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Improper Integrals



What Makes an Integral Improper?
Infinite Limits of Integration
Improper Integral Notation
Discontinuous Integrands
Convergence vs Divergence
The p-Test
Comparison Test
Limit Comparison Test
Summary: Tests for Improper Integral Convergence
Improper Integrals FAQ



Integrating Beyond Bounds


Standard definite integrals require finite intervals and bounded integrands. Improper integrals remove these restrictions, extending integration to infinite intervals and functions with vertical asymptotes.

The key idea: replace the problematic bound or point with a variable, compute the resulting proper integral, then take a limit. If the limit exists and is finite, the improper integral converges to that value. If not, it diverges.

Some infinite regions have finite area. The region under 1/x21/x^2 from 11 to ∞\infty has area exactly 11. Other infinite regions have infinite area—the region under 1/x1/x from 11 to ∞\infty diverges. The distinction matters throughout mathematics and physics.

Key Terms

Improper Integral— infinite interval or unbounded integrand, evaluated as a limit
Definite Integral— improper integrals extend the definite integral framework
Bounds of Integration— one or both bounds may be ±∞\pm\infty
Limit— convergence or divergence determined by whether the limit exists

See All Calculus Definitions →


Infinite Limits of Integration


Replace the infinite limit with a finite variable and take a limit.

Type 1: Upper limit infinite

∫a∞f(x) dx=lim⁡b→∞∫abf(x) dx\int_a^{\infty} f(x)\, dx = \lim_{b \to \infty} \int_a^b f(x)\, dx


Type 2: Lower limit infinite

∫−∞bf(x) dx=lim⁡a→−∞∫abf(x) dx\int_{-\infty}^b f(x)\, dx = \lim_{a \to -\infty} \int_a^b f(x)\, dx


Type 3: Both limits infinite

∫−∞∞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


Improper Integral (Infinite Limits)
∫a∞f(x) dx=lim⁡b→∞∫abf(x) dx\int_a^{\infty} f(x)\, dx = \lim_{b \to \infty} \int_a^b f(x)\, dx
Learn more about this formula: Improper Integral (Infinite Limits) →


Both integrals must converge independently. The choice of cc is arbitrary—any finite value works.

Improper Integral Notation

Notation

Improper Integral Notation

A symbol that quietly means a limit, the scaffold that makes the limit visible, and the verdict vocabulary that comes with it.
∫ab\int_a^b and its parts — definite integrals; ∞\infty and lim⁡=∞\lim = \infty — limits at infinity; one-sided arrows — one-sided limits.
∫a∞f(x) dx\int_a^{\infty} f(x)\, dx
The integral from a to infinity
∞\infty in the bound is not a point — the whole symbol is defined as lim⁡b→∞∫abf(x) dx\lim_{b \to \infty} \int_a^b f(x)\,dx, the scaffold of Infinite Limits of Integration above. The compact form hides the limit; the definition is the limit.
CasesLower bound −∞-\infty mirrors it. Both bounds infinite — split at an arbitrary cc, and each half must converge on its own.
Also written∫a+∞\int_a^{+\infty}, with the sign spelled out — matching the signed-infinity tradition of limits at infinity.
Do not confuseThe symmetric limit lim⁡R→∞∫−RR\lim_{R \to \infty} \int_{-R}^{R} — Cauchy's principal value, written PV ⁣∫\mathrm{PV}\!\int. Not the same object: ∫−∞∞x dx\int_{-\infty}^{\infty} x\,dx diverges, while its principal value is 00. Tying the two ends to one variable changes the definition.
lim⁡t→b−∫atf(x) dx\lim\limits_{t \to b^{-}} \int_a^t f(x)\, dx
The limit, as t approaches b from below, of the integral from a to t
The scaffold for an unbounded integrand: the bad endpoint is replaced by a variable that approaches it from inside the interval — the one-sided superscript carries the direction. Nothing in ∫011x dx\int_0^1 \frac{1}{\sqrt{x}}\,dx announces impropriety; the integrand has to be inspected, then the scaffold written.
CasesAsymptote at the left endpoint — t→a+t \to a^{+}; at the right — t→b−t \to b^{-}; at an interior cc — two scaffolds with independent letters tt and ss, as in Discontinuous Integrands below.
Also writtenε\varepsilon-style: lim⁡ε→0+∫ab−ε\lim_{\varepsilon \to 0^{+}} \int_a^{b-\varepsilon}, common in analysis texts, where the small excision is named directly.
Do not confuseOne letter for both halves of an interior split. That ties the two limits together — a hidden principal value again — and can declare a divergent integral finite.
converges · diverges · ∫<∞\int < \infty
The integral converges; the integral diverges
Verdicts about the defining limit, as in Convergence vs Divergence below. Only after “converges” is the equation form ∫1∞1x2 dx=1\int_1^{\infty} \frac{1}{x^2}\,dx = 1 legitimate — the value is the limit.
CasesDivergence comes in two species: =∞= \infty, borrowing the equals-abuse of infinite limits, or by oscillation — ∫0∞sin⁡x dx\int_0^{\infty} \sin x\,dx — where no verdict value exists at all.
Also written∫a∞∣f∣<∞\int_a^{\infty} |f| < \infty — the analyst's shorthand for “converges absolutely”, an inequality doing the work of a sentence.
Do not confuse“Diverges” with “equals infinity”. Every =∞= \infty diverges; not every divergence is an ∞\infty — the oscillating case fails without ever growing.

