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Improper Integrals - Type I

Calculus 2 · Chapter 2: Indeterminate Forms & Improper IntegralsSection 2.2 · lesson 9 of 5412 min · free

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In this lesson

Infinite intervals, and when they still enclose finite area.

  • Define an infinite-interval integral as a limit of ordinary ones
  • Classify integrals as convergent or divergent
  • Require each half of a doubly infinite integral to converge on its own
  • Use the p-integral as the benchmark for every later test

Transcript

Everything said in the video, section by section (1,720 words). Read along, or search it.

MAT-265 - Calculus 2 · Section 2.2

Improper Integrals - Type I

An engineer designing a radio antenna models the total energy radiated as an integral that runs forever. From the moment of transmission, out to t equals infinity. And intuition rebels at this. How can a region stretching infinitely far have any finite area at all? Yet some do. The decaying signal e to the minus t traces a curve that never touches the axis, extends without end — and still encloses an area of exactly one. Other curves, which look equally innocent, enclose infinite area. The difference between those two outcomes is the central question of this section. And it determines whether a physical quantity — radiated energy, total probability, accumulated cost — is actually meaningful.

The Question

An Infinite Shape With a Finite Area

Let's be precise about why the usual machinery breaks here. An ordinary definite integral is defined on a finite, closed interval. From a to b, both real numbers. An improper integral of Type One relaxes that restriction. At least one limit of integration is infinite. And because the area-under-a-curve machinery requires finite endpoints, such an integral cannot be evaluated directly. You can't just plug infinity in. Infinity isn't a number. So instead, we define it. As the limit of ordinary integrals over progressively larger finite intervals. That's the whole idea, and everything in this section follows from it.

Definition

Integrate to a Movable Boundary, Then Let It Run

Here's the strategy. Integrate over a finite interval ending at a movable boundary — call it t. Then let that boundary march off to infinity, and ask whether the accumulating area settles on a finite number. There are three forms, depending on which endpoint is infinite. Infinite upper limit. The integral from a to infinity is the limit, as t goes to infinity, of the integral from a to t. Infinite lower limit. Same idea, running the other direction. And doubly infinite. Split at any convenient real number c, and add the two pieces. In each case, if the limit exists as a finite number, the integral converges to that value. Otherwise it diverges. Now — for the doubly infinite case, there's a critical rule, and students break it constantly. The two halves must converge independently. It is not enough for the combined expression to balance out. If either half diverges, the whole integral diverges. Regardless of what the other half does. Regardless of any apparent cancellation between them. Hold onto that. We'll come back to why it matters.

Convergent vs Divergent

Two Curves That Look Alike

Every Type One evaluation follows the same three-step rhythm. Replace the infinity with a finite variable. Integrate normally. Then take the limit. And the value of that limit classifies the integral. Look at these two curves. On the left, e to the minus x. On the right, one over x. Both decay toward zero. Both never touch the axis. Visually, they're cousins. For e to the minus x, replace the upper limit with t and integrate. You get one minus e to the minus t. As t goes to infinity, e to the minus t goes to zero. So the limit is one. The integral converges. That infinitely long region encloses exactly one unit of area. Now one over x. Same setup. The antiderivative is log absolute x. You get log t minus log one. And log t goes to infinity. The integral diverges. Strikingly, this curve looks much like the first one. Both decay toward zero. But one over x decays too slowly. And its area is unbounded. Same shape. Opposite fate. That's the whole tension of this section in one picture.

Worked Examples

A Doubly Infinite Integral, and One Needing L'Hopital

Two more examples, each teaching something specific. First, a doubly infinite integral. The integral of one over one plus x squared, from minus infinity to infinity. Split at c equals zero into two independent improper integrals. The antiderivative is arctangent x. For the upper half, you get the limit of arctangent t minus zero, which is pi over two. For the lower half, by symmetry, also pi over two. Both halves converge. So the full integral converges to their sum. Pi. Second example, and this one shows you why the last section mattered. The integral from zero to infinity of x e to the minus x. This gives the average lifetime of a randomly decaying particle. Integrate by parts over zero to t, with u equals x and d v equals e to the minus x d x. You get minus t e to the minus t, minus e to the minus t, plus one. Now take the limit. The second term goes to zero directly. But the first term, t e to the minus t, is the indeterminate product infinity times zero. Rewrite it as t over e to the t. That's infinity over infinity. Apply L'Hopital's Rule and it goes to zero. So the integral converges to one — and evaluating it relied directly on the technique from section two point one. Now here's something to think about. Consider the integral of x, from minus infinity to infinity. If you evaluate it as a single symmetric limit — from minus t to t — every positive contribution is cancelled by a negative one, and you get zero. But the definition demands that each half converge on its own. And neither does. So ask yourself: why does the definition insist on independent convergence? And what goes wrong, physically, if we accept the convenient cancellation answer of zero?

The Benchmark

The p-Integral

There's one family of improper integrals that appears so often, and serves as the benchmark for so many later tests, that its behavior deserves to be established once and carefully. The p-integral. The integral from one to infinity of one over x to the p, with p positive. Its convergence hinges entirely on the size of p. And the derivation splits into two cases, because the antiderivative of x to the minus p takes a different form when p equals one. Case one: p not equal to one. The power rule gives x to the one minus p, over one minus p. Evaluate over one to t, take the limit, and everything now depends on the exponent one minus p. If p is greater than one, then one minus p is negative. So t to the one minus p is one over t to the p minus one, which goes to zero. The limit becomes one over p minus one. A finite number. The integral converges. If p is between zero and one, then one minus p is positive. So t to the one minus p goes to infinity. The integral diverges. Case two: p equals one. Here the antiderivative isn't a power at all. It's log x. And that's exactly the integral from Example two, which we already showed diverges.

