Improper Integral Convergence Calculator

Determine whether an improper integral converges or diverges using our calculator. This tool helps you analyze integrals with infinite limits or singularities in the integrand.

What is an Improper Integral?

An improper integral is an integral that has one or more infinite limits of integration or a discontinuity within the interval of integration. These integrals are called "improper" because they don't fit the standard definition of an integral.

There are three types of improper integrals:

Integrals with infinite limits of integration
Integrals with a finite limit but an infinite discontinuity in the integrand
Integrals with a finite limit but an infinite discontinuity at the limit point

To evaluate an improper integral, we take a limit of a proper integral. If the limit exists and is finite, the integral converges. If the limit does not exist or is infinite, the integral diverges.

Convergence Tests

Several tests can determine whether an improper integral converges or diverges. The most common tests include:

Comparison Test

The comparison test compares the integrand to a known integrable function. If the integral of the known function converges, then the original integral may also converge.

If \( 0 \leq f(x) \leq g(x) \) for \( x \geq a \), and \( \int_a^\infty g(x) \, dx \) converges, then \( \int_a^\infty f(x) \, dx \) converges.

Limit Comparison Test

The limit comparison test compares the integrand to another function by taking the limit of their ratio.

If \( \lim_{x \to \infty} \frac{f(x)}{g(x)} = L \) where \( L > 0 \) and \( \int_a^\infty g(x) \, dx \) converges, then \( \int_a^\infty f(x) \, dx \) converges.

Integral Test

The integral test applies to positive, decreasing functions. If the integral of the function from 1 to ∞ converges, then the series converges.

If \( f \) is continuous, positive, and decreasing on \( [1, \infty) \), then \( \sum_{n=1}^\infty f(n) \) converges if and only if \( \int_1^\infty f(x) \, dx \) converges.

Ratio Test

The ratio test compares the terms of a series to determine convergence.

If \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = L \), then the series converges if \( L < 1 \) and diverges if \( L > 1 \).

Worked Examples

Example 1: Infinite Limit

Determine whether \( \int_1^\infty \frac{1}{x^2} \, dx \) converges.

First, evaluate the integral from 1 to b:

\( \int_1^b \frac{1}{x^2} \, dx = \left[ -\frac{1}{x} \right]_1^b = -\frac{1}{b} + 1 \)

Now take the limit as b approaches ∞:

\( \lim_{b \to \infty} \left( -\frac{1}{b} + 1 \right) = 1 \)

Since the limit is finite, the integral converges to 1.

Example 2: Discontinuous Integrand

Determine whether \( \int_0^1 \frac{1}{\sqrt{x}} \, dx \) converges.

First, evaluate the integral from a to 1:

\( \int_a^1 \frac{1}{\sqrt{x}} \, dx = 2\sqrt{x} \Big|_a^1 = 2(1 - \sqrt{a}) \)

Now take the limit as a approaches 0:

\( \lim_{a \to 0^+} 2(1 - \sqrt{a}) = 2 \)

Since the limit is finite, the integral converges to 2.

FAQ

What does it mean for an integral to converge?

An integral converges if the limit of the integral exists and is finite. This means the area under the curve is finite.

What does it mean for an integral to diverge?

An integral diverges if the limit of the integral does not exist or is infinite. This means the area under the curve is infinite.

How do I know which convergence test to use?

The choice of test depends on the form of the integrand. The comparison test is often the simplest to apply, while the limit comparison test is more flexible.

Can an integral converge if the integrand is always positive?

Yes, an integral can converge even if the integrand is always positive, as long as the area under the curve is finite.