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Singular Integrands · Tutorial 312 of 998

Rational Functions Near Finite Poles

Classify a rational function near a finite pole, then determine the behavior and sign of the one-sided improper integrals.

Intermediate 9 min read

What You'll Learn

  • Define the order of a finite pole of a rational function in reduced form
  • Express the function locally as a nonzero factor divided by a power of the distance to the pole
  • Derive why every genuine pole makes each one-sided integral diverge
  • Distinguish the logarithmic behavior of a simple pole from higher-order power blow-ups
  • Determine whether each divergent one-sided integral tends to positive or negative infinity
  • Avoid treating an antiderivative evaluation across a pole as an ordinary definite integral

Why pole order matters

A rational function can become unbounded at a point where its denominator vanishes. For an improper integral, it is not enough to notice that the integrand is unbounded: we need to know how quickly it grows and what happens on each side of the singularity.

The key information is the order of the pole. Near a simple pole, the leading behavior is proportional to \(1/(x-c)\), whose antiderivative is logarithmic. At a pole of higher order, the leading behavior is proportional to \(1/(x-c)^m\) for \(m\ge 2\), whose antiderivative has a power blow-up. In both cases, each one-sided integral diverges, but the signs of the divergences can differ.

Finite poles and their orders

Write a rational function in reduced form as \(f(x)=P(x)/Q(x)\), where \(P\) and \(Q\) have no common polynomial factor. A real number \(c\) is a finite pole if \(Q(c)=0\). If \(Q\) has a zero of multiplicity \(m\) at \(c\), then \(c\) is a pole of order \(m\). The reduced-form condition matters: a denominator zero that cancels is not a pole of the rational function.

Definition: A rational function has a pole of order \(m\) at \(c\) if, near \(c\), it can be written as \(f(x)=g(x)/(x-c)^m\), where \(g\) is continuous near \(c\) and \(g(c)\ne 0\). The nonzero value \(g(c)\) determines the leading coefficient of the blow-up.

For a rational function in reduced form, this local form follows by factoring \(Q(x)=(x-c)^mS(x)\), where \(S(c)\ne0\). Then \(g(x)=P(x)/S(x)\) is continuous near \(c\), and \(g(c)\ne0\) because the numerator does not vanish at the pole in reduced form.

Consequently, as \(x\) approaches \(c\), the function has the same leading behavior as \(g(c)/(x-c)^m\). More precisely, their ratio tends to \(1\): \(f(x)\big/\big(g(c)/(x-c)^m\big)=g(x)/g(c)\to1\). This local description is what lets us decide convergence without finding a complete partial-fraction decomposition.

One-sided integrals at a pole

If \(c\) is an interior singularity of an interval \([a,b]\), the integral across \(c\) is defined by separating its two sides. The left-hand integral is the limit of integrals ending before \(c\); the right-hand integral is the limit of integrals beginning after \(c\). Both must converge to finite real numbers for the ordinary improper integral across \(c\) to converge.

$$ \int_a^c f(x)\,dx = \lim_{t\to c^-}\int_a^t f(x)\,dx, \qquad \int_c^b f(x)\,dx = \lim_{t\to c^+}\int_t^b f(x)\,dx. $$

The orientation in the right-hand definition is important: for \(t>c\), the integral runs from \(t\) to \(b\), not from \(b\) to \(t\). Opposite signs of divergence on the two sides do not make either one-sided limit finite.

Why every genuine pole causes one-sided divergence

Suppose \(f(x)=g(x)/(x-c)^m\) near \(c\), with \(g(c)=A\ne0\). By continuity, sufficiently close to \(c\), \(g(x)\) has the same sign as \(A\), and its absolute value is bounded below by \(|A|/2\). Therefore, on either side of \(c\),

$$ |f(x)| \ge \frac{|A|}{2|x-c|^m}. $$

The comparison power on the right has an explicitly divergent integral at \(c\). With \(u=|x-c|\), its relevant antiderivative is logarithmic when \(m=1\), and a negative power of \(u\) when \(m>1\):

$$ \int_\varepsilon^\delta u^{-m}\,du = \begin{cases} \ln(\delta)-\ln(\varepsilon), & m=1,\\[4pt] \dfrac{\delta^{1-m}-\varepsilon^{1-m}}{1-m}, & m>1. \end{cases} $$

Here \(\delta>0\) is fixed and \(\varepsilon\to0^+\). The first expression tends to \(+\infty\). In the second, \(1-m<0\), and the expression also tends to \(+\infty\). Since \(f\) has a fixed sign sufficiently close to \(c\) on each side, the one-sided integral diverges with a definite sign rather than oscillating.

