Why a Local Expansion May Not Work Across an Interval
A Taylor polynomial can approximate a function very well near its center. But integrating a Taylor series over an interval is a stronger demand: the series must converge suitably throughout that interval, and its sum must be the function being integrated there.
The first warning sign is the radius of convergence. A Taylor series centered at \(c\) may converge only when \(x\) is within a particular distance of \(c\). An integration interval can extend beyond that range even when the function itself is defined and well behaved throughout the interval. A second issue is uniform convergence: pointwise convergence at each point does not, by itself, justify exchanging an infinite sum and an integral.
Radius of Convergence and the Interval Check
A power series centered at \(c\) has the form \(\sum_{n=0}^{\infty} a_n(x-c)^n\). Its radius of convergence \(R\) describes the distances from \(c\) for which it converges: it converges absolutely when \(|x-c|<R\), and diverges when \(|x-c|>R\). At the boundary \(|x-c|=R\), each endpoint must be checked separately; the radius alone does not decide convergence there.
For an interval \([u,v]\), a useful sufficient test is to calculate the greatest distance from the center to the interval. If that distance is less than \(R\), the whole interval lies strictly inside the convergence region. For a real interval, that distance is \(\max(|u-c|,|v-c|)\). If it equals or exceeds \(R\), do not assume the series works throughout: check boundary behavior and whether the interval extends outside the convergence region.
Convergence is not the same as representation. The radius tells us where the series converges; it does not, by itself, prove that its sum equals a specified function. In familiar Taylor expansions, such as the geometric series for a rational function, the equality can be verified directly. In other cases, equality between the function and its Taylor series needs its own justification.
Why Uniform Convergence Allows Term-by-Term Integration
Suppose \(f_n\) are continuous on a closed interval \([u,v]\) and their series converges uniformly there to \(S\). Uniform convergence means that, after some index, the difference between a partial sum and \(S\) is small at every point of the interval, with one bound that works for the entire interval. Under these conditions, the integral of the sum is the limit of the integrals of the partial sums.
The reason is that uniform closeness bounds the difference of the integrals. If \(|S(x)-S_N(x)|<\varepsilon\) everywhere on \([u,v]\), then \(\left|\int_u^v S(x)\,dx-\int_u^v S_N(x)\,dx\right|\le (v-u)\varepsilon\). As the uniform error can be made arbitrarily small, the integral differences tend to zero.
For a power series, uniform convergence is guaranteed on every closed interval strictly inside its radius of convergence. To see why, let \(|x-c|\le r<R\), and choose \(s\) with \(r<s<R\). Since the power series converges at distance \(s\), its terms \(a_ns^n\) are bounded in magnitude by some constant \(M\). Therefore, throughout the smaller interval,
The right-hand side is the term of a convergent geometric series, so the tails of the power series are bounded uniformly by tails of that geometric series. Thus the power series converges uniformly on the smaller interval. Since each term is a polynomial, the theorem applies if its sum equals the integrand there.
Worked Example: A Function Defined Beyond Its Taylor Series
Worked Example: A Function Defined Beyond Its Taylor Series
Consider \(f(x)=\frac{1}{1+x^2}\) on \([0,2]\), and its Taylor series centered at \(0\). The geometric-series identity gives $$ \frac{1}{1+x^2}=\sum_{n=0}^{\infty}(-1)^n x^{2n} $$ when \(|x|<1\). The series has radius of convergence \(1\), while the interval extends to \(x=2\). Although \(f\) is defined across the whole interval, its Taylor series is not.
In fact, integrating the terms formally from \(0\) to \(2\) gives the partial sums $$ \sum_{n=0}^{N}(-1)^n\int_0^2 x^{2n}\,dx = \sum_{n=0}^{N}(-1)^n\frac{2^{2n+1}}{2n+1}. $$ The magnitude of the \(n\)-th term is \(\frac{2^{2n+1}}{2n+1}\), which does not tend to zero. Therefore these partial sums cannot converge, so this term-by-term integration does not produce the integral on \([0,2]\).
By contrast, the original function has the finite integral $$ \int_0^2\frac{1}{1+x^2}\,dx = \left[\arctan x\right]_0^2 = \arctan 2. $$ Differentiating \(\arctan x\) gives \(\frac{1}{1+x^2}\), confirming the antiderivative. The issue is not the existence of the integral; it is the attempt to use a series outside its convergence range.
Worked Example: Integrating Safely Inside the Radius
Worked Example: Integrating Safely Inside the Radius
Use the same function, \(f(x)=\frac{1}{1+x^2}\), but integrate over \([0,\frac12]\). Here \(|x|\le \frac12<1\), so the Taylor series converges uniformly on the entire closed interval.
