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A Rational Integral With Multiple Finite Poles

Use separate one-sided limits at both poles to determine whether a rational improper integral converges.

Intermediate 10 min read

What You'll Learn

  • Split an integral at each interior pole and identify every one-sided limit required.
  • Use partial fractions and an antiderivative to evaluate truncated pieces.
  • Determine convergence from the one-sided limits, not from a combined expression.
  • Distinguish a genuine pole from a removable hole in a rational expression.
  • Recognize why symmetric cancellation does not establish ordinary convergence.

Why Two Interior Poles Require Separate Limits

A rational function can become unbounded at several points inside an interval. Each such point divides the interval into pieces, and every approach to a pole from either side must be checked. An antiderivative can help evaluate the truncated integrals, but it cannot make a divergent piece acceptable by cancelling it against another divergent piece.

For an integral with two interior singularities \(a<b\) on \([A,B]\), where \(A<a<b<B\), the natural pieces are \([A,a)\), \((a,b)\), and \((b,B]\). The middle piece is itself improper at both ends. Its integral exists only if the limit at \(a\) from the right and the limit at \(b\) from the left are both finite.

Definition: An improper integral across an interior singularity converges in the ordinary sense only when the integrals on both sides of that point have finite limits separately. With multiple interior singularities, split at every singular point and require every resulting one-sided limit to be finite.

This is a statement about ordinary improper integrals, not about a Cauchy principal value. A principal value may use coordinated cutoffs to cancel divergences; ordinary convergence does not permit that cancellation.

Setting Up the Integral Correctly

Suppose \(f\) has singularities at \(a\) and \(b\). Choose a point \(c\) strictly between them to express the middle integral as two one-sided limits. Then define the full improper integral by the sum of the separate limits, provided each exists and is finite:

$$ \int_A^B f(x)\,dx = \int_A^a f(x)\,dx + \int_a^c f(x)\,dx + \int_c^b f(x)\,dx + \int_b^B f(x)\,dx. $$

Each term in this expression is interpreted as an appropriate one-sided limit. The point \(c\) is an ordinary interior point; it merely separates the two ends of the middle interval so they can be tested independently.

When \(f\) is rational, partial fractions can make the behavior visible. After finding an antiderivative on each nonsingular piece, evaluate it at truncated endpoints and take the limits. Do not substitute a pole into the antiderivative if the resulting endpoint expression is undefined or unbounded.

1
Locate the genuine poles.
Check the rational function after any valid algebraic simplification; denominator zeros that cancel are not genuine poles.
2
Split at every pole.
Write a separate one-sided limit for each approach to each singular point.
3
Find an antiderivative on each piece.
Partial fractions often reduce the calculation to logarithms and powers.
4
Take the limits independently.
The ordinary integral converges only if all the required limits are finite.

Worked Example: Two Simple Poles

Worked Example: Two Simple Poles

Determine whether the integral of \(f(x)=\frac{1}{x(x-2)}\) over \([-1,3]\) converges as an ordinary improper integral. The poles \(0\) and \(2\) both lie inside the interval.

First decompose into partial fractions. Writing \(\frac{1}{x(x-2)}=\frac{A}{x}+\frac{B}{x-2}\), the identity \(1=A(x-2)+Bx\) gives \(A=-\frac12\) and \(B=\frac12\). Thus

$$ f(x)=-\frac{1}{2x}+\frac{1}{2(x-2)},\qquad F(x)=-\frac12\ln|x|+\frac12\ln|x-2|. $$

Differentiating verifies the antiderivative: \(F'(x)=-\frac{1}{2x}+\frac{1}{2(x-2)}=\frac{1}{x(x-2)}\).

Split at both poles, and split the middle interval at \(1\). The left piece, truncated at \(-\varepsilon\), is

$$ \int_{-1}^{-\varepsilon}f(x)\,dx =F(-\varepsilon)-F(-1) =\frac12\ln(2+\varepsilon)-\frac12\ln\varepsilon-\frac12\ln3 \longrightarrow+\infty. $$

Here \(F(-1)=\frac12\ln3\), and the term \(-\frac12\ln\varepsilon\) tends to \(+\infty\).

On the right side of the first pole,

$$ \int_{\varepsilon}^{1}f(x)\,dx =F(1)-F(\varepsilon) =\frac12\ln\varepsilon-\frac12\ln(2-\varepsilon) \longrightarrow-\infty, $$

since \(F(1)=0\). The left side of the second pole also diverges:

$$ \int_{1}^{2-\varepsilon}f(x)\,dx =F(2-\varepsilon)-F(1) =\frac12\ln\varepsilon-\frac12\ln(2-\varepsilon) \longrightarrow-\infty. $$

Finally, on the right side of \(2\),

$$ \int_{2+\varepsilon}^{3}f(x)\,dx =F(3)-F(2+\varepsilon) =-\frac12\ln3-\frac12\ln\varepsilon+\frac12\ln(2+\varepsilon) \longrightarrow+\infty. $$

In this last calculation, \(F(3)=-\frac12\ln3\), not zero. The endpoint value is included explicitly. Since the one-sided pieces diverge, the original integral does not converge in the ordinary sense.

Why the Pieces Must Be Tested Separately

The need for separate limits follows from the definition of an improper integral. For example, around one pole \(a\), the integral across that point is defined by adding the two one-sided integrals only when each is finite. The two one-sided limits are independent: one is obtained by moving an endpoint toward \(a\) from the left, and the other by moving an endpoint toward \(a\) from the right.

Key principle: A sum of divergent one-sided integrals is not an ordinary improper integral, even if a coordinated truncation makes the sum appear finite. The separate limits must exist as finite numbers before they can be added.

