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Why The Fundamental Theorem Cannot Cross A Pole

Learn why an antiderivative’s endpoint difference is not valid across a pole, and how to evaluate the two sides separately.

Intermediate 11 min read

What You'll Learn

  • State the continuity conditions that allow the Fundamental Theorem of Calculus on intervals avoiding a pole.
  • Define an improper integral across an interior singularity using separate one-sided limits.
  • Compute both one-sided integrals for \(1/x\) and determine whether they converge.
  • Identify how a rational antiderivative can produce a false finite endpoint difference.
  • Check when antiderivative limits can be used on either side of a singularity.

Why a Familiar Formula Can Fail

An antiderivative can make an integral look simple: find \(F\) with \(F'=f\), then subtract endpoint values. But the Fundamental Theorem of Calculus requires suitable continuity on the whole interval where it is applied. If \(f\) has a pole between the endpoints, that condition fails. A finite value from subtracting \(F\) at the outer endpoints does not establish that the corresponding improper integral converges.

The issue is not that antiderivatives become useless near singularities. On each side of a pole, they can still evaluate ordinary integrals over intervals that stay away from it. The essential step is to take the two one-sided limits separately and check whether both exist as finite numbers.

Central principle: The Fundamental Theorem of Calculus applies on intervals where its hypotheses hold. It cannot be applied as one ordinary integral across a pole; any improper integral must instead be defined and evaluated by separate one-sided limits.

What the Fundamental Theorem Requires

One standard form of the Fundamental Theorem says that if \(f\) is continuous on a closed interval \([p,q]\), and \(F\) is an antiderivative of \(f\) there, then

$$ \int_p^q f(x)\,dx=F(q)-F(p). $$

Continuity is important: the theorem applies to a proper integral on the interval in question. If a function has a singularity at \(c\), it is not continuous on \([a,b]\) when \(a<c<b\). The formula therefore cannot be used directly with endpoints \(a\) and \(b\).

To make the safe use of antiderivatives precise, suppose \(f\) is continuous on every compact interval that avoids \(c\), and \(F'=f\) on each side of \(c\). For any \(u<c<v\), the Fundamental Theorem does apply on \([a,u]\) and \([v,b]\), provided \(a<u<c<v<b\): on these closed intervals, \(f\) is continuous. Thus

$$ \int_a^u f(x)\,dx=F(u)-F(a), \qquad \int_v^b f(x)\,dx=F(b)-F(v). $$

These formulas stop short of \(c\). To reach the singular point, their limits must be taken separately.

Definition: For a function with an interior singularity at \(c\), the ordinary improper integral \(\int_a^b f(x)\,dx\) converges only if both one-sided improper integrals converge to finite values. When they do, its value is their sum:
$$ \int_a^b f(x)\,dx = \lim_{u\to c^-}\int_a^u f(x)\,dx + \lim_{v\to c^+}\int_v^b f(x)\,dx. $$

Using antiderivatives on the separate intervals gives the more explicit test

$$ \int_a^b f(x)\,dx = \lim_{u\to c^-}\bigl(F(u)-F(a)\bigr) + \lim_{v\to c^+}\bigl(F(b)-F(v)\bigr), $$

only when both limits exist as finite numbers. The two limits need not agree, and one cannot replace them with \(F(b)-F(a)\) unless additional reasoning justifies that replacement.

Worked Example: The Pole of \(1/x\)

Worked Example: The Pole of \(1/x\)

Determine whether \(\int_{-1}^{1}\frac{1}{x}\,dx\) converges as an ordinary improper integral.

The singularity is at \(x=0\). An antiderivative on each interval avoiding zero is \(F(x)=\ln|x|\), since

$$ F'(x)=\frac{1}{x} \qquad (x\ne 0). $$

First evaluate the left-hand integral by stopping at \(-\varepsilon\), where \(0<\varepsilon<1\):

$$ \int_{-1}^{-\varepsilon}\frac{1}{x}\,dx = \ln|-\varepsilon|-\ln|-1| = \ln\varepsilon-\ln 1 = \ln\varepsilon. $$

As \(\varepsilon\to0^+\), \(\ln\varepsilon\to-\infty\). Therefore the left-hand improper integral does not converge to a finite number.