The finite geometric-series identity gives $$ \frac{1}{1+x^2} = \sum_{n=0}^{N}(-1)^n x^{2n} + \frac{(-1)^{N+1}x^{2N+2}}{1+x^2}. $$ On \([0,\frac12]\), the magnitude of the remainder is at most \((\frac12)^{2N+2}\), which tends to zero uniformly. Thus integrating the series term by term is justified. For \(N=2\), the integrated partial sum is $$ \int_0^{1/2}(1-x^2+x^4)\,dx = \left[x-\frac{x^3}{3}+\frac{x^5}{5}\right]_0^{1/2} = \frac12-\frac1{24}+\frac1{160} = \frac{223}{480}. $$ The integrated remainder has magnitude at most $$ \int_0^{1/2}\left(\frac12\right)^6\,dx = \frac{1}{128}. $$ As \(N\) increases, this bound tends to zero. The infinite series of integrated terms therefore equals the integral.
Finally, direct integration gives $$ \int_0^{1/2}\frac{1}{1+x^2}\,dx = \left[\arctan x\right]_0^{1/2} = \arctan\left(\frac12\right). $$ Again, differentiating \(\arctan x\) verifies the antiderivative. This is the value to which the integrated partial sums converge.
Recenter When the Original Center Does Not Cover the Interval
When a Taylor series fails to cover an integration interval, one option is to choose a different center. The new center can make every point of the interval sufficiently close for the corresponding series to converge uniformly. The function must still be represented by that series on the interval; in the example below, the geometric-series identity verifies this directly.
Worked Example: Recenter to Cover a Wider Interval
Integrate \(f(x)=\frac1x\) on \([\frac12,2]\). The series centered at \(1\) is $$ \frac1x = \frac{1}{1+(x-1)} = \sum_{n=0}^{\infty}(-1)^n(x-1)^n, $$ but it does not converge at \(x=2\), where \(x-1=1\). Instead, center at \(\frac32\). Put \(q=\frac{x-\frac32}{\frac32}\). Then $$ \frac1x = \frac23\frac{1}{1+q} = \frac23\sum_{n=0}^{\infty}(-1)^n q^n. $$ On the interval, \(q\) ranges from \(-\frac23\) to \(\frac13\), so \(|q|\le\frac23<1\). The geometric series converges uniformly there.
Since \(dx=\frac32\,dq\), the integral of the series becomes $$ \int_{1/2}^{2}\frac1x\,dx = \sum_{n=0}^{\infty}(-1)^n\int_{-2/3}^{1/3}q^n\,dq = \sum_{n=0}^{\infty}(-1)^n \frac{(1/3)^{n+1}-(-2/3)^{n+1}}{n+1}. $$ To check the resulting value, sum the geometric series as a function of \(q\) and integrate: $$ \int_{-2/3}^{1/3}\frac{1}{1+q}\,dq = \left[\ln(1+q)\right]_{-2/3}^{1/3} = \ln\left(\frac43\right)-\ln\left(\frac13\right) = \ln 4. $$ Directly, \(\int_{1/2}^{2}\frac1x\,dx=[\ln x]_{1/2}^{2}=\ln 2-\ln(\frac12)=\ln4\), confirming the result.
A Common Mistake: Checking Only That the Function Exists
A frequent error is to reason, “The function is continuous on the interval, so its Taylor series can be integrated there.” Continuity ensures the function has an integral on a closed interval; it does not ensure that a Taylor series centered at one point converges across that interval. The first worked example separates those two facts: \(\frac{1}{1+x^2}\) is continuous on \([0,2]\), but its series centered at zero does not converge there.
Another error is to treat the radius as though it guarantees equality with the function everywhere inside it. The radius describes convergence of the series. To use the series as an expansion of \(f\), establish that its sum is \(f\) on the relevant interval as well. For standard Taylor expansions, this may follow from a known identity or a suitable Taylor theorem; it should not be silently assumed from convergence alone.
Check Your Understanding
Use the radius of convergence, representation, and uniform convergence checks to assess each situation.
- A power series centered at \(0\) has radius \(3\). Does it converge uniformly on \([-2,2]\)? Explain which condition is relevant.
- Why does continuity of \(f\) on \([0,2]\) not justify integrating the Taylor series for \(\frac{1}{1+x^2}\) centered at \(0\) across that interval?
- For a series with radius \(R\), what extra care is needed if an integration interval includes a point at distance exactly \(R\) from the center?
- State the uniform-convergence condition that permits term-by-term integration of a series of continuous functions on a closed interval.
- For \(\frac1x\) on \([\frac12,2]\), why does centering at \(\frac32\) give better convergence coverage than centering at \(1\)?