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Improper Integral Calculator

Evaluate improper integrals with infinite bounds or unbounded integrands using AI step-by-step solutions

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∑Math Input
integral from 0 to infinity of e^(-x) dx
integral from 1 to infinity of 1/x^2 dx
integral from 0 to 1 of 1/sqrt(x) dx
integral from -infinity to infinity of 1/(1+x^2) dx

What is an Improper Integral?

An improper integral is a definite integral where either:

  1. The interval is infinite: e.g., ∫1∞f(x) dx\int_1^\infty f(x)\,dx or ∫−∞∞f(x) dx\int_{-\infty}^\infty f(x)\,dx
  2. The integrand has a vertical asymptote inside or at an endpoint of the interval: e.g., ∫011x dx\int_0^1 \frac{1}{\sqrt{x}}\,dx

In both cases, the standard Riemann integral is undefined, but we can sometimes assign a finite value using limits.

If the limit exists and is finite, the improper integral converges. If the limit is infinite or doesn't exist, the integral diverges.

Improper integrals are central to probability (normalization constants), Laplace and Fourier transforms, and series convergence tests.

How to Evaluate Improper Integrals

Type 1: Infinite Interval

Replace infinity with a limit:

∫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

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

For both bounds infinite, split at any convenient point cc:

∫−∞∞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

Both pieces must converge independently — otherwise the whole integral diverges.

Type 2: Unbounded Integrand

If ff is unbounded at x=cx = c inside [a,b][a, b], split and take limits:

∫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

If the singularity is at x=ax = a:

∫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

The pp-Test

∫1∞1xp dxconverges if p>1, diverges if p≤1\int_1^\infty \frac{1}{x^p}\,dx \quad \text{converges if } p > 1, \text{ diverges if } p \leq 1

∫011xp dxconverges if p<1, diverges if p≥1\int_0^1 \frac{1}{x^p}\,dx \quad \text{converges if } p < 1, \text{ diverges if } p \geq 1

The critical exponent is p=1p = 1. Note the opposite convergence rules for the two cases.

Comparison Test

If 0≤f(x)≤g(x)0 \leq f(x) \leq g(x) on the interval:

  • ∫g\int g converges ⇒∫f\Rightarrow \int f converges
  • ∫f\int f diverges ⇒∫g\Rightarrow \int g diverges

Useful when the integral itself is hard but the bound is easy.

Common Mistakes to Avoid

  • Treating ∞\infty as a number: You cannot 'plug in' ∞\infty. You must use a limit.
  • Missing internal singularities: ∫−111x dx\int_{-1}^1 \frac{1}{x}\,dx has a singularity at 00 inside the interval. Naïvely evaluating gives 00 (wrong) — the integral actually diverges.
  • Adding piecewise improper integrals that 'cancel': ∫−∞∞x dx\int_{-\infty}^\infty x\,dx — both halves diverge, so the integral diverges. The 'principal value' is a different (weaker) notion.
  • Wrong pp-test direction: At ∞\infty, 1/xp1/x^p converges for p>1p > 1. At 00, it converges for p<1p < 1. These are opposite — memorize both.
  • Forgetting to verify convergence before integrating: A divergent improper integral doesn't have a value. Always check convergence first.

Examples

Step 1: Replace bound with limit: lim⁡t→∞∫0te−x dx\lim_{t \to \infty} \int_0^t e^{-x}\,dx
Step 2: Compute antiderivative: ∫e−x dx=−e−x+C\int e^{-x}\,dx = -e^{-x} + C
Step 3: Apply bounds: lim⁡t→∞[−e−x]0t=lim⁡t→∞(−e−t+1)\lim_{t \to \infty} \left[-e^{-x}\right]_0^t = \lim_{t \to \infty}(-e^{-t} + 1)
Step 4: As t→∞t \to \infty, e−t→0e^{-t} \to 0, so the limit equals 11
Answer: 11 (converges)

Step 1: Apply pp-test with p=1p = 1: ∫1∞1/xp dx\int_1^\infty 1/x^p\,dx converges iff p>1p > 1
Step 2: Here p=1p = 1, so the integral diverges
Step 3: Verify by limit: lim⁡t→∞[ln⁡x]1t=lim⁡t→∞ln⁡t=∞\lim_{t \to \infty} [\ln x]_1^t = \lim_{t \to \infty} \ln t = \infty
Answer: Diverges

Step 1: Singularity at x=0x = 0. Use pp-test at 00: 1/xp1/x^p converges iff p<1p < 1
Step 2: Here p=1/2<1p = 1/2 < 1, so it converges
Step 3: Compute: lim⁡t→0+∫t1x−1/2 dx=lim⁡t→0+[2x]t1\lim_{t \to 0^+} \int_t^1 x^{-1/2}\,dx = \lim_{t \to 0^+} [2\sqrt{x}]_t^1
Step 4: =lim⁡t→0+(2−2t)=2= \lim_{t \to 0^+} (2 - 2\sqrt{t}) = 2
Answer: 22 (converges)

Frequently Asked Questions

An improper integral converges if the limit defining it is finite. Otherwise it diverges, meaning the area under the curve is either infinite or undefined.

The p-test applies to integrals of the form ∫1/x^p over [1, ∞) or (0, 1]. It's most useful as a comparison: if your integrand behaves asymptotically like 1/x^p, you can determine convergence quickly.

An improper integral converges absolutely if ∫|f| converges. It converges conditionally if ∫f converges but ∫|f| diverges. Absolute convergence is strictly stronger.

Yes — the area can be infinite. ∫_1^∞ 1/x dx is the canonical example: the curve y = 1/x is everywhere positive over [1, ∞), yet the area underneath is infinite (diverges).

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