Improper Integrals

Can we assign a finite area to the region under y=1/x2y = 1/x^2 from x=1x=1 to x=∞x=\infty? This region stretches out forever. Can we make sense of ∫011x dx\int_0^1 \frac{1}{\sqrt{x}}\,dx? Here, the integrand itself blows up as x→0+x \to 0^+. The ordinary Riemann Integral is built for bounded functions. It applies to bounded, closed intervals. Neither situation fits that mold. Improper integrals extend integration to these two settings. They use a single unifying device. Take a proper Riemann integral over a safely contained sub-domain. Then pass to a limit. That sub-domain must exhaust the full unbounded interval, or approach the singular point. This extension matters enormously. Many of the most important functions in analysis and probability are defined by improper integrals. These include the Gamma function, the Gaussian integral, and Laplace and Fourier transforms. Questions of convergence versus divergence recur throughout physics. Examples include energy integrals and potentials with 1/r1/r singularities. They also recur in probability, such as whether a density integrates to a finite total mass. The theory also foreshadows a deep structural parallel. Whether ∫1∞f(x) dx\int_1^\infty f(x)\,dx converges behaves exactly like whether ∑n=1∞an\sum_{n=1}^\infty a_n converges. This analogy becomes the precise Integral Test for Series. The central technical challenge is that limits do not always commute nicely with integration. So we must be careful about what is being defined. We must also track in what order limits are taken. We will define convergence rigorously via limits of proper integrals. We will prove the basic comparison tests. These let us decide convergence without computing an antiderivative. We will carefully distinguish convergence from absolute convergence. This article gives the precise definitions for both unbounded-interval and unbounded-integrand improper integrals. It proves the Direct and Limit Comparison Tests with full rigor. It works two nontrivial examples. It shows exactly where the standard hypotheses are needed. These hypotheses are nonnegativity and local integrability. It does this by exhibiting failures when they are dropped.

Key takeaways
  • An improper integral is defined as a limit of proper Riemann integrals, not as a new kind of Riemann integral.
  • Two independent sources of "impropriety": an unbounded interval of integration, or an unbounded integrand near an endpoint.
  • Convergence means the limit exists and is finite; otherwise the integral diverges (including diverging to ±∞\pm\infty).
  • The Comparison Test decides convergence of nonnegative integrands by squeezing against a known integral — no antiderivative needed.
  • Absolute convergence implies convergence, but not conversely — conditional convergence is a genuine phenomenon (e.g. ∫1∞sin⁡xxdx\int_1^\infty \frac{\sin x}{x}dx).
  • Key idea: replace "integrate over the whole bad domain" with "integrate over a safe truncation, then let the truncation limit to the bad domain."

Where this lives today

Improper integrals are the rigorous foundation for special functions defined by integral formulas that fail to be proper Riemann integrals: the Gamma function Γ(s)=∫0∞xs−1e−xdx\Gamma(s) = \int_0^\infty x^{s-1}e^{-x}dx (improper at both 00, when s<1s<1, and ∞\infty), the Gaussian integral ∫−∞∞e−x2dx=π\int_{-\infty}^\infty e^{-x^2}dx = \sqrt\pi underlying the normal distribution in statistics, and the Laplace and Fourier transforms ∫0∞f(t)e−stdt\int_0^\infty f(t)e^{-st}dt, ∫−∞∞f(t)e−iωtdt\int_{-\infty}^\infty f(t)e^{-i\omega t}dt used throughout engineering and signal processing. In physics, potentials like ∫r∞1x2dx\int_r^\infty \frac{1}{x^2}dx (gravitational/electrostatic potential energy at infinity) and singular densities near a point charge rely on exactly this convergence theory. In probability theory, whether a probability density integrates to a finite total mass, and whether moments ∫xkf(x) dx\int x^k f(x)\,dx exist, are improper-integral convergence questions decided by the same comparison techniques developed here. Most directly, the comparison and pp-integral results here are the exact analytic engine behind the Integral Test for Series, which converts convergence questions about series ∑an\sum a_n into convergence questions about ∫1∞f(x) dx\int_1^\infty f(x)\,dx for a decreasing ff with f(n)=anf(n)=a_n — turning, for instance, the divergence of the harmonic series into an immediate consequence of the p=1p=1 divergence proven above.

Ask the tutor: How does the Dirichlet test prove ∫1∞sin⁡xx dx\int_1^\infty \frac{\sin x}{x}\,dx converges even though it isn't absolutely convergent?

DefinitionImproper Integral

Type I (unbounded interval). Let ff be Riemann integrable on [a,t][a,t] for every t>at > a. The improper integral of ff over [a,∞)[a,\infty) is defined by

∫a∞f(x) dx  :=  lim⁡t→∞∫atf(x) dx,\int_a^\infty f(x)\,dx \;:=\; \lim_{t\to\infty} \int_a^t f(x)\,dx,

provided this limit exists in R\mathbb{R}; we then say the integral converges. If the limit is +∞+\infty, −∞-\infty, or does not exist, the integral diverges. The case ∫−∞bf dx\int_{-\infty}^b f\,dx is defined symmetrically using s→−∞s \to -\infty, and

∫−∞∞f(x) dx:=∫−∞cf(x) dx+∫c∞f(x) dx\int_{-\infty}^\infty f(x)\,dx := \int_{-\infty}^c f(x)\,dx + \int_c^\infty f(x)\,dx

for any fixed cc, provided both pieces converge (the value is independent of cc).

