12 Improper Integrals
12.1 Beyond Bounded Intervals
The Riemann integral, as developed in previous chapters, requires two constraints:
- The interval [a,b] must be finite
- The integrand f must be bounded on [a,b]
Both restrictions are essential. If f is unbounded, upper sums U(f,P) can diverge to infinity—no finite supremum exists. If the interval is unbounded, the length (b-a) itself is infinite, and partitions require infinitely many subintervals.
Yet we encounter integrals that violate these constraints naturally:
Infinite intervals: \int_1^{\infty} \frac{1}{x^2}\,dx, \quad \int_{-\infty}^{\infty} e^{-x^2}\,dx
Unbounded integrands: \int_0^1 \frac{1}{\sqrt{x}}\,dx, \quad \int_0^1 \ln x\,dx
In each case, the naive interpretation fails—we cannot apply the Fundamental Theorem directly to an infinite interval or an integrand with vertical asymptotes. Yet geometric intuition suggests these might represent finite areas. The region under y = 1/x^2 from x = 1 to infinity appears to taper rapidly enough to enclose finite area, while y = 1/\sqrt{x} near x = 0 grows slowly enough that the spike might contribute finite area.
We formalize this intuition by defining improper integrals as limits of proper integrals. This parallels our approach to infinite series: we defined \sum_{n=1}^{\infty} a_n as the limit of partial sums s_N = \sum_{n=1}^{N} a_n. Here, we’ll define \int_a^{\infty} f(x)\,dx as the limit of \int_a^{b} f(x)\,dx as b \to \infty.
The analogy runs deeper. Convergence tests for series—comparison, ratio, root—have direct analogs for improper integrals. The integral test, which we develop at the end of this chapter, makes the connection explicit: series and integrals converge or diverge together under suitable conditions.
12.2 Type I: Infinite Intervals
12.2.1 Definition and Motivation
Consider \int_1^{\infty} \frac{1}{x^2}\,dx. We cannot evaluate this directly—the upper limit is not a real number. But for any finite b > 1, the integral \int_1^b \frac{1}{x^2}\,dx = \left[-\frac{1}{x}\right]_1^b = 1 - \frac{1}{b} is well-defined. As b increases, the integral approaches 1. This motivates the definition.
Definition 12.1 (Improper Integral of Type I) Let f be integrable on [a,b] for every b > a. The improper integral \int_a^{\infty} f(x)\,dx is defined as \int_a^{\infty} f(x)\,dx = \lim_{b \to \infty} \int_a^b f(x)\,dx, provided the limit exists (as a finite real number). If the limit exists, the improper integral converges; otherwise, it diverges.
Similarly, for b < a, \int_{-\infty}^b f(x)\,dx = \lim_{a \to -\infty} \int_a^b f(x)\,dx.
For integrals over \mathbb{R}, we split at any point c: \int_{-\infty}^{\infty} f(x)\,dx = \int_{-\infty}^c f(x)\,dx + \int_c^{\infty} f(x)\,dx, and require both integrals on the right to converge.
Remark on the split. The choice of c is arbitrary—if both integrals converge for one choice of c, they converge for all choices, and the sum is independent of c. This follows from additivity of (proper) integrals: if a < c_1 < c_2, then \int_a^{c_2} f = \int_a^{c_1} f + \int_{c_1}^{c_2} f, so changing the split point just redistributes contributions between the two improper integrals.
Warning. The condition that both integrals converge is essential. Consider \int_{-\infty}^{\infty} x\,dx. If we evaluate “symmetrically,” \lim_{R \to \infty} \int_{-R}^R x\,dx = \lim_{R \to \infty} 0 = 0, which appears to converge. But this is false. The correct definition requires \lim_{a \to -\infty} \int_a^0 x\,dx \quad \text{and} \quad \lim_{b \to \infty} \int_0^b x\,dx to both exist. The first is \lim_{a \to -\infty} [x^2/2]_a^0 = \lim_{a \to -\infty} -a^2/2 = -\infty, which diverges. Thus \int_{-\infty}^{\infty} x\,dx diverges, despite the symmetric limit being zero.