On the right, stop at \(\varepsilon\):

$$ \int_{\varepsilon}^{1}\frac{1}{x}\,dx = \ln|1|-\ln|\varepsilon| = -\ln\varepsilon. $$

As \(\varepsilon\to0^+\), \(-\ln\varepsilon\to+\infty\). The right-hand improper integral also does not converge to a finite number. Since neither side has a finite limit, the ordinary improper integral diverges.

Check: Differentiating the proposed antiderivative gives \(\frac{d}{dx}\ln|x|=1/x\) for \(x\ne0\), as required. The failure is not an incorrect antiderivative; it is the attempt to apply the theorem across the point where \(f\) is undefined.

Why the Endpoint Difference Is Not Enough

For \(1/x\), subtracting the outer endpoint values would give \(\ln|1|-\ln|-1|=0\). That is a finite number, but it is not the value of the ordinary improper integral: the two required one-sided limits diverge. The calculation omits the behavior at the pole.

Here is the general reason. On intervals stopped at \(u<c\) and \(v>c\), the valid endpoint differences are \(F(u)-F(a)\) and \(F(b)-F(v)\). The inner endpoint contributions involve \(F(u)\) and \(F(v)\), not a single value \(F(c)\) supplied by the Fundamental Theorem. If those terms diverge, there is no finite sum of ordinary improper integrals. Even when some cancellation appears under a particular way of approaching \(c\), that does not prove that the two one-sided limits exist separately.

Important distinction: A finite result from \(F(b)-F(a)\) is justified by the Fundamental Theorem only when its hypotheses hold on the whole interval. At an interior pole, evaluate the two one-sided limits first; do not infer convergence from the outer endpoint values.

Worked Example: A Simple Rational Pole

Worked Example: A Simple Rational Pole

Consider the rational function \(f(x)=\frac{x+1}{x-2}\) on \([1,3]\). Determine whether its improper integral converges, and compare the one-sided calculation with the tempting outer endpoint subtraction.

The pole is at \(x=2\). A useful decomposition is

$$ \frac{x+1}{x-2} = \frac{(x-2)+3}{x-2} = 1+\frac{3}{x-2}. $$

An antiderivative on either side of \(2\) is

$$ F(x)=x+3\ln|x-2|, \qquad F'(x)=1+\frac{3}{x-2}=\frac{x+1}{x-2}. $$

For \(0<\varepsilon<1\), the left-hand integral is

$$ \begin{aligned} \int_1^{2-\varepsilon}\frac{x+1}{x-2}\,dx &=F(2-\varepsilon)-F(1)\\ &=\bigl(2-\varepsilon+3\ln\varepsilon\bigr) -\bigl(1+3\ln 1\bigr)\\ &=1-\varepsilon+3\ln\varepsilon. \end{aligned} $$

As \(\varepsilon\to0^+\), this tends to \(-\infty\). The left-hand integral does not converge finitely.

On the right,

$$ \begin{aligned} \int_{2+\varepsilon}^{3}\frac{x+1}{x-2}\,dx &=F(3)-F(2+\varepsilon)\\ &=\bigl(3+3\ln 1\bigr) -\bigl(2+\varepsilon+3\ln\varepsilon\bigr)\\ &=1-\varepsilon-3\ln\varepsilon. \end{aligned} $$

As \(\varepsilon\to0^+\), this tends to \(+\infty\). The right-hand integral also fails to converge finitely, so the ordinary improper integral across \(2\) diverges.

The misleading calculation: The outer endpoint difference is \(F(3)-F(1)=3-1=2\). It is not the improper integral, because the limits at the pole do not converge on either side. The derivative check above confirms the antiderivative, but it does not make the theorem applicable across \(x=2\).