Key Result

The p-Integral Test

Collecting all three sub-cases gives the complete result. And this is worth memorizing. The p-integral from one to infinity converges if and only if p is greater than one. And when it converges, its value is one over p minus one. In every other case — p equal to one, or p less than one — it diverges. This is the canonical reference for the threshold at which one-over-x-to-the-p decay is fast enough to enclose finite area. And later chapters, including the series p-test in Chapter Seven, cite this directly rather than re-deriving it. So let's use it. Classify three integrals without doing any work. One over x squared: here p is two, which is greater than one. So it converges. And its value is one over two minus one, which is one. One over root x: rewrite as x to the minus one half, so p is one half, which is at most one. It diverges. And one over x to the one point zero zero one: here p is one point zero zero one, just barely greater than one. So it converges. To one over zero point zero zero one. Which is one thousand. That razor-thin margin above p equals one is enough. Barely — but enough.

Why p = 1

The Tipping Point, and Two Pitfalls

So why should p equals one be the magic value, rather than two, or one half? The answer lies in the antiderivative, and it's genuinely elegant. For every p not equal to one, integrating x to the minus p produces another power of x. And a power either shrinks to zero or grows without bound, depending purely on the sign of its exponent. Only at p equals one does the antiderivative change species entirely. It becomes the logarithm. And the logarithm is a strange animal. It grows to infinity — but so slowly that no power of x can match its sluggishness. That qualitative shift, from power to logarithm, is what makes p equals one the precise tipping point. The logarithm grows without bound, which forces divergence. Yet it grows slower than any positive power — which is why pushing p even slightly above one restores convergence. Now, two pitfalls. The most frequent error in Type One problems is treating infinity as a number to be substituted directly into the antiderivative. Writing something like the bracket of minus e to the minus x, evaluated from zero to infinity, and plugging in infinity. That may stumble onto the right number for simple cases. But it is not valid mathematics, and it fails the moment a limit is subtle. The limit notation is not decoration. It is the definition. The second pitfall is specific to doubly infinite integrals. Collapsing both endpoints into a single symmetric limit. As the think-about-it prompt showed, that can manufacture a finite answer for an integral that genuinely diverges.

Key Takeaways

What to Carry Forward

Five things to carry forward. Type One integrals are defined as limits of ordinary definite integrals over finite intervals, with the boundary sent to plus or minus infinity. Convergence means the defining limit is finite. Divergence means it's infinite, or doesn't exist. Doubly infinite integrals split into two halves that must each converge independently. Symmetric cancellation is not allowed. The p-integral from one to infinity converges if and only if p is greater than one, with value one over p minus one. That's the canonical reference cited by later convergence tests. Learn it now and you'll use it for the rest of the course. And never substitute infinity directly. Always pass through the finite variable and take the limit. Next up is section two point three, Improper Integrals of Type Two — where the interval is finite, but the integrand blows up somewhere inside it.

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FAQ

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About Improper Integrals - Type I

What is a Type I improper integral?
A Type I improper integral is one where at least one limit of integration is infinite, and it is defined as a limit of ordinary definite integrals. For example, \int_a^\infty f(x)\,dx = \lim_{t\to\infty}\int_a^t f(x)\,dx. If that limit is a finite number, the integral converges to it. If the limit is infinite or does not exist, the integral diverges. So \int_0^\infty e^{-x}\,dx = 1 converges, while \int_1^\infty \frac{1}{x}\,dx diverges.
How do you evaluate an integral from negative infinity to infinity?
Split it at any convenient real number c and evaluate each half as its own limit. The integral converges only if both halves converge independently. For \int_{-\infty}^{\infty} \frac{1}{1+x^2}\,dx, splitting at 0 and using \arctan x gives \frac{\pi}{2} for each half, so the total is \pi. A single symmetric limit is not allowed. For example, \int_{-t}^{t} x\,dx = 0, yet \int_{-\infty}^{\infty} x\,dx diverges because neither half converges.
When does the p-integral converge?
The p-integral \int_1^\infty \frac{1}{x^p}\,dx converges if and only if p > 1, and then its value is \frac{1}{p-1}. For p \le 1 it diverges. At p = 1 the antiderivative is \ln x, which grows without bound. So \int_1^\infty \frac{1}{x^2}\,dx = 1, \int_1^\infty \frac{1}{\sqrt{x}}\,dx diverges, and \int_1^\infty x^{-1.001}\,dx = 1000. This result is the benchmark for later convergence tests, including the series p-test.
Can you just plug infinity into the antiderivative of an improper integral?
No, substituting infinity directly into an antiderivative is not valid, because infinity is not a number. Instead, replace the infinite limit with a variable t, integrate over the finite interval, and then take the limit. For \int_0^\infty xe^{-x}\,dx, integration by parts gives -te^{-t} - e^{-t} + 1. The term te^{-t} is an indeterminate \infty\cdot 0 form. Rewriting it as \frac{t}{e^t} and applying L'Hôpital's Rule shows it goes to 0, so the integral equals 1.

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