Discontinuous Integrands


When ff has a vertical asymptote at cc within [a,b][a, b], split the integral and use limits.

Asymptote at left endpoint:

∫abf(x) dx=lim⁡t→a+∫tbf(x) dx\int_a^b f(x)\, dx = \lim_{t \to a^+} \int_t^b f(x)\, dx


Asymptote at right endpoint:

∫abf(x) dx=lim⁡t→b−∫atf(x) dx\int_a^b f(x)\, dx = \lim_{t \to b^-} \int_a^t f(x)\, dx


Asymptote at interior point:

∫abf(x) dx=lim⁡t→c−∫atf(x) dx+lim⁡s→c+∫sbf(x) dx\int_a^b f(x)\, dx = \lim_{t \to c^-} \int_a^t f(x)\, dx + \lim_{s \to c^+} \int_s^b f(x)\, dx


Improper Integral (Discontinuous Integrand)
∫abf(x) dx=lim⁡t→b−∫atf(x) dxwhen f has an asymptote at b\int_a^b f(x)\, dx = \lim_{t \to b^-} \int_a^t f(x)\, dx \quad \text{when } f \text{ has an asymptote at } b
Learn more about this formula: Improper Integral (Discontinuous Integrand) →


Both limits must exist independently.
Improper feature Setup as a limit Convergence requirement
Upper limit = ∞ ∫a∞ f(x) dx = limb → ∞ ∫ab f(x) dx single limit must exist and be finite
Lower limit = −∞ ∫−∞b f(x) dx = lima → −∞ ∫ab f(x) dx single limit must exist and be finite
Both limits infinite ∫−∞∞ f(x) dx = ∫−∞c f + ∫c∞ f  (any finite c) both pieces must converge independently
Asymptote at left endpoint a ∫ab f(x) dx = limt → a⁺ ∫tb f(x) dx single one-sided limit must exist and be finite
Asymptote at right endpoint b ∫ab f(x) dx = limt → b⁻ ∫at f(x) dx single one-sided limit must exist and be finite
Asymptote at interior point c split at c: limt → c⁻ ∫at f  +  lims → c⁺ ∫sb f both one-sided limits must converge independently

Convergence vs Divergence


An improper integral converges if the defining limit exists and is finite. It diverges if the limit is infinite or fails to exist.

Convergent example:

∫1∞1x2 dx=lim⁡b→∞[−1x]1b=lim⁡b→∞(−1b+1)=1\int_1^{\infty} \frac{1}{x^2}\, dx = \lim_{b \to \infty} \left[-\frac{1}{x}\right]_1^b = \lim_{b \to \infty} \left(-\frac{1}{b} + 1\right) = 1


Divergent example:

∫1∞1x dx=lim⁡b→∞[ln⁡x]1b=lim⁡b→∞ln⁡b=∞\int_1^{\infty} \frac{1}{x}\, dx = \lim_{b \to \infty} [\ln x]_1^b = \lim_{b \to \infty} \ln b = \infty


Convergence depends on how fast the integrand decays. The 1/x21/x^2 decays fast enough; 1/x1/x does not.

The p-Test


The integrals of 1/xp1/x^p serve as benchmarks.

At infinity:

∫1∞1xp dx{convergesp>1divergesp≤1\int_1^{\infty} \frac{1}{x^p}\, dx \quad \begin{cases} \text{converges} & p > 1 \\ \text{diverges} & p \leq 1 \end{cases}


At zero:

∫011xp dx{convergesp<1divergesp≥1\int_0^1 \frac{1}{x^p}\, dx \quad \begin{cases} \text{converges} & p < 1 \\ \text{diverges} & p \geq 1 \end{cases}


p-Test for Improper Integrals
∫1∞1xp dx{convergesp>1divergesp≤1\int_1^{\infty} \frac{1}{x^p}\, dx \quad \begin{cases} \text{converges} & p > 1 \\ \text{diverges} & p \leq 1 \end{cases}
Learn more about this formula: p-Test for Improper Integrals →


The boundary case p=1p = 1 always diverges—∫1/x dx=ln⁡∣x∣\int 1/x\, dx = \ln|x|, which is unbounded both as x→∞x \to \infty and as x→0+x \to 0^+.
Form p > 1 p = 1 p < 1
∫1∞ (1 / xp) dx converges diverges diverges
∫01 (1 / xp) dx diverges diverges converges

Comparison Test


Compare an unknown integral to one with known behavior.

