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Improper Integrals Online Calculator

Improper-integral convergence checker. Enter p to test ∫₁^∞ 1/x^p dx and compute its value when convergent—with p-series conclusion.

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∫₁^∞ 1/x^p dx = {1/(p−1), p>1; diverges, p≤1}
Exponent p of ∫₁^∞ 1/x^p dx

📖 Tutorial | Improper Integrals

1. Definition
An improper integral has an infinite limit or an unbounded integrand. For the p-integral ∫₁^∞ 1/x^p dx: it converges to 1/(p−1) if p>1, and diverges if p≤1.
2. Symbols
SymbolsMeaning
∞Infinite limit
pPower exponent
ConvergentLimit is finite
DivergentLimit is infinite/nonexistent
3. How it works
  • Applies the analytic p-integral result;
  • For p>1 returns 1/(p−1);
  • p=1 is the logarithmic divergence; p<1 decays slower and also diverges.
4. Steps
  1. Enter p;
  2. Click Calculate;
  3. Read the convergence verdict and value.
5. Example
Example: p=2 gives ∫₁^∞ 1/x² dx = 1. p=1 gives ln x→∞, divergent.
6. Pitfalls
p=1 diverges logarithmically; it is not a finite boundary value;
Lowering the lower limit to 0 is another type (unbounded point);
Convergent does not mean small, only finite.

❓ FAQ | Improper Integrals

When does an improper integral converge?
For ∫₁^∞ 1/x^p: p>1 converges, p≤1 diverges.
Why does p=1 diverge?
∫1/x dx=ln x, which tends to ∞, not finite.
What is the convergent value?
1/(p−1) for p>1.
Does it only handle 1/x^p?
Yes; it is the classic p-test. General improper integrals need separate limit analysis.
What if the lower limit is 0?
That is the unbounded-integrand type, discussed separately.