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Improper Integrals - Type II

Calculus 2 · Chapter 2: Indeterminate Forms & Improper IntegralsSection 2.3 · lesson 10 of 5411 min · free

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In this lesson

When the integrand blows up inside the interval.

  • Define an integral with an endpoint singularity as a limit
  • Split at an interior discontinuity and test each piece
  • Decide convergence at a glance with the p-test
  • Avoid the most dangerous error in the topic

Transcript

Everything said in the video, section by section (1,602 words). Read along, or search it.

MAT-265 - Calculus 2 · Section 2.3

Improper Integrals - Type II

An engineer computing the work needed to separate two charged particles. A physicist finding the time for an object to fall into a gravitational well. Both of them arrive at a definite integral that looks perfectly ordinary. Finite limits. A clean formula. Nothing suspicious about it. And yet it hides a trap. Somewhere on the interval of integration, the integrand shoots off to infinity. Now, the familiar Fundamental Theorem of Calculus assumes a continuous function on a closed interval. When that assumption fails, plugging the limits into an antiderivative can produce a confident-looking number that is completely wrong. These integrals — where the integrand itself becomes unbounded — are called Type Two improper integrals. And handling them correctly requires the careful use of limits.

The Trap

A Definite Integral That Hides a Singularity

Section two point two treated Type One. Where the interval of integration runs to infinity. Type Two is the complementary case, and it's worth seeing the symmetry. Here the interval is finite. But the function has an infinite discontinuity — a vertical asymptote — either at one of the endpoints, or somewhere inside. A standard definite integral is guaranteed a finite value whenever its integrand is continuous on a closed, bounded interval. Type One breaks the bounded requirement. Type Two breaks continuity. But both repairs use exactly the same idea. Replace the problem point with a finite variable, and take a limit. So the limit-based definition from the last section transfers almost verbatim. And geometrically the question is identical in both cases. Does an unbounded region nonetheless trap a finite amount of area?

Definition

Approach the Bad Point, Never Reach It

Here's the definition. When an integrand f is continuous on an interval except for an infinite discontinuity at one endpoint, the definite integral is defined not by direct evaluation, but as a limit. You approach the bad point, but you never reach it. Discontinuity at the upper endpoint. If f is continuous on a to b, open at b, then the integral is the limit as t approaches b from below, of the integral from a to t. Discontinuity at the lower endpoint. Same idea, mirrored. The limit as t approaches a from above. In each case, if the limit exists as a finite number, the integral converges to that value. If the limit is infinite, or fails to exist, the integral diverges. Notice how careful that is. You never evaluate the function at the singularity. You only ever approach it.

Worked Examples 1 and 2

Infinite Height, Finite Area

Let's see the two outcomes side by side. First, the integral from zero to one of one over root x. The integrand is continuous on zero to one, but it has a vertical asymptote at x equals zero — the lower limit. So replace that endpoint with a variable t approaching zero from the right. The antiderivative of x to the minus one half is two x to the one half. Evaluate from t to one, and you get two minus two root t. As t goes to zero, that goes to two. The limit is finite. The integral converges, to two. And look at what that means. The region under the curve has infinite height, yet finite area. The curve grows without bound near zero — but it does so slowly enough that the accumulated area stays finite. That's the defining surprise of convergent Type Two integrals. Now the second one. The integral from zero to one of one over x. Again the trouble is at zero. Replace the lower limit with t. The antiderivative is log absolute x. You get zero minus log t. As t approaches zero from the right, log t tends to minus infinity. So minus log t tends to plus infinity. The integral diverges. And that contrast is instructive. One over root x and one over x both blow up at the origin. Yet one encloses a finite area and the other does not. The rate at which the integrand grows near the singularity is what decides the outcome.

Interior Discontinuities

Split at the Singularity

Now the most error-prone situation. A discontinuity that lies strictly inside the interval, rather than at an endpoint. Here the integral cannot be evaluated as a single limit — because the antiderivative isn't valid across the break. Instead the integral must be split at the singularity, into two separate improper integrals. Each one handled by the endpoint definition. And there's a strict rule attached. The original integral converges if and only if both pieces converge. If either piece diverges, the whole integral diverges. Even if the other piece is perfectly finite. Let's do one that works. The integral from zero to eight of one over x minus one, to the two thirds. The integrand has an infinite discontinuity at x equals one, which lies inside zero to eight. So split there. The antiderivative is three times x minus one, to the one third. For the first piece, the discontinuity is at the upper endpoint. Take the limit from below, and you get three. For the second piece, it's at the lower endpoint. Take the limit from above, and you get three times seven to the one third. Both pieces converge. So the original converges, to their sum. About eight point seven four.