Type II (unbounded integrand). Let ff be Riemann integrable on [a+ε,b][a+\varepsilon, b] for every ε∈(0,b−a)\varepsilon \in (0, b-a), and suppose ff is unbounded near aa (a singularity at the left endpoint). Define

∫abf(x) dx:=lim⁡ε→0+∫a+εbf(x) dx,\int_a^b f(x)\,dx := \lim_{\varepsilon \to 0^+} \int_{a+\varepsilon}^b f(x)\,dx,

provided the limit exists in R\mathbb{R}; symmetrically for a singularity at bb. If ff has a singularity at an interior point c∈(a,b)c \in (a,b), define

∫abf(x) dx:=∫acf(x) dx+∫cbf(x) dx,\int_a^b f(x)\,dx := \int_a^c f(x)\,dx + \int_c^b f(x)\,dx,

provided both pieces (each itself possibly improper) converge. In every case, "the improper integral converges" means the relevant limit(s) exist and are finite; the value of the improper integral is that limit.

Intuition

The unifying picture is exhaustion by safe truncations. In the Type I case, we cannot integrate all the way to ∞\infty directly — the Riemann integral requires a bounded interval — so we integrate up to a large finite tt, compute F(t)=∫atfF(t) = \int_a^t f, and ask what happens to F(t)F(t) as tt grows without bound. In the Type II case, we cannot integrate all the way down to the singular endpoint, so we back off by a small ε\varepsilon, compute ∫a+εbf\int_{a+\varepsilon}^b f, and ask what happens as ε\varepsilon shrinks to 00. Both constructions reduce a genuinely new question ("does this improper integral converge?") to a familiar one from single-variable calculus ("does this limit exist?"), applied to the function t↦∫atf(x) dxt \mapsto \int_a^t f(x)\,dx, whose properties (continuity, monotonicity when f≥0f \ge 0) come from the Second Fundamental Theorem of Calculus and the Monotonicity of the Integral. A key qualitative fact, visible already in the two canonical examples ∫1∞x−p dx\int_1^\infty x^{-p}\,dx and ∫01x−p dx\int_0^1 x^{-p}\,dx, is that convergence is a battle between the rate of decay of the integrand and the length of the domain being exhausted. Slow decay over an infinite domain, or a mild singularity over a domain shrinking to a point, can still yield finite area; too slow a decay, or too wild a singularity, makes the area infinite.

Figure
The two faces of impropriety: unbounded domain (left) vs. unbounded integrand (right)
x y a=1 t → ∞ y = 1/x² ∫₁ᵗ x⁻² dx → 1 x y ε → 0⁺ b=1 y = 1/√x ∫_ε¹ x⁻¹ᐟ² dx → 2
Theorempp-integral benchmarks

Let p∈Rp \in \mathbb{R}.

(a) ∫1∞dxxp\displaystyle \int_1^\infty \frac{dx}{x^p} converges if and only if p>1p > 1, in which case its value is 1p−1\dfrac{1}{p-1}.

(b) ∫01dxxp\displaystyle \int_0^1 \frac{dx}{x^p} converges if and only if p<1p < 1, in which case its value is 11−p\dfrac{1}{1-p}.

Proof

Technique: direct computation of the truncated integral via the Second Fundamental Theorem of Calculus, followed by evaluation of the resulting limit.

(a) Fix t>1t > 1. The function x↦x−px \mapsto x^{-p} is continuous, hence Riemann integrable, on [1,t][1,t]. Case p≠1p \ne 1. An antiderivative of x−px^{-p} on (0,∞)(0,\infty) is x1−p1−p\dfrac{x^{1-p}}{1-p}. By the Second Fundamental Theorem of Calculus,

∫1tx−p dx=t1−p1−p−11−p.\int_1^t x^{-p}\,dx = \frac{t^{1-p}}{1-p} - \frac{1}{1-p}.

If p>1p > 1, then 1−p<01-p < 0, so t1−p=t−(p−1)→0t^{1-p} = t^{-(p-1)} \to 0 as t→∞t \to \infty (since p−1>0p - 1 > 0). Hence

lim⁡t→∞∫1tx−p dx=0−11−p=1p−1,\lim_{t\to\infty}\int_1^t x^{-p}\,dx = 0 - \frac{1}{1-p} = \frac{1}{p-1},

a finite limit, so the integral converges to 1p−1\frac{1}{p-1}. If p<1p < 1, then 1−p>01 - p > 0, so t1−p→∞t^{1-p} \to \infty as t→∞t\to\infty, and the limit above is +∞+\infty: the integral diverges. Case p=1p = 1. Here ∫1tx−1 dx=ln⁡t−ln⁡1=ln⁡t→∞\int_1^t x^{-1}\,dx = \ln t - \ln 1 = \ln t \to \infty as t→∞t \to \infty: divergence. Combining the three cases: convergence holds exactly when p>1p>1, with value 1p−1\frac{1}{p-1}. This proves (a).