Worked Example: A Pole of Order Two

Worked Example: A Pole of Order Two

Examine \(\int_0^2\frac{1}{(x-1)^2}\,dx\), with the singularity at \(x=1\).

On either side of \(1\), an antiderivative is \(F(x)=-\frac{1}{x-1}\), since

$$ F'(x)=\frac{1}{(x-1)^2}. $$

For \(0<\varepsilon<1\), the left-hand integral is

$$ \begin{aligned} \int_0^{1-\varepsilon}\frac{1}{(x-1)^2}\,dx &=F(1-\varepsilon)-F(0)\\ &=\frac{1}{\varepsilon}-1. \end{aligned} $$

This tends to \(+\infty\) as \(\varepsilon\to0^+\). The right-hand integral is

$$ \begin{aligned} \int_{1+\varepsilon}^{2}\frac{1}{(x-1)^2}\,dx &=F(2)-F(1+\varepsilon)\\ &=-1+\frac{1}{\varepsilon}. \end{aligned} $$

This also tends to \(+\infty\). Thus the ordinary improper integral diverges. Yet the outer endpoint subtraction gives \(F(2)-F(0)=-1-1=-2\), another false finite answer. Differentiating \(F\) verifies the antiderivative on each interval; the separate one-sided limits show why the outer subtraction is invalid.

A Reliable Procedure

When an interior pole lies between the limits of integration, use the following sequence. The continuity condition in the first step ensures that each application of the Fundamental Theorem is justified on the truncated intervals.

1
Locate the singularity.
Identify the point \(c\) where the integrand is undefined or unbounded. Confirm that the integrand is continuous on compact intervals that stay away from \(c\).
2
Split the integral.
Write the improper integral as a left-hand limit plus a right-hand limit. Both limits must be finite for ordinary convergence.
3
Use antiderivatives only on truncated intervals.
If \(F'=f\) on each side, apply the Fundamental Theorem where the endpoints remain away from \(c\).
4
Take the limits separately.
Find each one-sided limit. If either is infinite or fails to exist, the ordinary improper integral does not converge.

Common Pitfall: Cancelling Divergent Sides

A frequent error is to add the two sides while using the same cutoff distance from the pole, then treat a finite cancellation as proof of ordinary convergence. For \(1/x\), that symmetric truncation gives

$$ \int_{-1}^{-\varepsilon}\frac{1}{x}\,dx + \int_{\varepsilon}^{1}\frac{1}{x}\,dx = \ln\varepsilon-\ln\varepsilon = 0. $$

But this computation pairs two cut-off integrals before taking limits. The ordinary improper integral requires the left and right limits to exist separately; as shown above, one tends to \(-\infty\) and the other to \(+\infty\). Their cancellation under a symmetric cutoff does not turn either divergent one-sided integral into a finite one. This distinction is the reason a principal value is not the same as ordinary convergence.

When Antiderivative Limits Do Work

The lesson is not to avoid antiderivatives near singularities. Use them on each side, under the theorem’s hypotheses, and then examine the endpoint limits. If both one-sided limits are finite, their sum gives the ordinary improper integral. If either fails, there is no ordinary integral value to obtain by subtracting the outer values of an antiderivative.

In particular, knowing a formula for \(F\) and checking \(F'=f\) away from \(c\) establishes only a local fact on those intervals. It does not establish continuity of \(f\) at \(c\), convergence of the improper integral, or permission to apply the Fundamental Theorem on \([a,b]\). The singular point must be treated explicitly.

Check Your Understanding

Answer these questions using separate one-sided limits whenever an interior singularity is present.

  1. What continuity condition allows the Fundamental Theorem to be applied to a truncated interval that avoids a pole?
  2. For an improper integral with an interior singularity, which two limits must be finite for ordinary convergence?
  3. For \(1/x\) on \([-1,1]\), what are the limits of the left-hand and right-hand truncated integrals?
  4. Why does a finite value of \(F(b)-F(a)\) not prove convergence when \(F'=f\) only on either side of an interior pole?
  5. Why does a symmetric cancellation of divergent terms fail to establish ordinary convergence?

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