Improper Integral Divergence Calculator

Decide convergence for common improper integral patterns today. Analyze endpoints and tails clearly. Export results for reports and assignments quickly.

Calculator inputs

Pick a model, enter parameters, and get a decision.
CSV + PDF export Session history Advanced tests

These templates cover the most-used tests.
Tip: compare with a simple benchmark near 0 or infinity.
Output includes:
  • Convergent or divergent decision
  • Short test-based explanation
  • Value for select closed-form cases
  • Saved history with exports

Example data table

Sample inputs with expected decisions.
Model Parameters Expected Reason
Integral from 1 to infinity of 1/x^p dx p=2 Convergent Tail p-test, p>1.
Integral from 1 to infinity of 1/x^p dx p=1 Divergent Harmonic tail, logarithmic blow-up.
Integral from 0 to 1 of 1/x^p dx p=0.5 Convergent Endpoint p-test, p<1.
Integral from e to infinity of 1/(x (ln x)^q) dx q=2 Convergent Log-test: q>1.
Integral from 0 to 1 of 1/(x (ln(1/x))^q) dx q=3 Divergent Becomes integral t^{-q} dt on (0, infinity).
Integral from 0 to infinity of x^2 e^{-x} dx n=2, k=1 Convergent Exponential dominates polynomial.

Formula used

This calculator applies standard tests for improper integrals.

How to use this calculator

  1. Select the model that matches your integral form.
  2. Enter the parameters shown in the form.
  3. Click Submit to compute the decision.
  4. Read the notes to understand the applied test.
  5. Export saved rows using the CSV or PDF buttons.

Why improper integrals fail

Improper integrals diverge when the area near an endpoint or the tail cannot be bounded. In practice, divergence is driven by slow decay at infinity or strong blow‑ups near zero. This calculator focuses on benchmark families so you can classify behavior quickly and consistently.

Power tails and thresholds

For tail forms like integral from a to infinity of 1/x^p dx, the threshold is p=1. When p is 1.20, the computed finite value equals a^(1-p)/(p-1). When p is 0.80, partial areas grow without bound as the upper limit increases, signaling divergence.

Endpoint singularities

Near zero, integral from 0 to a of 1/x^p dx flips the rule: convergence requires p<1. Small changes matter. If a=1 and p=0.90, the finite result equals 1/(1-p)=10. If p=1.10, the integral diverges because the singularity is too strong. For p=0.50 and a=0.25, the value is 2*sqrt(a), illustrating how shrinking the interval can reduce the total area even when the curve spikes.

Log corrections in the tail

Log factors refine borderline cases. The model 1/(x (ln x)^q) converges only when q>1. With q=2, the tail area shrinks fast enough; with q=1, it behaves like 1/(x ln x) and diverges. The calculator reports this decision without requiring symbolic integration. Treat q=1.01 as different from q=0.99, because the test is strict.

Log structure near zero

For 1/(x^p (ln(1/x))^q) on (0,1), the exponential substitution shows that p controls convergence first. When p<1, the integrand becomes integrable after transformation even if q is small. When p=1, the transformed integral is t^{-q} over (0,infinity), which always diverges. A practical workflow is to test p, then use q to judge how quickly the logarithmic factor changes the curve’s height near the endpoint.

Exponential dominance and safe decay

Exponential tails are typically safe. For x^n e^{-k x} from a to infinity, any k>0 forces rapid decay, so polynomial growth cannot break convergence. When n is a small integer, the calculator also estimates the finite value using a stable recurrence, which is useful for quick checks. Increasing k from 1 to 2 roughly halves the decay length, so the plotted curve collapses faster and the area shrinks. If k=0 or negative, the tool flags divergence because the integrand fails to decay.

FAQs

1) What makes an integral “improper”?

It is improper when a limit is infinite or the integrand is unbounded at an endpoint or interior point. The integral is defined using a limit process instead of a direct area calculation.

2) Does “divergent” mean the function has no antiderivative?

No. Divergence means the limit defining the integral does not produce a finite number. Many divergent improper integrals still have antiderivatives; the issue is the boundary behavior.

3) Why is p=1 special in power models?

For tails, 1/x is the boundary between enough decay and too little decay. For endpoint singularities, 1/x is exactly strong enough to cause logarithmic blow‑up at zero.

4) When do log factors change the decision?

Log factors matter mainly in borderline cases, such as 1/(x (ln x)^q). The decision switches at q>1, because the added logarithmic decay becomes integrable.

5) Is the Plotly graph a proof of convergence?

No. The graph visualizes the integrand shape and decay, but convergence depends on limits and comparison arguments. Use the plotted behavior to build intuition, then rely on the stated tests.

6) Can I export multiple runs?

Yes. Each calculation is stored in the session table. Use the CSV or PDF export buttons to download the saved rows for sharing, reporting, or practice problem documentation.

Saved results

Latest 20 rows shown. Export for all session rows.
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Timestamp Model Parameters Decision Notes Value
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