(b) Fix ε∈(0,1)\varepsilon \in (0,1). Since pp may be negative, zero, or positive, note x−px^{-p} is continuous on [ε,1][\varepsilon, 1] regardless (it may be unbounded only as x→0+x \to 0^+), so it is Riemann integrable there. Case p≠1p \ne 1. As above,

∫ε1x−p dx=11−p−ε1−p1−p.\int_\varepsilon^1 x^{-p}\,dx = \frac{1}{1-p} - \frac{\varepsilon^{1-p}}{1-p}.

If p<1p < 1, then 1−p>01 - p > 0, so ε1−p→0\varepsilon^{1-p} \to 0 as ε→0+\varepsilon \to 0^+ (a positive power of ε→0\varepsilon \to 0). Hence

lim⁡ε→0+∫ε1x−p dx=11−p−0=11−p,\lim_{\varepsilon \to 0^+} \int_\varepsilon^1 x^{-p}\,dx = \frac{1}{1-p} - 0 = \frac{1}{1-p},

so the integral converges to 11−p\frac{1}{1-p}. If p>1p > 1, then 1−p<01 - p < 0, so ε1−p=ε−(p−1)→∞\varepsilon^{1-p} = \varepsilon^{-(p-1)} \to \infty as ε→0+\varepsilon \to 0^+ (since p−1>0p-1>0 and ε→0+\varepsilon \to 0^+ makes a negative power blow up); the limit is −∞-\infty times a fixed sign that works out to +∞+\infty overall (concretely: −ε1−p1−p-\frac{\varepsilon^{1-p}}{1-p} with 1−p<01-p<0 gives −(→∞)(negative)=+∞-\frac{(\to\infty)}{(\text{negative})} = +\infty): divergence. Case p=1p=1. Here ∫ε1x−1dx=ln⁡1−ln⁡ε=−ln⁡ε→+∞\int_\varepsilon^1 x^{-1}dx = \ln 1 - \ln \varepsilon = -\ln \varepsilon \to +\infty as ε→0+\varepsilon \to 0^+: divergence. So convergence holds exactly when p<1p < 1, with value 11−p\frac{1}{1-p}. This proves (b). ■\blacksquare

∎
TheoremComparison Test for Improper Integrals

Let f,gf, g be functions that are Riemann integrable on [a,t][a,t] for every t>at > a (the Type I setting; the Type II statement is analogous with ε→0+\varepsilon \to 0^+ in place of t→∞t \to \infty), and suppose

0≤f(x)≤g(x)for all x≥a.0 \le f(x) \le g(x) \quad \text{for all } x \ge a.

Then:

(i) If ∫a∞g(x) dx\displaystyle\int_a^\infty g(x)\,dx converges, then ∫a∞f(x) dx\displaystyle\int_a^\infty f(x)\,dx converges, and

∫a∞f(x) dx≤∫a∞g(x) dx.\int_a^\infty f(x)\,dx \le \int_a^\infty g(x)\,dx.

(ii) If ∫a∞f(x) dx\displaystyle\int_a^\infty f(x)\,dx diverges (to +∞+\infty), then ∫a∞g(x) dx\displaystyle\int_a^\infty g(x)\,dx diverges (to +∞+\infty).

Proof

Technique: monotone convergence of a bounded, monotone real-valued function of tt (the Monotone Bounded Sequence/Function Criterion for limits). Define F(t)=∫atf(x) dxF(t) = \int_a^t f(x)\,dx and G(t)=∫atg(x) dxG(t) = \int_a^t g(x)\,dx for t≥at \ge a; both are well-defined real numbers for each tt since f,gf,g are Riemann integrable on [a,t][a,t].

Step 1: FF and GG are nondecreasing on [a,∞)[a,\infty). For a≤t1<t2a \le t_1 < t_2, additivity of the integral over adjacent intervals gives

F(t2)−F(t1)=∫t1t2f(x) dx≥0,F(t_2) - F(t_1) = \int_{t_1}^{t_2} f(x)\,dx \ge 0,

using f≥0f \ge 0 and the Monotonicity of the Integral (a nonnegative integrable function has nonnegative integral). Hence F(t1)≤F(t2)F(t_1) \le F(t_2); FF is nondecreasing. The same argument applies to GG since g≥0g \ge 0 (as g≥f≥0g \ge f \ge 0).

Step 2: F(t)≤G(t)F(t) \le G(t) for every t≥at \ge a. Since 0≤f(x)≤g(x)0 \le f(x) \le g(x) on [a,t][a,t], monotonicity of the integral gives ∫atf(x) dx≤∫atg(x) dx\int_a^t f(x)\,dx \le \int_a^t g(x)\,dx, i.e. F(t)≤G(t)F(t) \le G(t).

Step 3 (proof of (i)): if G(t)→L∈RG(t) \to L \in \mathbb{R} as t→∞t \to \infty, then F(t)→F(t) \to some finite limit ≤L\le L. Since GG is nondecreasing (Step 1) and G(t)→LG(t) \to L, we have G(t)≤LG(t) \le L for all t≥at \ge a (a nondecreasing function converging to LL never exceeds LL: if G(t0)>LG(t_0) > L for some t0t_0, then for all t≥t0t \ge t_0, G(t)≥G(t0)>LG(t) \ge G(t_0) > L by monotonicity, contradicting G(t)→LG(t)\to L). Combined with Step 2,

F(t)≤G(t)≤Lfor all t≥a.F(t) \le G(t) \le L \quad \text{for all } t \ge a.