Direct comparison: For f(x)≥0f(x) \geq 0 and g(x)≥0g(x) \geq 0:

If f(x)≤g(x)f(x) \leq g(x) and ∫g\int g converges, then ∫f\int f converges.

If f(x)≥g(x)f(x) \geq g(x) and ∫g\int g diverges, then ∫f\int f diverges.

Example: Does ∫1∞1x2+1 dx\int_1^{\infty} \frac{1}{x^2 + 1}\, dx converge?

Since 1x2+1<1x2\dfrac{1}{x^2 + 1} < \dfrac{1}{x^2} and ∫1∞1x2 dx\int_1^{\infty} \dfrac{1}{x^2}\, dx converges, the given integral converges by comparison.

Limit Comparison Test


When direct comparison is awkward, use limits.

For f(x)>0f(x) > 0 and g(x)>0g(x) > 0, if:

lim⁡x→∞f(x)g(x)=Lwhere 0<L<∞\lim_{x \to \infty} \frac{f(x)}{g(x)} = L \quad \text{where } 0 < L < \infty


then ∫f\int f and ∫g\int g both converge or both diverge.

Example: Does ∫1∞xx3+5 dx\int_1^{\infty} \frac{x}{x^3 + 5}\, dx converge?

Compare to g(x)=1/x2g(x) = 1/x^2:

lim⁡x→∞x/(x3+5)1/x2=lim⁡x→∞x3x3+5=1\lim_{x \to \infty} \frac{x/(x^3 + 5)}{1/x^2} = \lim_{x \to \infty} \frac{x^3}{x^3 + 5} = 1


Since ∫1∞1/x2 dx\int_1^{\infty} 1/x^2\, dx converges, so does the given integral.

Summary: Tests for Improper Integral Convergence


Determining whether an improper integral converges or diverges typically uses one of a small set of standard tests. The table below collects them in one place, pairing each test with the kind of integrand it suits, the conclusion it produces, and a worked benchmark from the sections above. Read the &quot;When to use&quot; column first when scanning an unfamiliar integral; the right row points directly to the test that settles convergence.
Test When to use Conclusion Example or benchmark
Direct evaluation an antiderivative is available exact value if the limit is finite; otherwise diverges ∫1∞ 1/x² dx = 1
p-test at ∞ integrand behaves like 1/xp as x → ∞ converges iff p > 1 ∫1∞ 1/xp dx (benchmark)
p-test near 0 integrand behaves like 1/xp as x → 0⁺ converges iff p < 1 ∫01 1/√x dx (p = 1/2)
Direct comparison f ≥ 0 and bounded above by convergent g, or below by divergent g f inherits the convergence behavior of the comparator 1/(x² + 1) < 1/x²  ⇒  ∫ 1/(x² + 1) converges
Limit comparison f, g > 0 and limx → ∞ f/g = L with 0 < L < ∞ ∫ f and ∫ g both converge or both diverge x/(x³ + 5) ~ 1/x²  ⇒  ∫ x/(x³ + 5) converges

Improper Integrals FAQ

How do you know an integral is improper if it looks normal?

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Inspect the integrand, because nothing in the notation warns you. ∫₀¹ (1/√x) dx looks like an ordinary definite integral, but the integrand blows up at x = 0, so it is improper and needs a limit. Applying the Fundamental Theorem straight to it happens to work here; on a divergent example it would produce a confident, wrong number.Read more →

Why must an integral from −∞ to ∞ be split into two separate limits?

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Because tying both ends to one variable changes the definition. Split at any finite c and require each half to converge on its own. The symmetric version, letting −R and R run out together, is Cauchy's principal value — a different object. The integral of x over the whole line diverges, yet its principal value is 0.Read more →

Does an integral that diverges always equal infinity?

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No. Divergence comes in two species. One is growth without bound, written = ∞ by the same equals-sign abuse used for infinite limits. The other is oscillation: ∫₀^∞ sin x dx never settles and never grows, so no value exists at all. Every integral equal to infinity diverges, but not every divergent integral is infinite.Read more →

Why do the two sides of an interior asymptote need separate limits?

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Because one variable for both sides secretly cancels the two infinities against each other. With an asymptote at c inside [a, b], write the left limit as t → c⁻ and the right as s → c⁺, using independent letters. Both must converge on their own; a shared letter can declare a divergent integral finite.Read more →

Why does the p-test flip between infinity and zero?

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Because the danger sits at opposite ends. Out at infinity, ∫₁^∞ (1/xᵖ) dx converges when p > 1 — the integrand has to decay fast enough. Near zero, ∫₀¹ (1/xᵖ) dx converges when p < 1 — the blow-up has to be mild enough. Larger p helps at infinity and hurts at zero. At p = 1 both diverge.Read more →