Theorem: At a real pole of any positive integer order \(m\), both one-sided improper integrals diverge. The sign of \(f\) near the pole is the sign of \(A/(x-c)^m\), where \(A=g(c)\). Thus the right-hand divergence has the sign of \(A\); the left-hand divergence has the sign of \((-1)^mA\).

For a simple pole, the antiderivative of the leading term \(A/(x-c)\) is \(A\ln|x-c|\), so the divergence is logarithmic. For \(m\ge2\), the leading term has antiderivative \(A(x-c)^{1-m}/(1-m)\), giving a power blow-up. The theorem shows that lower-order terms cannot remove the divergence: sufficiently close to the pole, the nonzero leading factor controls the sign and magnitude.

Worked Example: A simple pole on both sides

Worked Example: A simple pole on both sides

Determine the one-sided improper integrals of \(f(x)=(2x+1)/(x-1)\) on \([0,1]\) and \([1,2]\).

The denominator has a simple zero at \(x=1\), and the numerator there is \(2(1)+1=3\ne0\). Thus \(x=1\) is a simple pole. Rewrite the function by expressing its numerator in terms of \(x-1\):

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

An antiderivative on either interval that does not cross \(1\) is \(F(x)=2x+3\ln|x-1|\), since

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

For the left side, let \(t\to1^-\). Since \(F(0)=0\),

$$ \int_0^1 f(x)\,dx = \lim_{t\to1^-}\big(F(t)-F(0)\big) = \lim_{t\to1^-}\big(2t+3\ln|t-1|\big) = -\infty. $$

For the right side, let \(t\to1^+\). Since \(F(2)=4\),

$$ \int_1^2 f(x)\,dx = \lim_{t\to1^+}\big(F(2)-F(t)\big) = \lim_{t\to1^+}\big(4-2t-3\ln|t-1|\big) = +\infty. $$

The left-hand divergence is negative and the right-hand divergence is positive, matching the sign of \(3/(x-1)\) on each side. The antiderivative check confirms the decomposition and the endpoint limits confirm that neither one-sided integral converges.

Worked Example: A pole of order two

Worked Example: A pole of order two

Classify the pole and determine the one-sided behavior of \(f(x)=(3x+1)/(x-2)^2\) on \([1,3]\).

The numerator at \(2\) is \(3(2)+1=7\ne0\), so \(x=2\) is a pole of order two. Rewrite the numerator as \(3x+1=3(x-2)+7\), giving

$$ f(x)=\frac{3}{x-2}+\frac{7}{(x-2)^2}. $$

An antiderivative on either side of \(2\) is \(F(x)=3\ln|x-2|-7/(x-2)\), because

$$ F'(x)=\frac{3}{x-2}+\frac{7}{(x-2)^2} =\frac{3(x-2)+7}{(x-2)^2} =\frac{3x+1}{(x-2)^2}. $$

On the left, \(F(1)=7\), and for \(t\to2^-\),

$$ \int_1^2 f(x)\,dx = \lim_{t\to2^-}\big(F(t)-7\big) = \lim_{t\to2^-} \left(3\ln|t-2|-\frac{7}{t-2}-7\right) = +\infty. $$

On the right, \(F(3)=-7\), and for \(t\to2^+\),

$$ \int_2^3 f(x)\,dx = \lim_{t\to2^+}\big(-7-F(t)\big) = \lim_{t\to2^+} \left(-7-3\ln|t-2|+\frac{7}{t-2}\right) = +\infty. $$

Both divergences are positive. This also follows directly because \(3x+1>0\) on \([1,3]\) and \((x-2)^2>0\) away from the pole. The term with the squared denominator gives the dominant power blow-up on each side.

