series?
No. When you integrate a power series term by term, the resulting series has the same radius of convergence as the original. However, convergence at the endpoints of the interval may change. For instance, a series that diverges at an endpoint might converge there after integration, because the coefficients become smaller (divided by n+1n + 1).
Why can you integrate a power series term by term?
A theorem from analysis guarantees that a power series converges uniformly on any closed subinterval strictly inside its interval of convergence. Uniform convergence allows you to swap the integral and the infinite sum. This is not true for all infinite series of functions, but it always holds for power series within their radius of convergence.
How do you find the constant of integration when integrating a power series?
You determine the constant CC by substituting a specific value of xx (usually x=cx = c, the center) into both sides. At the center, every term of the power series is zero, so CC equals the value of the antiderivative at that point. For example, if the antiderivative is ln⁡(1+x)\ln(1+x), then setting x=0x=0 gives C=ln⁡(1)=0C = \ln(1) = 0.

Integral of a Power Series vs. Derivative of a Power Series

Integral of a Power SeriesDerivative of a Power Series
OperationIntegrate each term: anxn→ann+1xn+1a_n x^n \to \frac{a_n}{n+1} x^{n+1}Differentiate each term: anxn→n⋅anxn−1a_n x^n \to n \cdot a_n x^{n-1}
Effect on coefficientsDivides the coefficient by n+1n+1 (makes terms smaller)Multiplies the coefficient by nn (makes terms larger)
Constant termAdds a constant of integration CCThe constant term a0a_0 disappears
Radius of convergenceSame as the original seriesSame as the original series
Endpoint convergenceMay improve (series may converge at endpoints where original did not)May worsen (series may diverge at endpoints where original converged)

Why It Matters

Term-by-term integration of power series is one of the main tools for building new series from known ones in Calculus II and beyond. It lets you derive the series for ln⁡(1+x)\ln(1+x), arctan⁡(x)\arctan(x), and many other functions that are difficult to expand directly using Taylor's formula. This technique also appears in differential equations, where integrating a series solution produces the general solution with an arbitrary constant.

Common Mistakes

Mistake: Forgetting the constant of integration CC.
Correction: An indefinite integral always includes +C+C. When building a specific function (like ln⁡(1+x)\ln(1+x)), you must determine CC by evaluating at a known point, typically the center of the series.
Mistake: Assuming endpoint convergence stays the same after integration.
Correction: The radius of convergence is preserved, but behavior at the endpoints can change. After integration the coefficients shrink by a factor of 1n+1\frac{1}{n+1}, which can turn a divergent endpoint into a convergent one. Always recheck endpoints separately.

Related Terms