How to Tell If an Integral Diverges or Converges: 5 Tests

An improper integral converges if its limit exists as a finite number, and diverges if that limit is infinite or fails to exist. To tell which, check the integrand’s behavior and match it to a known test: p-test, comparison, limit comparison, ratio, or geometric/integral test. If the limit of the partial integrals settles on a real, finite value, it converges; otherwise it diverges.

Convergence vs. Divergence

An improper integral is a definite integral with either an infinite limit of integration or an integrand that becomes unbounded somewhere on the interval. Because it involves infinity, you cannot just plug in endpoints — you must rewrite it as a limit and evaluate that limit.

Formally, for a function continuous on [a, ∞):

∫ₐ^∞ f(x) dx = lim_{b→∞} ∫ₐ^b f(x) dx

If this limit equals a finite number L, the integral converges to L. If the limit is ±∞ or does not exist, the integral diverges.

The distinction matters because a divergent integral has no meaningful finite area. In applications, a finite value is often required for the model to make sense. A probability density over an infinite range, for instance, must integrate to exactly 1; if it diverges, it isn’t a valid density at all. These limit-based definitions are standard in any calculus sequence; Paul’s Online Math Notes presents the improper integral limit definition the same way.

The 5 Convergence Tests

Most convergence questions reduce to matching the integrand against a known benchmark. Here are the five tests that cover the overwhelming majority of problems.

1. The p-test

For integrals of the form ∫₁^∞ 1/xᵖ dx:

• Converges if p > 1
• Diverges if p ≤ 1

So ∫₁^∞ 1/x² dx converges (p = 2), while the harmonic integral ∫₁^∞ 1/x dx diverges (p = 1). This is the single most useful benchmark, because other integrals are often compared to a p-integral.

2. Direct Comparison Test

If 0 ≤ f(x) ≤ g(x) for all x ≥ a:

• If ∫ g converges, then ∫ f converges.
• If ∫ f diverges, then ∫ g diverges.

You bound a messy integrand by a simpler one whose behavior you already know. The inequalities only need to hold eventually (for large x), not everywhere.

3. Limit Comparison Test

Compute lim_{x→∞} f(x)/g(x) = c, where g is a known benchmark. If 0 < c < ∞, then f and g either both converge or both diverge. This is often easier than direct comparison because you don’t need to prove an inequality — just evaluate one limit. MIT OpenCourseWare covers both comparison tests together, since they’re typically applied in tandem.

4. Ratio Test

Mostly used for series, but in the integral setting it applies to integrands with factorials or exponentials. For a term aₙ, compute lim_{n→∞} |a_{n+1}/aₙ| = r. If r < 1, it converges; if r > 1, it diverges; if r = 1, the test is inconclusive and you switch methods.

5. Geometric / Integral Test

Geometric-type integrals ∫ e^{-kx} dx (k > 0) over [a, ∞) converge, because the exponential decays faster than any power. The integral test links series to integrals: for a positive, decreasing, continuous f, the series Σf(n) and the integral ∫ f(x) dx converge or diverge together. This lets you reuse integral results to settle series. Khan Academy walks through this connection with worked series examples.

Choosing the Right Test

Matching the form of the integrand to a test saves time and avoids dead ends. Use this as a decision guide:

• See 1/xᵖ? Go straight to the p-test.
• See a rational function (polynomial over polynomial)? Compare its leading behavior to a p-integral, usually via limit comparison.
• See a bounded function times something you recognize? Try direct comparison.
• See factorials, n!, or xⁿ with variable exponent? Try the ratio test.
• See e^{-kx}, or need to settle a series instead? Use the geometric/integral test.

When in doubt, start with limit comparison against the dominant term. It rarely fails and requires the least algebraic work. If the limit comes out 0 or ∞, reach for direct comparison instead.

Worked Examples

Example 1: ∫₁^∞ 1/x³ dx. Here p = 3 > 1, so by the p-test it converges. Evaluating gives 1/(p−1) = 1/2.

Example 2: ∫₁^∞ (x + 1)/(x³ + 2) dx. Compare to 1/x². Limit comparison: lim (x+1)/(x³+2) · x² = lim (x³+x²)/(x³+2) = 1. Since 1/x² converges (p = 2) and the limit is a finite positive number, the original integral converges.

Example 3: ∫₂^∞ 1/(x ln x) dx. Using the integral test pattern, substitute u = ln x, du = dx/x. The integral becomes ∫ 1/u du = ln(ln x), which grows without bound. It diverges — a classic case where the p-test alone doesn’t apply.

Absolute vs. Conditional Convergence

When an integrand changes sign, a new subtlety appears. An integral ∫ f(x) dx is absolutely convergent if ∫ |f(x)| dx converges. It is conditionally convergent if ∫ f(x) dx converges but ∫ |f(x)| dx diverges.

The key fact: absolute convergence implies convergence. If you can show ∫ |f(x)| dx is finite, the original integral converges automatically. This gives you license to ignore sign changes and test the absolute value instead.

A classic illustration is ∫₁^∞ sin(x)/x dx. It converges conditionally by an integration-by-parts argument, yet ∫₁^∞ |sin x|/x dx diverges. Sign changes can rescue an integral that would otherwise diverge.

Common Pitfalls to Avoid

Even experienced students lose points on a few recurring mistakes.

• Forgetting the limit. You cannot evaluate an improper integral by naively substituting ∞. The limit notation is the whole method, not decoration.
• Splitting at a discontinuity incorrectly. If a singularity sits inside the interval, split the integral at that point and test each piece separately. If either piece diverges, the whole thing diverges.
• Misapplying comparison when the limit is 0 or ∞. The limit comparison test requires 0 < c < ∞. If c = 0, convergence of the benchmark tells you nothing about your integral.
• Assuming a small integrand guarantees convergence. The function 1/(x ln x) is smaller than 1/x for large x, but both diverge. Shrinking the integrand is not enough — you need it to shrink fast enough.

Key Takeaways

• An improper integral converges to a finite limit and diverges otherwise; always rewrite it as a limit before judging.
• The p-test is the fastest check: ∫₁^∞ 1/xᵖ dx converges only when p > 1.
• Direct comparison uses inequalities with a known benchmark; limit comparison uses a ratio to a benchmark and works when the limit is finite and positive.
• Use the ratio test for factorials and exponentials, and the geometric/integral test for decaying exponentials or to connect series to integrals.
• Match the integrand’s form to the test — starting with limit comparison against the dominant term is usually the most efficient path.