The symmetric limit \lim_{R \to \infty} \int_{-R}^R f(x)\,dx is called the Cauchy principal value and is a distinct concept. We do not use it as the definition of convergence.
12.2.2 Examples
Example 1. Evaluate \int_1^{\infty} \frac{1}{x^2}\,dx.
By definition, \int_1^{\infty} \frac{1}{x^2}\,dx = \lim_{b \to \infty} \int_1^b \frac{1}{x^2}\,dx = \lim_{b \to \infty} \left[-\frac{1}{x}\right]_1^b = \lim_{b \to \infty} \left(1 - \frac{1}{b}\right) = 1.
The integral converges to 1.
Example 2. Determine whether \int_1^{\infty} \frac{1}{x}\,dx converges.
We have \int_1^{\infty} \frac{1}{x}\,dx = \lim_{b \to \infty} \int_1^b \frac{1}{x}\,dx = \lim_{b \to \infty} [\ln x]_1^b = \lim_{b \to \infty} \ln b = \infty.
The integral diverges.
Example 3. Evaluate \int_0^{\infty} e^{-x}\,dx.
\int_0^{\infty} e^{-x}\,dx = \lim_{b \to \infty} \int_0^b e^{-x}\,dx = \lim_{b \to \infty} [-e^{-x}]_0^b = \lim_{b \to \infty} (1 - e^{-b}) = 1.
The integral converges to 1.
Example 4. Show that \int_{-\infty}^{\infty} \frac{1}{1+x^2}\,dx converges and find its value.
Split at 0: \int_{-\infty}^{\infty} \frac{1}{1+x^2}\,dx = \int_{-\infty}^0 \frac{1}{1+x^2}\,dx + \int_0^{\infty} \frac{1}{1+x^2}\,dx.
For the second integral, \int_0^{\infty} \frac{1}{1+x^2}\,dx = \lim_{b \to \infty} [\arctan x]_0^b = \lim_{b \to \infty} \arctan b = \frac{\pi}{2}.
By symmetry (or direct computation), the first integral also equals \pi/2. Thus \int_{-\infty}^{\infty} \frac{1}{1+x^2}\,dx = \frac{\pi}{2} + \frac{\pi}{2} = \pi.
12.2.3 The p-Integral
The behavior of \int_1^{\infty} \frac{1}{x^p}\,dx determines convergence of many improper integrals via comparison. This is the continuous analog of the p-series \sum \frac{1}{n^p}.
Theorem 12.1 (p-Integral Test) The integral \int_1^{\infty} \frac{1}{x^p}\,dx converges if and only if p > 1.
Case 1: p \neq 1. We have \int_1^b \frac{1}{x^p}\,dx = \left[\frac{x^{1-p}}{1-p}\right]_1^b = \frac{1}{1-p}\left(b^{1-p} - 1\right).
As b \to \infty:
If p > 1, then 1-p < 0, so b^{1-p} \to 0. The limit is \frac{1}{p-1}.
If p < 1, then 1-p > 0, so b^{1-p} \to \infty. The integral diverges.
Case 2: p = 1. We showed in Example 2 that \int_1^{\infty} \frac{1}{x}\,dx = \infty, so the integral diverges. \square
The critical value p = 1 marks the boundary between convergence and divergence, exactly as for series. The integral \int_1^{\infty} \frac{1}{x^{1.001}}\,dx converges (slowly), while \int_1^{\infty} \frac{1}{x^{0.999}}\,dx diverges (slowly). This threshold arises from the logarithmic growth of \int \frac{1}{x}\,dx = \ln x, which is intermediate between power-law decay and power-law growth.
12.3 Type II: Unbounded Integrands
12.3.1 Definition and Examples
An integrand with a vertical asymptote within the interval of integration is unbounded. The Riemann integral does not apply directly—upper sums diverge. We again use limits to assign meaning.