So FF is nondecreasing (Step 1) and bounded above by LL. By the Monotone Convergence Criterion for functions of a real variable (the function analogue of the Monotone Bounded Sequence Theorem, applied along t→∞t \to \infty: a nondecreasing function bounded above converges to its supremum), F(t)F(t) converges as t→∞t \to \infty to

M:=sup⁡t≥aF(t)≤L.M := \sup_{t \ge a} F(t) \le L.

This means ∫a∞f(x) dx\int_a^\infty f(x)\,dx converges (the defining limit exists and is finite, equal to MM), and ∫a∞f dx=M≤L=∫a∞g dx\int_a^\infty f\,dx = M \le L = \int_a^\infty g\,dx. This proves (i). (Justification of the Monotone Convergence Criterion used above, for completeness: let M=sup⁡t≥aF(t)<∞M = \sup_{t\ge a} F(t) < \infty. Given η>0\eta > 0, by definition of supremum there is t0t_0 with F(t0)>M−ηF(t_0) > M - \eta. For all t≥t0t \ge t_0, monotonicity gives F(t)≥F(t0)>M−ηF(t) \ge F(t_0) > M-\eta, while F(t)≤MF(t) \le M always. So ∣F(t)−M∣<η|F(t) - M| < \eta for all t≥t0t \ge t_0, which is precisely lim⁡t→∞F(t)=M\lim_{t\to\infty} F(t) = M.)

Step 4 (proof of (ii), the contrapositive-flavored statement): if F(t)→∞F(t) \to \infty as t→∞t\to\infty, then G(t)→∞G(t) \to \infty. Since F(t)≤G(t)F(t) \le G(t) for all tt (Step 2), and F(t)→∞F(t) \to \infty means: for every N>0N > 0 there is t0t_0 such that F(t)>NF(t) > N for all t≥t0t \ge t_0. Then for the same t0t_0, G(t)≥F(t)>NG(t) \ge F(t) > N for all t≥t0t \ge t_0. This is exactly the statement G(t)→∞G(t) \to \infty. Since GG is nondecreasing, this divergence is genuinely "diverges to +∞+\infty," not oscillatory. This proves (ii). ■\blacksquare

∎
CorollaryLimit Comparison Test

Let f,gf, g be Riemann integrable on [a,t][a,t] for every t>at > a, with f(x)≥0f(x) \ge 0 and g(x)>0g(x) > 0 for all x≥ax \ge a. Suppose

lim⁡x→∞f(x)g(x)=L,0<L<∞.\lim_{x \to \infty} \frac{f(x)}{g(x)} = L, \qquad 0 < L < \infty.

Then ∫a∞f(x) dx\int_a^\infty f(x)\,dx converges if and only if ∫a∞g(x) dx\int_a^\infty g(x)\,dx converges.

Proof

Technique: reduce to the Direct Comparison Test using the definition of the limit with ε=L/2\varepsilon = L/2. Since f(x)/g(x)→Lf(x)/g(x) \to L with L>0L > 0, apply the definition of limit with tolerance η=L/2>0\eta = L/2 > 0: there exists M≥aM \ge a such that for all x≥Mx \ge M,

∣f(x)g(x)−L∣<L2,i.e.L2<f(x)g(x)<3L2.\left| \frac{f(x)}{g(x)} - L \right| < \frac{L}{2}, \quad \text{i.e.} \quad \frac{L}{2} < \frac{f(x)}{g(x)} < \frac{3L}{2}.

Since g(x)>0g(x) > 0, this rearranges to

L2g(x)<f(x)<3L2g(x)for all x≥M.\frac{L}{2} g(x) < f(x) < \frac{3L}{2} g(x) \qquad \text{for all } x \ge M.

(⇐\Leftarrow) Suppose ∫a∞g\int_a^\infty g converges. Then ∫M∞g\int_M^\infty g converges too (they differ by the finite proper integral ∫aMg\int_a^M g, which does not affect convergence of the limit as t→∞t \to \infty), so ∫M∞3L2g(x) dx\int_M^\infty \frac{3L}{2} g(x)\,dx converges (finite constant multiple of a convergent improper integral, by linearity of limits). Since 0≤f(x)<3L2g(x)0 \le f(x) < \frac{3L}{2} g(x) on [M,∞)[M,\infty), the Comparison Test (i) gives that ∫M∞f\int_M^\infty f converges. Adding back the finite proper integral ∫aMf\int_a^M f (finite since ff is Riemann integrable there), ∫a∞f\int_a^\infty f converges. (⇒\Rightarrow) Suppose ∫a∞f\int_a^\infty f converges. Then ∫M∞f\int_M^\infty f converges. Since L2g(x)<f(x)\frac{L}{2} g(x) < f(x) for x≥Mx \ge M, i.e. g(x)<2Lf(x)g(x) < \frac{2}{L} f(x), and 2L>0\frac{2}{L} > 0 is a finite constant, comparison Test (i) (applied with the roles played by gg as the "smaller" function and 2Lf\frac{2}{L}f as the convergent "larger" one) gives that ∫M∞g\int_M^\infty g converges. Adding back ∫aMg\int_a^M g, we conclude ∫a∞g\int_a^\infty g converges. Both directions hold, establishing the equivalence. ■\blacksquare

∎
TheoremAbsolute Convergence Implies Convergence

Let ff be Riemann integrable on [a,t][a,t] for every t>at > a. If

∫a∞∣f(x)∣ dxconverges,\int_a^\infty |f(x)|\,dx \quad \text{converges},

then ∫a∞f(x) dx\int_a^\infty f(x)\,dx converges as well, and

∣∫a∞f(x) dx∣≤∫a∞∣f(x)∣ dx.\left| \int_a^\infty f(x)\,dx \right| \le \int_a^\infty |f(x)|\,dx.

