Improper Riemann Integrals

Iniitally Riemann integrals are defined only for bounded functions defined on a set of compact intervals.

Type 1

A function that is not integrable at one endpoint of the interval.

Suppose f:(a,b]→Rf:(a,b]\to\mathbb{R} is integrable on [c,b]  ∀c∈(a,b)[c,b]\;\forall c\in (a,b).

∫abf=lim⁡ϵ→0  ∫a+ϵbf\int_a^b f = \lim_{\epsilon \to 0}\; \int_{a+\epsilon}^b f

Can be similarly defined on the other endpoint. The above integral converges iff the limit exists and finite. Otherwise diverges.

Type 2

A function defined on unbounded interval (including ∞\infty).

Suppose f:[a,∞)→Rf:[a,\infty)\to\mathbb{R} is integrable on [a,r]∀r>a[a,r] \forall r > a.

∫a∞f=lim⁡r→∞  ∫arf\int_a^\infty f = \lim_{r \to \infty}\; \int_a^r f

Can be similarly defined on the other endpoint. The above integral converges iff the limit exists and finite. Otherwise diverges.

Type 3

A function that is undefined at finite number of points. The integral can be split into multiple integrals of type 1. Similarly integrals from −∞-\infty to ∞\infty can be defined.

Convergence of improper integrals is similar to the convergence of series.

Absolute Convergence Test

∫ab∣f∣  converges  ⟹  ∫abf  converges\int_a^b \lvert f \rvert\;\text{converges} \implies \int_a^b f \;\text{converges}

Common Integrals

∫011xp dx                  ∫1∞1xp dx\int_0^1 \frac{1}{x^p}\,\text{d}x \;\;\; \;\;\; \;\;\; \int_1^\infty \frac{1}{x^p}\,\text{d}x

The above integrals converge iff pp is in the integrating (open) interval. Converges to 1p−1\frac{1}{p-1} in that case.

∫01sin⁡2xx2 dx                  ∫1∞sin⁡2xx2 dx\int_0^1 \frac{\sin^2 x}{x^2}\,\text{d}x \;\;\; \;\;\; \;\;\; \int_1^\infty \frac{\sin^2 x}{x^2}\,\text{d}x

Both of the above integrals converges. Direct comparison test can be used.

  • For the 1st integral, x−1\sqrt{x^{-1}} can be used
  • For the 2nd integral, x−2x^{-2} can be used
Written by September 29, 2026 2 min read
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