Definition 12.2 (Improper Integral of Type II) Let f be continuous on (a,b] but unbounded as x \to a^+. The improper integral \int_a^b f(x)\,dx is defined as \int_a^b f(x)\,dx = \lim_{c \to a^+} \int_c^b f(x)\,dx, provided the limit exists.
Similarly, if f is unbounded as x \to b^-, define \int_a^b f(x)\,dx = \lim_{c \to b^-} \int_a^c f(x)\,dx.
If f is unbounded at an interior point c \in (a,b), split: \int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx, and require both improper integrals to converge.
Example 1. Evaluate \int_0^1 \frac{1}{\sqrt{x}}\,dx.
The integrand is unbounded as x \to 0^+. By definition, \int_0^1 \frac{1}{\sqrt{x}}\,dx = \lim_{c \to 0^+} \int_c^1 \frac{1}{\sqrt{x}}\,dx = \lim_{c \to 0^+} [2\sqrt{x}]_c^1 = \lim_{c \to 0^+} (2 - 2\sqrt{c}) = 2.
The integral converges to 2.
Example 2. Determine whether \int_0^1 \frac{1}{x}\,dx converges.
\int_0^1 \frac{1}{x}\,dx = \lim_{c \to 0^+} \int_c^1 \frac{1}{x}\,dx = \lim_{c \to 0^+} [\ln x]_c^1 = \lim_{c \to 0^+} (0 - \ln c) = \infty.
The integral diverges.
Example 3. Evaluate \int_0^1 \ln x\,dx.
The integrand \ln x \to -\infty as x \to 0^+. By definition, \int_0^1 \ln x\,dx = \lim_{c \to 0^+} \int_c^1 \ln x\,dx.
Using integration by parts (from the previous chapter), \int \ln x\,dx = x\ln x - x + C. Thus \int_c^1 \ln x\,dx = [x\ln x - x]_c^1 = (0 - 1) - (c\ln c - c) = -1 - c\ln c + c.
As c \to 0^+, we have c \to 0 and c\ln c \to 0 (by L’Hôpital’s rule or series expansion). Therefore, \int_0^1 \ln x\,dx = \lim_{c \to 0^+} (-1 - c\ln c + c) = -1.
The integral converges to -1.
12.3.2 The q-Integral at Zero
Just as the p-integral governs behavior at infinity, the q-integral governs behavior near a singularity.
Theorem 12.2 (q-Integral Test) Let a < b. The integral \int_a^b \frac{1}{(x-a)^q}\,dx converges if and only if q < 1.
Set u = x - a, so the integral becomes \int_0^{b-a} \frac{1}{u^q}\,du. It suffices to consider \int_0^1 \frac{1}{u^q}\,du.
Case 1: q \neq 1. We have \int_c^1 \frac{1}{u^q}\,du = \left[\frac{u^{1-q}}{1-q}\right]_c^1 = \frac{1}{1-q}(1 - c^{1-q}).
As c \to 0^+:
If q < 1, then 1-q > 0, so c^{1-q} \to 0. The limit is \frac{1}{1-q}.
If q > 1, then 1-q < 0, so c^{1-q} \to \infty. The integral diverges.
Case 2: q = 1. We have \int_c^1 \frac{1}{u}\,du = [\ln u]_c^1 = -\ln c \to \infty \quad \text{as } c \to 0^+.
The integral diverges. \square
Note the reversed threshold: for Type I integrals, p > 1 ensures convergence; for Type II, q < 1 ensures convergence. The exponent q = 1 remains the critical boundary in both cases.
12.4 Comparison Tests
12.4.1 Direct Comparison
Just as for series, if 0 \leq f(x) \leq g(x) on [a,\infty) and \int_a^{\infty} g(x)\,dx converges, then \int_a^{\infty} f(x)\,dx must converge—the area under f is trapped below the area under g. Conversely, if \int_a^{\infty} f(x)\,dx diverges, so does \int_a^{\infty} g(x)\,dx.