(The analogous statement holds for Type II improper integrals.)

Proof

Technique: split into positive and negative parts, apply the Comparison Test to each, then recombine via linearity of the limit. Define f+(x)=max⁡(f(x),0)f^+(x) = \max(f(x), 0) and f−(x)=max⁡(−f(x),0)f^-(x) = \max(-f(x),0), so that

f=f+−f−,∣f∣=f++f−,0≤f+≤∣f∣,0≤f−≤∣f∣.f = f^+ - f^-, \qquad |f| = f^+ + f^-, \qquad 0 \le f^+ \le |f|, \qquad 0 \le f^- \le |f|.

Since ff is Riemann integrable on each [a,t][a,t], so are f+f^+ and f−f^- (both are continuous functions of ff, namely f+=f+∣f∣2f^+ = \frac{f+|f|}{2}, f−=∣f∣−f2f^- = \frac{|f|-f}{2}, and ∣f∣|f| is Riemann integrable whenever ff is — a standard closure property of the Riemann integral). By hypothesis ∫a∞∣f∣ dx\int_a^\infty |f|\,dx converges. Since 0≤f+(x)≤∣f(x)∣0 \le f^+(x) \le |f(x)| for all xx, the Comparison Test (i) gives that ∫a∞f+(x) dx\int_a^\infty f^+(x)\,dx converges. Since 0≤f−(x)≤∣f(x)∣0 \le f^-(x) \le |f(x)|, likewise ∫a∞f−(x) dx\int_a^\infty f^-(x)\,dx converges. Now write, for t>at > a,

∫atf(x) dx=∫atf+(x) dx−∫atf−(x) dx.\int_a^t f(x)\,dx = \int_a^t f^+(x)\,dx - \int_a^t f^-(x)\,dx.

Both terms on the right converge to finite limits as t→∞t \to \infty (by the previous paragraph), so by the algebraic limit law for limits of functions (limit of a difference is the difference of limits, when both exist),

lim⁡t→∞∫atf(x) dx=∫a∞f+(x) dx−∫a∞f−(x) dx,\lim_{t\to\infty}\int_a^t f(x)\,dx = \int_a^\infty f^+(x)\,dx - \int_a^\infty f^-(x)\,dx,

and this limit is a finite real number. Hence ∫a∞f(x) dx\int_a^\infty f(x)\,dx converges. For the inequality: for every t>at > a, the Triangle Inequality for Integrals gives

∣∫atf(x) dx∣≤∫at∣f(x)∣ dx.\left| \int_a^t f(x)\,dx \right| \le \int_a^t |f(x)|\,dx.

Both sides converge as t→∞t \to \infty (left side because ∫a∞f\int_a^\infty f converges, as just shown; right side by hypothesis). Taking t→∞t \to \infty on both sides preserves the (non-strict) inequality, by the order limit theorem: if A(t)≤B(t)A(t) \le B(t) for all tt and both converge, lim⁡A(t)≤lim⁡B(t)\lim A(t) \le \lim B(t). This gives

∣∫a∞f(x) dx∣≤∫a∞∣f(x)∣ dx.■\left| \int_a^\infty f(x)\,dx \right| \le \int_a^\infty |f(x)|\,dx. \qquad \blacksquare
∎

Worked examples

Example 1 (Type I, requiring both comparison and a direct computation piece). Determine whether

∫1∞x+sin⁡xx3+2 dx\int_1^\infty \frac{x + \sin x}{x^3 + 2}\,dx

converges, and if so bound it above. Solution. Let f(x)=x+sin⁡xx3+2f(x) = \dfrac{x+\sin x}{x^3+2}. First, f(x)≥0f(x) \ge 0 for x≥1x \ge 1: indeed x+sin⁡x≥x−1≥0x + \sin x \ge x - 1 \ge 0 for x≥1x \ge 1 (since sin⁡x≥−1\sin x \ge -1), and x3+2>0x^3 + 2 > 0 always. Also ff is continuous on [1,∞)[1,\infty) (a quotient of continuous functions with nonvanishing denominator), hence Riemann integrable on every [1,t][1,t]. For an upper comparison, use x+sin⁡x≤x+1≤2xx + \sin x \le x + 1 \le 2x for x≥1x \ge 1, and x3+2>x3x^3 + 2 > x^3, so

0≤f(x)=x+sin⁡xx3+2≤2xx3=2x2for all x≥1.0 \le f(x) = \frac{x+\sin x}{x^3+2} \le \frac{2x}{x^3} = \frac{2}{x^2} \qquad \text{for all } x \ge 1.

