Improper Integral Calculator
An improper integral calculator computes integrals with infinite limits or discontinuities. It determines convergence/divergence and calculates values for applications in physics, engineering, and probability. Essential for solving real-world problems involving unbounded regions or singularities, it automates complex limit-based computations, ensuring accuracy and saving time.
Formula
For ∫a∞f(x)dx = limb→∞∫abf(x)dx. For discontinuities at c ∈ [a,b], split into limt→c⁻∫atf(x)dx + limt→c⁺∫tbf(x)dx.
How to Use
1. Enter the function (e.g., 1/x^2). 2. Specify limits (use 'inf' for ∞). 3. Define the integration variable. 4. Click "Compute". The tool evaluates convergence/divergence and provides the result. Interpret results carefully: finite values indicate convergence; ∞ or undefined implies divergence.
Derivation Process
Improper integrals extend Riemann integrals to infinite intervals or discontinuous integrands. Derived by replacing infinite limits with variables approaching infinity (Type I) or integrating near discontinuities (Type II). Convergence is determined if limits exist; otherwise, the integral diverges. Techniques include comparison tests and substitution.
Common Improper Integrals
Integral | Result | Converges? |
---|---|---|
∫₁∞1/x² dx | 1 | Yes |
∫₀11/√x dx | 2 | Yes |
∫₁∞1/x dx | ∞ | No |
FAQs
What is an improper integral?
An improper integral involves infinite limits or integrand discontinuities. It evaluates via limits to determine convergence. If the limit exists, it converges; otherwise, it diverges. Used in physics for unbounded regions.
How does this calculator handle infinity?
It substitutes infinity with a large number (1e8) and computes numerically. For discontinuities, it splits the integral and approaches the limit from both sides, checking for finite results.
Can it check convergence?
Yes. If the result is finite, it converges; if infinite or NaN, it diverges. However, always verify analytically for critical applications, as numerical methods can approximate but not prove convergence.