Theorem 12.3 (Comparison Test for Improper Integrals) Let f, g be continuous on [a,\infty) with 0 \leq f(x) \leq g(x) for all x \geq a.
If \int_a^{\infty} g(x)\,dx converges, then \int_a^{\infty} f(x)\,dx converges.
If \int_a^{\infty} f(x)\,dx diverges, then \int_a^{\infty} g(x)\,dx diverges.
The analogous statements hold for Type II integrals.
Let F(b) = \int_a^b f(x)\,dx and G(b) = \int_a^b g(x)\,dx. Since f, g \geq 0, both F and G are increasing functions of b.
(1) Suppose \int_a^{\infty} g(x)\,dx converges. Then G(b) is bounded as b \to \infty. Since f \leq g, we have F(b) \leq G(b) for all b. Thus F(b) is increasing and bounded, hence converges by monotonicity.
(2) This is the contrapositive of (1). \square
Example 1. Show that \int_1^{\infty} \frac{e^{-x}}{x}\,dx converges.
For x \geq 1, we have \frac{e^{-x}}{x} \leq e^{-x}. Since \int_1^{\infty} e^{-x}\,dx converges (Example 3 in Section 12.2), the comparison test gives convergence of \int_1^{\infty} \frac{e^{-x}}{x}\,dx.
Example 2. Show that \int_2^{\infty} \frac{1}{\sqrt{x^2-1}}\,dx diverges.
For x \geq 2, we have x^2 - 1 < x^2, so \frac{1}{\sqrt{x^2-1}} > \frac{1}{x}. Since \int_2^{\infty} \frac{1}{x}\,dx diverges, the comparison test gives divergence of \int_2^{\infty} \frac{1}{\sqrt{x^2-1}}\,dx.
12.4.2 Limit Comparison
When direct inequalities are difficult to establish, comparing asymptotic behavior often suffices.
Theorem 12.4 (Limit Comparison Test) Let f, g be positive continuous functions on [a,\infty). If \lim_{x \to \infty} \frac{f(x)}{g(x)} = L where 0 < L < \infty, then \int_a^{\infty} f(x)\,dx and \int_a^{\infty} g(x)\,dx either both converge or both diverge.
The analogous statement holds for Type II integrals, with the limit taken as x approaches the singularity.
Since \lim_{x \to \infty} \frac{f(x)}{g(x)} = L with 0 < L < \infty, there exists x_0 \geq a such that for all x \geq x_0, \frac{L}{2} < \frac{f(x)}{g(x)} < 2L.
This gives \frac{L}{2}g(x) < f(x) < 2L\,g(x) for x \geq x_0.
If \int_a^{\infty} g(x)\,dx converges, then \int_{x_0}^{\infty} 2L\,g(x)\,dx converges, so by comparison, \int_{x_0}^{\infty} f(x)\,dx converges. Since \int_a^{x_0} f(x)\,dx is finite (proper integral on a finite interval), \int_a^{\infty} f(x)\,dx converges.
If \int_a^{\infty} f(x)\,dx converges, then \int_{x_0}^{\infty} \frac{L}{2}g(x)\,dx converges by comparison, hence \int_a^{\infty} g(x)\,dx converges. \square
Example 1. Determine convergence of \int_1^{\infty} \frac{2x^2 + 3}{x^4 + 5x + 1}\,dx.
For large x, the integrand behaves like \frac{2x^2}{x^4} = \frac{2}{x^2}. Compute \lim_{x \to \infty} \frac{(2x^2+3)/(x^4+5x+1)}{2/x^2} = \lim_{x \to \infty} \frac{x^2(2x^2+3)}{2(x^4+5x+1)} = \lim_{x \to \infty} \frac{2x^4 + 3x^2}{2x^4 + 10x + 2} = 1.
Since \int_1^{\infty} \frac{2}{x^2}\,dx converges (by Theorem 12.1 with p = 2 > 1), the limit comparison test gives convergence.