Worked Example: A pole of odd order

Worked Example: A pole of odd order

Determine the pole order and one-sided integrals for \(f(x)=(x+3)/(x+1)^3\) on \([-2,0]\).

The denominator has a zero of order three at \(x=-1\), while the numerator there is \(-1+3=2\ne0\). Thus the pole has order three. Since \(x+3=(x+1)+2\),

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

An antiderivative is \(F(x)=-1/(x+1)-1/(x+1)^2\), as differentiation verifies:

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

Since \(F(-2)=0\), the left-hand integral is

$$ \int_{-2}^{-1}f(x)\,dx = \lim_{t\to-1^-}F(t) = \lim_{t\to-1^-} \left(-\frac{1}{t+1}-\frac{1}{(t+1)^2}\right) = -\infty. $$

Since \(F(0)=-2\), the right-hand integral is

$$ \int_{-1}^0f(x)\,dx = \lim_{t\to-1^+}\big(-2-F(t)\big) = \lim_{t\to-1^+} \left(-2+\frac{1}{t+1}+\frac{1}{(t+1)^2}\right) = +\infty. $$

The signs differ because an odd power of \(x+1\) changes sign across \(-1\). The one-sided calculations agree with the fact that the numerator is positive throughout the interval, while the cubed denominator is negative to the left and positive to the right.

A common mistake: evaluating across the pole

A tempting but invalid calculation is to use the antiderivative \(F(x)=\ln|x|\) to claim

$$ \int_{-1}^{1}\frac{1}{x}\,dx \stackrel{\text{incorrect}}{=} F(1)-F(-1) = \ln 1-\ln 1 = 0. $$

The Fundamental Theorem of Calculus cannot be applied across \(x=0\), where the integrand is undefined. The integral must first be split into one-sided limits:

$$ \int_{-1}^{0}\frac{1}{x}\,dx = \lim_{t\to0^-}\big(\ln|t|-\ln 1\big) = -\infty, \qquad \int_{0}^{1}\frac{1}{x}\,dx = \lim_{t\to0^+}\big(\ln 1-\ln|t|\big) = +\infty. $$

Neither one-sided integral converges. The finite result from the incorrect calculation comes from substituting endpoints on opposite sides into a formula without respecting the singularity between them.

Using pole order efficiently

For a rational function near a finite singularity, a reliable first pass is to locate the zero of the denominator, determine its multiplicity, and check that the numerator does not vanish there in reduced form. The local factor then identifies the order and the leading coefficient. That information alone determines whether either one-sided integral can converge and predicts its sign of divergence.

1
Locate the pole.
Find a denominator zero \(c\) and work with the rational function in reduced form.
2
Find its order.
Factor out the highest power \((x-c)^m\) from the denominator.
3
Read the local behavior.
Write \(f(x)=g(x)/(x-c)^m\) with \(g(c)\ne0\), and use the sign and magnitude of \(g(c)\).
4
Evaluate one side at a time.
Use separate limits at \(c^-\) and \(c^+\); a genuine pole makes both one-sided integrals divergent.

This method matters because a rational expression may look complicated away from its singularity, while its local factorization gives a decisive convergence test. An antiderivative can provide the exact divergence in an example, but the pole order already explains the general result.

Check Your Understanding

For each question, distinguish the pole order, the one-sided behavior, and the ordinary improper-integral conclusion.

  1. What condition on \(g\) defines a pole of order \(m\) at \(c\) in the local form \(f(x)=g(x)/(x-c)^m\)?
  2. Why does the integral of \(|x-c|^{-m}\) diverge on either side for every positive integer \(m\)?
  3. For a simple pole with leading coefficient \(A>0\), what are the signs of the left- and right-hand divergences?
  4. For a pole of order two with positive leading coefficient, what are the signs of the two one-sided divergences?
  5. Why is it invalid to evaluate an antiderivative at endpoints on opposite sides of a pole and subtract?

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