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What is an improper integral?

An improper integral occurs when the interval of integration is unbounded or the integrand has an infinite discontinuity on the interval. To evaluate, we rewrite it as a limit and check for convergence.

Improper integral formula

∫a∞f(x) dx=lim⁡R→∞∫aRf(x) dx\int_{a}^{\infty} f(x)\,dx = \lim_{R\to\infty}\int_{a}^{R} f(x)\,dx
∫abf(x) dx  where f has a singularity at c∈(a,b)=lim⁡t→c− ⁣∫atf(x) dx  +  lim⁡u→c+ ⁣∫ubf(x) dx\int_{a}^{b} f(x)\,dx \;\text{where }f\text{ has a singularity at }c\in(a,b) = \lim_{t\to c^-}\!\int_{a}^{t}f(x)\,dx \;+\; \lim_{u\to c^+}\!\int_{u}^{b}f(x)\,dx

How to solve an improper integral step by step

Step 1: Identify whether the issue is an infinite bound or an integrand singularity.

Step 2: Replace the problematic endpoint with a limit variable (e.g., R or t).

Step 3: Compute the integral with the variable bound.

Step 4: Evaluate the limit; if it exists, the integral converges, otherwise it diverges.

Example: ∫1∞1x2 dx\int_{1}^{\infty} \frac{1}{x^2}\,dx

∫1∞1x2 dx=lim⁡R→∞∫1Rx−2 dx=lim⁡R→∞[−x−1]1R=lim⁡R→∞(−R−1+1)=1 \int_{1}^{\infty} \frac{1}{x^2}\,dx = \lim_{R\to\infty} \int_{1}^{R} x^{-2}\,dx = \lim_{R\to\infty} \bigl[-x^{-1}\bigr]_{1}^{R} = \lim_{R\to\infty} \bigl(-R^{-1} + 1\bigr) = 1

Tips & Tricks

  • Use the p-test for ∫₁⁽∞⁾ 1/xᵖ dx: it converges if p > 1.
  • Apply the Comparison Test when f(x) is difficult to integrate directly.
  • For singularities, split at the discontinuity and evaluate each one‐sided limit separately.

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