By the pp-integral theorem with p=2>1p=2>1, ∫1∞x−2 dx\int_1^\infty x^{-2}\,dx converges (to value 11). By linearity, ∫1∞2x−2 dx\int_1^\infty 2x^{-2}\,dx converges to 22. By the Direct Comparison Test (i), ∫1∞f(x) dx\int_1^\infty f(x)\,dx converges, and

0≤∫1∞x+sin⁡xx3+2 dx≤2.0 \le \int_1^\infty \frac{x+\sin x}{x^3+2}\,dx \le 2.

So the integral converges, with value in [0,2][0,2]; we have not computed it exactly (no elementary antiderivative is expected), but convergence and an explicit bound are fully established rigorously.

Example 2 (Type II, singularity plus a subtlety at the other endpoint requiring care). Determine whether

∫01ln⁡xx dx\int_0^1 \frac{\ln x}{\sqrt{x}}\,dx

converges, and if so find its value. Solution. The integrand f(x)=ln⁡xxf(x) = \dfrac{\ln x}{\sqrt x} is continuous on (0,1](0,1], but f(x)→−∞f(x) \to -\infty as x→0+x \to 0^+ (since ln⁡x→−∞\ln x \to -\infty and 1/x→∞1/\sqrt x \to \infty), so this is a genuine Type II improper integral with singularity at 00. Note the integrand is not of one sign near 00... actually here ln⁡x<0\ln x < 0 and x>0\sqrt x > 0 for x∈(0,1)x\in(0,1), so f(x)<0f(x) < 0 throughout (0,1)(0,1) — it is one-signed, just negative rather than positive; we handle this by considering −f≥0-f \ge 0 or by direct antiderivative computation, the latter being cleaner here since ff has an elementary antiderivative. For ε∈(0,1)\varepsilon \in (0,1), compute ∫ε1x−1/2ln⁡x dx\int_\varepsilon^1 x^{-1/2}\ln x\,dx by integration by parts, with u=ln⁡xu = \ln x, dv=x−1/2dxdv = x^{-1/2}dx, so du=dx/xdu = dx/x, v=2xv = 2\sqrt x:

∫ε1x−1/2ln⁡x dx=[2xln⁡x]ε1−∫ε12x⋅1x dx=(2⋅1⋅ln⁡1−2εln⁡ε)−2∫ε1x−1/2 dx.\int_\varepsilon^1 x^{-1/2} \ln x\,dx = \Big[2\sqrt x \ln x\Big]_\varepsilon^1 - \int_\varepsilon^1 2\sqrt x \cdot \frac{1}{x}\,dx = \Big(2\cdot 1 \cdot \ln 1 - 2\sqrt\varepsilon \ln \varepsilon\Big) - 2\int_\varepsilon^1 x^{-1/2}\,dx.

Since ln⁡1=0\ln 1 = 0, and ∫ε1x−1/2 dx=[2x]ε1=2−2ε\int_\varepsilon^1 x^{-1/2}\,dx = \big[2\sqrt x\big]_\varepsilon^1 = 2 - 2\sqrt\varepsilon, this becomes

∫ε1x−1/2ln⁡x dx=−2εln⁡ε−2(2−2ε)=−2εln⁡ε−4+4ε.\int_\varepsilon^1 x^{-1/2}\ln x\,dx = -2\sqrt\varepsilon \ln\varepsilon - 2(2 - 2\sqrt\varepsilon) = -2\sqrt\varepsilon\ln\varepsilon - 4 + 4\sqrt\varepsilon.

Now let ε→0+\varepsilon \to 0^+. We need lim⁡ε→0+εln⁡ε\lim_{\varepsilon\to 0^+} \sqrt\varepsilon \ln \varepsilon. Substituting ε=e−u\varepsilon = e^{-u} with u→∞u \to \infty as ε→0+\varepsilon\to 0^+: εln⁡ε=e−u/2(−u)=−ue−u/2→0\sqrt\varepsilon \ln\varepsilon = e^{-u/2}(-u) = -u e^{-u/2} \to 0 as u→∞u \to \infty (exponential decay dominates linear growth — a standard limit, provable via L'Hôpital or the known fact uk/ecu→0u^k/e^{cu} \to 0 for c>0c>0). Also 4ε→04\sqrt\varepsilon \to 0. Therefore

lim⁡ε→0+∫ε1x−1/2ln⁡x dx=−2⋅0−4+0=−4.\lim_{\varepsilon\to 0^+}\int_\varepsilon^1 x^{-1/2}\ln x\,dx = -2\cdot 0 - 4 + 0 = -4.

So the improper integral converges:

∫01ln⁡xx dx=−4.\int_0^1 \frac{\ln x}{\sqrt x}\,dx = -4.

(The negative sign is consistent with f≤0f \le 0 throughout the domain of integration.)

CounterexampleWhy nonnegativity cannot be dropped from the Comparison Test

The Comparison Test's hypothesis 0≤f(x)≤g(x)0 \le f(x) \le g(x) is essential — not merely for convenience but for the truth of the conclusion. Consider

f(x)=sin⁡xx,g(x)=1x,x≥1.f(x) = \frac{\sin x}{x}, \qquad g(x) = \frac{1}{x}, \qquad x \ge 1.