Example 2. Determine convergence of \int_0^1 \frac{\sin x}{\sqrt{x}}\,dx.
The integrand is unbounded as x \to 0^+. Near x = 0, \sin x \approx x, so \frac{\sin x}{\sqrt{x}} \approx \frac{x}{\sqrt{x}} = \sqrt{x}. Compute \lim_{x \to 0^+} \frac{(\sin x)/\sqrt{x}}{\sqrt{x}} = \lim_{x \to 0^+} \frac{\sin x}{x} = 1.
Since \int_0^1 \sqrt{x}\,dx = \int_0^1 x^{1/2}\,dx converges (by Theorem 12.2 with q = -1/2 < 1), the limit comparison test gives convergence.
12.5 Absolute Convergence
12.5.1 Definition
When the integrand changes sign, convergence becomes more subtle. Just as for series, we distinguish absolute from conditional convergence.
Definition 12.3 (Absolutely Convergent Integral) An improper integral \int_a^{\infty} f(x)\,dx converges absolutely if \int_a^{\infty} |f(x)|\,dx converges.
Absolute convergence implies convergence, exactly as for series.
Theorem 12.5 (Absolute Convergence Implies Convergence) If \int_a^{\infty} |f(x)|\,dx converges, then \int_a^{\infty} f(x)\,dx converges.
Write f = f^+ - f^- where f^+(x) = \max(f(x), 0) and f^-(x) = \max(-f(x), 0). Then 0 \leq f^+, f^- \leq |f|.
Since \int_a^{\infty} |f(x)|\,dx converges, by comparison, both \int_a^{\infty} f^+(x)\,dx and \int_a^{\infty} f^-(x)\,dx converge. Therefore, \int_a^{\infty} f(x)\,dx = \int_a^{\infty} f^+(x)\,dx - \int_a^{\infty} f^-(x)\,dx converges. \square
Example. The integral \int_1^{\infty} \frac{\sin x}{x^2}\,dx converges absolutely because \left|\frac{\sin x}{x^2}\right| \leq \frac{1}{x^2}, and \int_1^{\infty} \frac{1}{x^2}\,dx converges.
12.6 The Integral Test
12.6.1 Connecting Series and Integrals
The integral test establishes a precise connection between convergence of series and convergence of improper integrals. This is the link promised at the chapter’s beginning.
Consider the series \sum_{n=1}^{\infty} a_n where a_n = f(n) for some function f. If f is positive, continuous, and decreasing on [1,\infty), we can relate the partial sums to the integral.
Theorem 12.6 (Integral Test) Let f : [1,\infty) \to \mathbb{R} be positive, continuous, and decreasing. Then the series \sum_{n=1}^{\infty} f(n) converges if and only if the improper integral \int_1^{\infty} f(x)\,dx converges.
Since f is decreasing, for n \leq x \leq n+1, we have f(n+1) \leq f(x) \leq f(n). Integrating over [n, n+1]: f(n+1) \leq \int_n^{n+1} f(x)\,dx \leq f(n).
Summing from n = 1 to N-1: \sum_{n=2}^{N} f(n) \leq \int_1^N f(x)\,dx \leq \sum_{n=1}^{N-1} f(n).
(\Leftarrow) Suppose \int_1^{\infty} f(x)\,dx converges. Then the sequence \left\{\int_1^N f(x)\,dx\right\} is bounded. By the left inequality, \sum_{n=1}^{N} f(n) = f(1) + \sum_{n=2}^{N} f(n) \leq f(1) + \int_1^N f(x)\,dx, which is bounded. Since partial sums are increasing (all terms positive) and bounded, the series converges.
