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Improper Integrals



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



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.

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 $c$:

∫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