Here ∣f(x)∣≤g(x)|f(x)| \le g(x) for all x≥1x \ge 1 (since ∣sin⁡x∣≤1|\sin x|\le 1), but ff takes both signs, so 0≤f≤g0 \le f \le g fails. If one incorrectly tried to run the comparison argument on ff itself (rather than on ∣f∣|f|), one might hope to conclude something about ∫1∞g dx=∫1∞x−1dx\int_1^\infty g\,dx = \int_1^\infty x^{-1}dx, which diverges (it is the p=1p=1 boundary case, shown to diverge in the pp-integral theorem). A naive "comparison" gives no information here because the hypothesis doesn't apply — and indeed the true behaviors of ff and gg are opposite: it is a classical fact (provable via integration by parts and the Dirichlet test for improper integrals) that

∫1∞sin⁡xx dxconverges (conditionally),\int_1^\infty \frac{\sin x}{x}\,dx \quad \text{converges (conditionally)},

even though g(x)=1/xg(x) = 1/x produces a divergent integral. So knowing ∣f∣≤g|f|\le g with gg's integral divergent tells us nothing about ff: the Comparison Test only transmits information from a convergent bounding integral downward, or from a divergent nonnegative integral upward — and both directions strictly require the nonnegativity hypothesis f≥0f \ge 0 to make the monotonicity argument (Step 1 of the proof) valid. Without it, F(t)=∫1tfF(t) = \int_1^t f need not be monotone, and the entire "bounded monotone function converges" argument collapses. This also shows why ∫1∞sin⁡xxdx\int_1^\infty \frac{\sin x}{x}dx converges but ∫1∞∣sin⁡xx∣dx\int_1^\infty \left|\frac{\sin x}{x}\right|dx diverges — it is a genuine example of conditional but not absolute convergence, confirming that the converse of the Absolute Convergence Theorem is false.

Scope, edge cases & common traps

SituationWhat to doCommon error
Both endpoints improper (∫−∞∞\int_{-\infty}^\infty, or ∫0∞\int_0^\infty with singularity at 00)Split at an interior point; both pieces must converge independentlyTreating it as one limit lim⁡t→∞∫−tt\lim_{t\to\infty}\int_{-t}^t (the symmetric/Cauchy principal value), which can exist even when the true improper integral diverges
Integrand changes signComparison Test needs f≥0f\ge 0; use ∣f∣|f| for absolute convergence, or a specialized test (Dirichlet/Abel) for conditional convergenceApplying Direct Comparison directly to a sign-changing ff
Singularity exactly at an endpoint of a finite domainStill genuinely improper — Riemann integral is undefined if ff is unboundedAssuming "the interval is bounded" alone is enough for properness
p=1p=1 boundary casesAlways diverges (logarithmic growth) — check separately, don't extrapolate from p≠1p\ne 1 formulaPlugging p=1p=1 into 1p−1\frac{1}{p-1} or 11−p\frac{1}{1-p} (division by zero, undefined)
Limit Comparison with L=0L=0 or L=∞L=\inftyTest as stated doesn't apply; one-directional comparison may still workAssuming the "iff" conclusion holds when L∉(0,∞)L \notin (0,\infty)

The mathematician behind it

Origins · 1 minBernhard Riemann and how this was discovered
BR
Bernhard Riemann
1826–1866 · Göttingen, mid-19th-century analysis

Riemann formalized the modern integral in his 1854 habilitation, giving the definition later called the Riemann Integral, built for bounded functions on bounded intervals via limits of Riemann sums. He did not himself develop "improper integrals" as a separate named theory — that systematic extension, treating unbounded domains and unbounded integrands via limiting truncations, crystallized over the following decades as mathematicians (including Cauchy before Riemann, and later analysts refining rigor) needed a careful, limit-based way to assign meaning to integrals like the Gamma function or the Dirichlet integral that fall outside the strict Riemann framework. The improper integral is thus a direct descendant of Riemann's definition: it is precisely the limiting procedure one is forced into once the underlying integral is required to be a genuine, finite Riemann integral on every safely truncated sub-interval.

History
- 1823 — Cauchy gives an early rigorous treatment of definite integrals and limiting integrals over infinite intervals, in the spirit of what become improper integrals.
- 1829 — Dirichlet studies convergence conditions for integrals and series, prefiguring tests like the one now bearing his name for conditionally convergent integrals.
- 1854 — Riemann's habilitation lecture formalizes the integral of a bounded function via Riemann sums, the base object that improper integrals extend.
- 1875–1890s — Systematic treatment of improper integrals (convergence tests, comparison principles) becomes standard in analysis texts as the Riemann integral's limitations near unbounded domains/integrands are worked out precisely.
- 1902 — Lebesgue's thesis introduces a strictly more general integral, in which many (but not all) improper Riemann integrals become honest Lebesgue integrals — clarifying exactly which improper integrals are "absolutely convergent" (Lebesgue integrable) versus merely "conditionally convergent" (a phenomenon invisible to the Lebesgue theory).
Why can't we just use the ordinary Riemann integral definition for unbounded intervals or integrands?