The p-Test

Deciding at a Glance

The examples so far hint at a general rule governing power-type singularities. And it's worth stating precisely, because it lets you decide convergence at a glance — without computing the full integral. The p-test for Type Two integrals. The integral from zero to one of one over x to the p converges if p is less than one, and diverges if p is at least one. In words: near a singularity, a milder blow-up — a smaller p — keeps the area finite. A blow-up as steep as one over x, or worse, makes the area infinite. And the same threshold applies to any singularity of the form one over x minus c, to the p. Notice this explains every example we've done. In Example one, one over root x is one over x to the one half, so p is one half, less than one. Converges. In Example two, one over x has p equals one. Diverges. In Example three, the exponent was two thirds, less than one. So each piece converged. One thing worth flagging. Compare this to the Type One p-test from the last section, which converged when p was greater than one. The thresholds are mirror images. Near a singularity you want a gentle blow-up. Out at infinity you want fast decay. Now — here's a genuine puzzle. Both one over root x and one over x rise to infinity as x approaches zero. And at every single point in zero to one, the value of one over x is larger than one over root x. It seems impossible that the larger function could enclose less total area. Yet the integral of one over root x is finite, and the integral of one over x is infinite. Resolve that apparent paradox. And ask what it tells you about reasoning from pointwise size alone, when a function is unbounded.

The Critical Pitfall

A Famously Wrong Evaluation

Now the defining hazard of Type Two integrals. Treating one as if it were proper. When a discontinuity hides inside the interval, blindly applying the Fundamental Theorem can yield a clean, finite, and entirely false answer. Here's the classic trap. Consider the integral from minus one to one of one over x squared. Apply the antiderivative minus one over x carelessly. You get minus one, minus one, which is minus two. Now stop and look at that answer. It is not merely inaccurate. It is absurd. The integrand one over x squared is positive everywhere it's defined. So any area it encloses must be positive. A negative answer is impossible. The error is that one over x squared has an infinite discontinuity at x equals zero, which lies inside minus one to one. And the antiderivative, minus one over x, is itself discontinuous there. So the Fundamental Theorem simply does not apply across the gap. Done correctly, you split at zero. Examine the second piece alone. Here p is two, at least one, so you should expect divergence. And indeed the limit is plus infinity. Because that piece diverges, the entire integral diverges. There's no need to even evaluate the first piece. Look at the picture. Both halves of the region under one over x squared have infinite area. The integral cannot possibly equal a finite negative number.

Key Takeaways

The Defensive Habit

So here's the defensive habit that prevents all of this. Before evaluating any definite integral, check whether the integrand is defined and continuous at every point of the interval. Including the interior. Look especially for denominators that vanish. Logarithms of zero. Even roots of negatives. If you find a discontinuity inside the limits, split there and treat each piece as improper. That one check prevents the most common wrong answer in this entire topic. Five things to carry forward. A Type Two integral has a finite interval but an infinite discontinuity on it. It's defined through a limit approaching the bad point. Endpoint singularities replace that endpoint with a variable and take a one-sided limit. Interior singularities must be split, and the original converges only if both pieces converge. The p-test: the integral of x to the minus p from zero to one converges for p less than one, and diverges for p at least one. That's the mirror image of the Type One threshold. And never integrate across an asymptote. A positive integrand can never give a negative result — if it does, you crossed one. Next up is section two point four, the Comparison Theorem — for integrals whose antiderivative you cannot find at all.

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FAQ

Questions students ask us

About Improper Integrals - Type II

What is a Type II improper integral?
A Type II improper integral is an integral over a finite interval where the integrand has an infinite discontinuity, or vertical asymptote, at an endpoint or inside the interval. It is defined as a limit. If f blows up at b, then \int_a^b f(x)\,dx = \lim_{t\to b^-}\int_a^t f(x)\,dx. If the limit is finite, the integral converges. If it is infinite or does not exist, the integral diverges.
How do you evaluate an improper integral with a discontinuity inside the interval?
Split the integral at the singularity and evaluate each piece as a separate improper integral using one-sided limits. The original converges only if both pieces converge. For example, \int_0^8 (x-1)^{-2/3}\,dx breaks at x=1. Using the antiderivative 3(x-1)^{1/3}, the left piece gives 3 and the right piece gives 3\cdot 7^{1/3}, so the integral converges to about 8.74.
What is the p-test for Type II improper integrals?
The p-test says \int_0^1 \frac{1}{x^p}\,dx converges when p<1 and diverges when p\ge 1. The same threshold applies to singularities of the form \frac{1}{(x-c)^p}. For instance, \int_0^1 x^{-1/2}\,dx converges to 2, while \int_0^1 \frac{1}{x}\,dx diverges. This is the mirror image of the Type I p-test, where integrals out to infinity converge only when p>1.
Why doesn't the integral of 1/x^2 from -1 to 1 equal -2?
The integral doesn't equal -2 because \frac{1}{x^2} has an infinite discontinuity at x=0, inside the interval, so the Fundamental Theorem of Calculus cannot be applied across it. Plugging the endpoints into -\frac{1}{x} gives -2, which is impossible for a positive integrand. Splitting at 0, the piece \int_0^1 \frac{1}{x^2}\,dx has p=2\ge 1 and diverges, so the whole integral diverges. Always check for interior discontinuities first.

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