Improper integrals: infinite region, finite area
About this app
The limit z runs up to one million while the area A(z) is plotted live: under 1/x² it approaches 1, under 1/x it grows like ln z without bound. Free exponent p (convergent exactly for p > 1), side-by-side comparison, the pole case ∫₀¹ with ε → 0 and a three-step calculation instead of “plugging in ∞”.
With explain mode: an AI tutor walks you through the app, points out what matters in the scene and answers your questions.
The idea
An improper integral has an infinite limit or a pole. It is defined as a limit: ∫₁^∞ f(x) dx = lim (z → ∞) ∫₁^z f(x) dx. For f(x) = 1/x², ∫₁^z 1/x² dx = 1 − 1/z, which tends to 1. For 1/x the integral is ln z, which grows without bound, although 1/x also tends to 0. In general ∫₁^∞ 1/xᵖ dx is finite exactly for p > 1, with value 1/(p − 1).
What the app shows
At the top you see f(x) = 1/xᵖ with the area from 1 to the limit z and the remaining area beyond it. Below, the app plots the area A as a function of z, on a logarithmic or linear axis, with a line at the limit value. In comparison mode 1/x² and 1/x sit side by side. The mode “open upwards” treats ∫₀¹ with a pole at 0 and a lower limit ε. The “Calculation” panel shows the antiderivative, the value at the current limit and the limit.
Things to try
Start “z → ∞” and watch A for 1/x² settle against the line 1 while A for 1/x keeps rising. Switch the axis to linear: there ln z looks flat, on the logarithmic axis it becomes a straight line. Change the exponent p and find the threshold p = 1. The pole case shows the counterpart: ∫₀¹ 1/√x dx = 2 is finite, ∫₀¹ 1/x² dx is not.
Frequently asked questions
- What is an improper integral?
- An integral over an unbounded interval or across a pole. It is not computed directly but as a limit of ordinary integrals, for example ∫₁^∞ f(x) dx = lim (z → ∞) ∫₁^z f(x) dx. If this limit exists, the integral is called convergent, otherwise divergent.
- Why does ∫₁^∞ 1/x² dx equal 1?
- An antiderivative of 1/x² is −1/x, so ∫₁^z 1/x² dx = 1 − 1/z. As z → ∞, 1/z tends to 0, so the limit is 1. The region reaches infinitely far but has area 1.
- Why does ∫₁^∞ 1/x dx diverge although 1/x tends to 0?
- Because ∫₁^z 1/x dx = ln z, and ln z grows without bound, just very slowly: every tenfold increase of z adds about 2.3. It is not enough for f to tend to 0; it has to do so fast enough, which for 1/xᵖ means p > 1.
- What can I try in the app?
- You run the limit up to one million, choose the exponent p, compare 1/x² with 1/x, switch between logarithmic and linear axes, and explore integrals with a pole such as ∫₀¹ 1/√x dx in the “open upwards” mode.
Subject: Analysis (one-dimensional)
More from Analysis (one-dimensional)
- Derivative as a sequence of secants
- Bounded growth
- Units of an integral: rate times time
- Exponential growth and decay
- Optimisation: the largest rectangle
- Area between curves
- Area function is an antiderivative
- Sequential continuity
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