The Riemann integral is defined via Riemann sums over a partition of a bounded interval, requiring the integrand to be bounded on that interval — both are needed for the upper and lower sums to make sense and for their infimum/supremum to coincide. If the interval is unbounded, there is no way to partition it into finitely many subintervals. If the integrand is unbounded, the upper Riemann sums become infinite regardless of partition, so the standard supremum/infimum construction breaks down entirely. The improper integral sidesteps this by never actually asking the Riemann integral to handle the bad behavior directly: it only ever integrates over safely bounded, safely non-singular sub-domains, and studies the resulting limit as a separate, purely limit-theoretic question.

Is every improper integral that "converges" actually equal to what a calculator or symbolic solver reports?

Not automatically — you must verify convergence before trusting a formally computed antiderivative evaluation. A classic trap: naively writing ∫−11x−2dx=[−x−1]−11=−1−1=−2\int_{-1}^1 x^{-2}dx = [-x^{-1}]_{-1}^1 = -1-1 = -2 is simply wrong, because x−2x^{-2} has an unbounded (in fact non-integrable) singularity at x=0∈(−1,1)x=0 \in (-1,1); splitting at 00, both ∫−10x−2dx\int_{-1}^0 x^{-2}dx and ∫01x−2dx\int_0^1 x^{-2}dx individually diverge to +∞+\infty (this is the p=2>1p=2>1 case of the Type II pp-integral, which diverges since 2≮12 \not< 1), so the correct answer is "diverges," not −2-2. The Second Fundamental Theorem of Calculus applies to proper Riemann integrals; using it blindly across a singularity gives nonsense.

What is the difference between "diverges" and "diverges to infinity"?

"Diverges" is the general statement that the defining limit fails to exist as a finite real number. "Diverges to +∞+\infty" (or −∞-\infty) is a specific, stronger kind of divergence, meaning the limit exists in the extended sense, growing without bound. There is a third possibility: the limit may simply oscillate and fail to approach anything, finite or infinite — for instance ∫0tcos⁡x dx=sin⁡t\int_0^t \cos x\,dx = \sin t, which oscillates between −1-1 and 11 forever as t→∞t\to\infty, never settling and never blowing up. So ∫0∞cos⁡x dx\int_0^\infty \cos x\,dx diverges, but not "to ∞\infty" — it diverges by oscillation.

Does the Comparison Test require knowing an antiderivative of gg?

No — and this is precisely its practical value. You only need to know, independently, whether ∫a∞g(x) dx\int_a^\infty g(x)\,dx converges or diverges, which is typically established once and for all for a small library of benchmark functions (the pp-integrals x−px^{-p} being the most important, along with e−xe^{-x}, e−x2e^{-x^2}, and a handful of others). Once that library is in hand, the Comparison Test lets you decide convergence for a much wider class of complicated ff by bounding ff above or below by a suitable constant multiple of a benchmark gg, entirely avoiding the need to find an antiderivative of ff itself — which frequently does not exist in closed form.

Can an improper integral converge conditionally but not absolutely, the way series can?

Yes, and this is one of the deepest analogies between improper integrals and infinite series. The standard example is ∫1∞sin⁡xx dx\int_1^\infty \frac{\sin x}{x}\,dx, which converges (by the Dirichlet test for integrals, an integration-by-parts argument exploiting the bounded oscillation of sin⁡x\sin x's antiderivative and the monotone decay of 1/x1/x), while ∫1∞∣sin⁡xx∣dx\int_1^\infty \left|\frac{\sin x}{x}\right|dx diverges (comparable in size, on a positive-density set of intervals, to a divergent harmonic-type sum). This mirrors the alternating harmonic series ∑(−1)n/n\sum (-1)^n/n, which converges conditionally while ∑1/n\sum 1/n diverges. The parallel is not a coincidence — it is the same phenomenon underlying the Integral Test for Series.

How does this connect to the Integral Test for series?

The Integral Test compares a series ∑n=1∞an\sum_{n=1}^\infty a_n, with an=f(n)a_n = f(n) for a nonnegative, nonincreasing continuous ff, to the improper integral ∫1∞f(x) dx\int_1^\infty f(x)\,dx. Because ff is nonincreasing, one can sandwich f(n)f(n) between ∫nn+1f(x)dx\int_n^{n+1}f(x)dx and ∫n−1nf(x)dx\int_{n-1}^n f(x)dx, so that partial sums of the series are trapped between truncated improper integrals. This gives: the series converges if and only if the improper integral converges — literally the same convergence/divergence dichotomy, transported from the continuous to the discrete setting via exactly the comparison and monotonicity machinery built here.

Why does the value ∫1∞x−1dx\int_1^\infty x^{-1} dx diverge exactly at p=1p=1, and why does that number matter so much?

At p=1p=1 the antiderivative of x−px^{-p} changes character: for p≠1p\ne 1 it is a power x1−p/(1−p)x^{1-p}/(1-p), but at p=1p=1 it becomes ln⁡x\ln x, which grows to infinity more slowly than any positive power of tt but still without bound. This makes p=1p=1 the exact threshold — any p>1p>1 decays just fast enough for the tail to have finite area, while p=1p=1 and below do not. This threshold is exactly why the harmonic series ∑1/n\sum 1/n diverges while ∑1/n1+δ\sum 1/n^{1+\delta} converges for any δ>0\delta>0, a fact discovered independently via the Integral Test using precisely this integral.

Before this