Improper Integrals with Infinite Discontinuities

An improper integral with an infinite discontinuity has an integrand that blows up at an endpoint or inside the interval, like \int_0^1 \dfrac{1}{\sqrt{x}}\,dx. Replace the bad point with t, integrate, and take a one-sided limit. A finite limit means the integral converges; otherwise it diverges.

Before this lesson: Improper Integrals, One-Sided Limits and Infinite Limits

Area next to a vertical asymptote

The graph of y = 1/√x shoots up as x gets close to 0. The region under it from 0 to 1 is infinitely tall, yet its area is exactly 2. The region is tall but very thin near the asymptote.

The danger is that these integrals look normal. \int_{-1}^{1} \dfrac{1}{x^2}\,dx has ordinary-looking limits, but 1/x² blows up at x = 0. Always check where the integrand is undefined before you integrate.

Definitions for infinite discontinuities

If f is continuous on [a, b) and blows up at b, or continuous on (a, b] and blows up at a:

\int_a^b f(x)\,dx = \lim_{t \to b^-} \int_a^t f(x)\,dx \qquad \int_a^b f(x)\,dx = \lim_{t \to a^+} \int_t^b f(x)\,dx

If f blows up at a point c inside [a, b], split the integral at c. It converges only if both pieces converge.

\int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx

The p-rule near 0: \int_0^1 \dfrac{1}{x^p}\,dx converges to \dfrac{1}{1 - p} if p is less than 1 and diverges if p ≥ 1. This is the reverse of the rule for \int_1^{\infty}.

Improper integral examples with infinite discontinuities

Example 1: Blowing up at the left end

Evaluate \int_0^1 \dfrac{1}{\sqrt{x}}\,dx.

\int_0^1 x^{-1/2}\,dx = \lim_{t \to 0^+} \int_t^1 x^{-1/2}\,dx

= \lim_{t \to 0^+} \Big[2\sqrt{x}\Big]_t^1

= \lim_{t \to 0^+} (2 - 2\sqrt{t})

= 2 - 0 = 2

Answer: the integral converges to 2.

Example 2: A divergent one

Evaluate \int_0^1 \dfrac{1}{x^2}\,dx.

\int_0^1 x^{-2}\,dx = \lim_{t \to 0^+} \Big[-\dfrac{1}{x}\Big]_t^1

= \lim_{t \to 0^+} \left(-1 + \dfrac{1}{t}\right)

= \infty

Answer: the integral diverges. Here p = 2 ≥ 1.

Example 3: A discontinuity inside the interval

Evaluate \int_0^3 \dfrac{1}{(x - 1)^{2/3}}\,dx.

(x - 1)^{2/3} = 0 \text{ at } x = 1

\int (x - 1)^{-2/3}\,dx = 3(x - 1)^{1/3} + C

\int_0^1 = \lim_{t \to 1^-} \Big[3(x - 1)^{1/3}\Big]_0^t = 0 - 3(-1) = 3

\int_1^3 = \lim_{t \to 1^+} \Big[3(x - 1)^{1/3}\Big]_t^3 = 3\sqrt[3]{2} - 0

3 + 3\sqrt[3]{2} \approx 6.780

Answer: the integral converges to 3 + 3\sqrt[3]{2} \approx 6.780.

Example 4: A logarithm at 0

Evaluate \int_0^1 \ln x\,dx.

\int \ln x\,dx = x\ln x - x + C

\int_0^1 \ln x\,dx = \lim_{t \to 0^+} \Big[x\ln x - x\Big]_t^1

= \lim_{t \to 0^+} \big[(0 - 1) - (t\ln t - t)\big]

\lim_{t \to 0^+} t\ln t = \lim_{t \to 0^+} \dfrac{\ln t}{1/t} = \lim_{t \to 0^+} \dfrac{1/t}{-1/t^2} = \lim_{t \to 0^+} (-t) = 0

= -1 - (0 - 0) = -1

Answer: the integral converges to −1. It is negative because ln x is negative on (0, 1).

Example 5: A tank that drains faster and faster

Water drains from a tank at a rate of r(t) = \dfrac{6}{\sqrt{4 - t}} liters per minute for t from 0 up to 4. The rate blows up as t nears 4 minutes. How much water drains in those 4 minutes?

\int_0^4 6(4 - t)^{-1/2}\,dt = \lim_{b \to 4^-} \int_0^b 6(4 - t)^{-1/2}\,dt

\int 6(4 - t)^{-1/2}\,dt = -12\sqrt{4 - t} + C

= \lim_{b \to 4^-} \left(-12\sqrt{4 - b} + 12\sqrt{4}\right)

= 0 + 24 = 24

Answer: 24 liters drain, even though the rate becomes infinite.