This principle also explains why evaluating an antiderivative at the far endpoints and ignoring the poles is invalid. The Fundamental Theorem of Calculus applies on intervals where the integrand is continuous; a pole divides the interval into separate pieces. It does not justify carrying one evaluation across a singularity.

Worked Example: A Double Pole and a Simple Pole

Worked Example: A Double Pole and a Simple Pole

Consider \(g(x)=\frac{x+1}{(x-1)^2(x-3)}\) on \([0,4]\), with poles at \(1\) and \(3\). Partial fractions give

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

Indeed, combining the right-hand side over the common denominator produces numerator \(-(x-1)(x-3)-(x-3)+(x-1)^2=x+1\). An antiderivative on each nonsingular interval is

$$ G(x)=-\ln|x-1|+\frac{1}{x-1}+\ln|x-3|. $$

Its derivative is \(-\frac{1}{x-1}-\frac{1}{(x-1)^2}+\frac{1}{x-3}=g(x)\).

At the double pole \(x=1\), \(G(0)=\ln3-1\) and \(G(2)=1\). The two truncated pieces adjacent to this pole are

$$ \int_0^{1-\varepsilon}g(x)\,dx =G(1-\varepsilon)-G(0) =-\ln\varepsilon-\frac1\varepsilon+\ln(2+\varepsilon)-(\ln3-1) \longrightarrow-\infty, $$
$$ \int_{1+\varepsilon}^{2}g(x)\,dx =G(2)-G(1+\varepsilon) =1+\ln\varepsilon-\frac1\varepsilon-\ln(2-\varepsilon) \longrightarrow-\infty. $$

Both limits diverge because the term \(-1/\varepsilon\) dominates the logarithmic terms.

At \(x=3\), the antiderivative has a logarithmic divergence. Since \(G(2)=1\), \(G(4)=\frac13-\ln3\), and

$$ G(3-\varepsilon)=-\ln(2-\varepsilon)+\frac{1}{2-\varepsilon}+\ln\varepsilon\longrightarrow-\infty, $$
$$ G(3+\varepsilon)=-\ln(2+\varepsilon)+\frac{1}{2+\varepsilon}+\ln\varepsilon\longrightarrow-\infty, $$

we obtain

$$ \int_2^{3-\varepsilon}g(x)\,dx=G(3-\varepsilon)-G(2)\longrightarrow-\infty, \qquad \int_{3+\varepsilon}^{4}g(x)\,dx=G(4)-G(3+\varepsilon)\longrightarrow+\infty. $$

Thus the integral fails to converge at both poles. The behavior at \(x=1\) alone is already enough to decide this, but evaluating the pieces at \(x=3\) confirms that this second singularity must be treated independently as well.

A Hole Is Not Necessarily a Pole

A rational expression may have a denominator zero that cancels with its numerator. Such a point is not a genuine pole of the simplified function. This distinction matters when deciding whether the integral is improper and what limits are required.

Worked Example: Two Removable Holes

Consider \(h(x)=\frac{x^2-1}{(x-1)(x+1)}\) on \([-2,2]\), initially undefined at \(x=-1\) and \(x=1\). For all other \(x\) in the interval,

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

Both apparent singularities are removable, and setting \(h(-1)=h(1)=1\) gives a continuous extension. The integral is therefore the integral of \(1\):

$$ \int_{-2}^{2}h(x)\,dx =\int_{-2}^{-1}1\,dx+\int_{-1}^{1}1\,dx+\int_{1}^{2}1\,dx =1+2+1=4. $$

The antiderivative \(H(x)=x\) checks the result: \(H(2)-H(-2)=2-(-2)=4\), and \(H'(x)=1\) on the extended interval. The original expression has holes, but it has no genuine pole at either point.

Common Pitfall: Cancelling Opposite Divergences

In the first example, restrict attention to \([-1,1]\), where the only pole is \(0\). The left-hand integral tends to \(+\infty\), while the right-hand integral tends to \(-\infty\). It is incorrect to declare their sum convergent because the divergences have opposite signs: \(+\infty+(-\infty)\) is not a defined sum of finite integral values.

A symmetric cutoff does produce a finite limit in this particular case. With the same \(\varepsilon\) on each side,

$$ \int_{-1}^{-\varepsilon}f(x)\,dx+\int_{\varepsilon}^{1}f(x)\,dx =\frac12\ln(2+\varepsilon)-\frac12\ln(2-\varepsilon)-\frac12\ln3 \longrightarrow-\frac12\ln3. $$

This coordinated limit is a Cauchy principal value. It does not change the ordinary conclusion, because the left and right one-sided integrals still diverge separately. Always decide which notion of convergence is being asked for before combining any truncated expressions.

Summary of the Convergence Test

For a rational integral with two interior poles, first determine which denominator zeros are genuine poles, then split at each such point. Use partial fractions or another valid antiderivative on the separate intervals, and evaluate every one-sided limit. The integral converges ordinarily only if every one-sided piece has a finite limit. Divergences at different sides or different poles cannot cancel to establish ordinary convergence.

Check Your Understanding

For each question, pay attention to which limits must be taken separately.

  1. If an integrand has poles at \(a\) and \(b\) inside an interval, how many one-sided approaches to poles must be checked?
  2. What condition must hold for the integral over the interval between two poles to converge?
  3. Why is it invalid to add a one-sided limit of \(+\infty\) to one of \(-\infty\) when testing ordinary convergence?
  4. In the second worked example, which term in the antiderivative drives the divergence near \(x=1\)?
  5. How can you distinguish a genuine pole from a removable hole in a rational expression?

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