(\Rightarrow) Suppose \sum_{n=1}^{\infty} f(n) converges. By the right inequality, \int_1^N f(x)\,dx \leq \sum_{n=1}^{N-1} f(n), which is bounded. Since \int_1^N f(x)\,dx is increasing in N and bounded, it converges as N \to \infty. \square
Geometric interpretation. The partial sum \sum_{n=1}^{N} f(n) represents the total area of rectangles of width 1 and heights f(1), f(2), \ldots, f(N). When f is decreasing, these rectangles sit above or below the curve y = f(x), depending on whether we use left or right endpoints. The integral \int_1^N f(x)\,dx is the area under the curve. Convergence of one implies boundedness of the other.
12.6.2 Application to p-Series
The integral test immediately gives the convergence behavior of p-series, which we stated without proof in earlier chapters.
Corollary 12.1 (p-Series Test) The series \sum_{n=1}^{\infty} \frac{1}{n^p} converges if and only if p > 1.
Apply Theorem 12.6 with f(x) = \frac{1}{x^p}, which is positive, continuous, and decreasing on [1,\infty) for p > 0. By Theorem 12.1, \int_1^{\infty} \frac{1}{x^p}\,dx converges if and only if p > 1. Therefore, \sum_{n=1}^{\infty} \frac{1}{n^p} converges if and only if p > 1. \square
Example. The harmonic series \sum_{n=1}^{\infty} \frac{1}{n} diverges because p = 1. The series \sum_{n=1}^{\infty} \frac{1}{n^2} converges because p = 2 > 1.
12.6.3 The Cauchy Condensation Test
A consequence of the integral test is the Cauchy condensation test, which provides another method for testing series of the form \sum f(n) where f decreases.
Theorem 12.7 (Cauchy Condensation Test) Let \{a_n\} be a positive decreasing sequence. Then \sum_{n=1}^{\infty} a_n \quad \text{and} \quad \sum_{k=0}^{\infty} 2^k a_{2^k} converge or diverge together.
We omit the proof, which uses geometric grouping of terms and the integral test. The condensation test is particularly useful for series involving logarithms.
Example. Test convergence of \sum_{n=2}^{\infty} \frac{1}{n(\ln n)^p}.
Let a_n = \frac{1}{n(\ln n)^p}, which is positive and decreasing for n \geq 2. The condensed series is \sum_{k=1}^{\infty} 2^k a_{2^k} = \sum_{k=1}^{\infty} 2^k \cdot \frac{1}{2^k(\ln 2^k)^p} = \sum_{k=1}^{\infty} \frac{1}{(k\ln 2)^p}.
This is a p-series in k, which converges if and only if p > 1. By the condensation test, \sum_{n=2}^{\infty} \frac{1}{n(\ln n)^p} converges if and only if p > 1.
12.7 Summary
Improper integrals extend integration to unbounded domains and unbounded integrands via limits of proper integrals. The analogy with infinite series is:
| Series | Improper Integral |
|---|---|
| \sum_{n=1}^{\infty} a_n = \lim_{N \to \infty} \sum_{n=1}^{N} a_n | \int_a^{\infty} f(x)\,dx = \lim_{b \to \infty} \int_a^b f(x)\,dx |
| p-series: \sum \frac{1}{n^p} converges iff p > 1 | p-integral: \int_1^{\infty} \frac{1}{x^p}\,dx converges iff p > 1 |
| Comparison test | Comparison test |
| Limit comparison test | Limit comparison test |
| Absolute convergence implies convergence | Absolute convergence implies convergence |
Tests:
Direct evaluation: If an antiderivative is known, compute the limit explicitly
p-integral test: Compare to \int \frac{1}{x^p}\,dx (Type I) or \int \frac{1}{(x-a)^q}\,dx (Type II)
Comparison: If 0 \leq f \leq g and \int g converges, then \int f converges
Limit comparison: If \lim f(x)/g(x) = L \in (0,\infty), then \int f and \int g have the same behavior
Absolute convergence: If \int |f| converges, then \int f converges
The techniques developed in this chapter—substitution, comparison, limit comparison—mirror those for series because both involve infinite processes. Improper integrals accumulate contributions over infinite domains or near singularities; series sum infinitely many terms. The shared structure explains why the same convergence tests apply to both.