Common mistakes

Related methods

Both kinds at once

Some integrals are improper in two ways, like \int_0^{\infty} \dfrac{1}{\sqrt{x}(1 + x)}\,dx. Split it at a point such as x = 1, so each piece has only one problem, and check each piece. This one converges to π.

Deciding without an antiderivative

If you can’t integrate, compare. Near the bad point, an integrand smaller than a convergent p-integral also converges. That is the comparison test, the next lesson.

The p-rule near 0

For p ≠ 1, \int_t^1 x^{-p}\,dx = \dfrac{1 - t^{1-p}}{1 - p}. If p is less than 1, t¹⁻ᵖ → 0 and the integral approaches 1/(1 − p). If p is more than 1, t¹⁻ᵖ → ∞. For p = 1, you get −ln t → ∞.

Practice problems

Decide whether each integral converges or diverges. If it converges, find its value.

1. \int_0^4 \dfrac{1}{\sqrt{x}}\,dx
2. \int_0^1 \dfrac{1}{x}\,dx
3. \int_0^8 x^{-1/3}\,dx
4. \int_1^2 \dfrac{1}{x - 1}\,dx
5. \int_0^2 \dfrac{1}{\sqrt{2 - x}}\,dx
6. \int_{-1}^{1} \dfrac{1}{x^{2/3}}\,dx
7. \int_0^{\pi/2} \tan x\,dx
8. \int_0^1 x\ln x\,dx
9. \int_0^1 \dfrac{1}{\sqrt{1 - x^2}}\,dx
10. Use the p-rule to find \int_0^1 x^{-0.9}\,dx.

\lim_{t \to 0^+} \Big[2\sqrt{x}\Big]_t^4

= \lim_{t \to 0^+} (4 - 2\sqrt{t}) = 4

\lim_{t \to 0^+} \Big[\ln x\Big]_t^1 = \lim_{t \to 0^+} (0 - \ln t)

= \infty

\lim_{t \to 0^+} \Big[\tfrac{3}{2}x^{2/3}\Big]_t^8

= \tfrac{3}{2}(8)^{2/3} - 0 = \tfrac{3}{2}(4) = 6

\lim_{t \to 1^+} \Big[\ln(x - 1)\Big]_t^2 = \lim_{t \to 1^+} \big(0 - \ln(t - 1)\big)

= \infty

\lim_{t \to 2^-} \Big[-2\sqrt{2 - x}\Big]_0^t

= \lim_{t \to 2^-} \left(-2\sqrt{2 - t} + 2\sqrt{2}\right) = 2\sqrt{2}

\int x^{-2/3}\,dx = 3x^{1/3} + C

\int_{-1}^{0} = 0 - 3(-1) = 3, \quad \int_0^1 = 3 - 0 = 3

3 + 3 = 6

\int \tan x\,dx = -\ln(\cos x) + C

\lim_{t \to \pi/2^-} \big(-\ln(\cos t) + \ln 1\big) = \infty

\int x\ln x\,dx = \dfrac{x^2}{2}\ln x - \dfrac{x^2}{4} + C

\lim_{t \to 0^+} \left[\left(0 - \dfrac{1}{4}\right) - \left(\dfrac{t^2}{2}\ln t - \dfrac{t^2}{4}\right)\right]

= -\dfrac{1}{4} - 0 = -\dfrac{1}{4}

\lim_{t \to 1^-} \Big[\arcsin x\Big]_0^t = \lim_{t \to 1^-} \arcsin t - 0

= \dfrac{\pi}{2}

p = 0.9, \text{ which is less than } 1

\dfrac{1}{1 - 0.9} = \dfrac{1}{0.1} = 10

Infinite discontinuities FAQ

What is an infinite discontinuity in an integral?

It is a point where the integrand blows up to ±∞, usually at a vertical asymptote. If it lies in the interval, the integral is improper.

How do you evaluate an improper integral with a vertical asymptote?

Replace the bad point with t, integrate, and take a one-sided limit toward it. If the asymptote is inside the interval, split there and do each side.

Does the integral of 1/x from 0 to 1 converge?

No. It equals the limit of −ln t as t → 0⁺, which is infinite, so it diverges.

Can you integrate 1/x² from −1 to 1?

Not as a normal integral, because 1/x² blows up at 0. Split at 0. Each piece diverges, so the integral diverges. The answer −2 from plugging in is wrong.

Summary

• Check where the integrand blows up before integrating.
• Replace the bad point with t and take a one-sided limit.
• Split at a bad point inside the interval; both pieces must converge.
• ∫ from 0 to 1 of 1/xᵖ converges only when p is less than 1, to 1/(1 − p).
• An infinitely tall region can still have a finite area.

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Next lesson: Comparison Test for Improper Integrals

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