Calculus · Grade 12 and undergraduate

Improper Integrals

Quick answer

An improper integral has an infinite limit of integration or an integrand that blows up on the interval. It is defined as a limit of ordinary integrals, such as ∫₁^∞ f(x) dx = lim as b → ∞ of ∫₁ᵇ f(x) dx. If the limit exists, the integral converges; otherwise it diverges. The integral of 1/xᵖ from 1 to ∞ converges exactly when p > 1, and from 0 to 1 exactly when p < 1. An infinitely long region can have a finite area, if its height falls off fast enough.

What you'll learn

  • Write an improper integral as a limit of ordinary integrals
  • Decide whether an improper integral converges, and find its value
  • Use the p-integrals as benchmarks
  • Spot an integrand that is unbounded inside the interval

Integrals without an end

A definite integral is built from Riemann sums over a finite interval, with a bounded integrand. Drop either condition and the definition needs a limit.

Definition

If ff is continuous on [a,∞)[a, \infty),

∫a∞f(x) dx=lim⁡b→∞∫abf(x) dx\int_a^{\infty} f(x)\,dx = \lim_{b \to \infty} \int_a^b f(x)\,dx

If the limit exists, the integral converges to that value. Otherwise it diverges. An integral over (−∞,∞)(-\infty, \infty) is split at any point into two such integrals, and converges only if both do.

The tails of y = 1/x² and y = 1/x Both curves fall toward the x-axis as x grows. The region under y = 1/x² from x = 1 onward is shaded; it stretches without end but has area 1. The dashed curve y = 1/x stays higher, and the area under it from 1 onward is infinite. 24681xy
  • y = 1/x²
  • y = 1/x
The tails of y = 1/x² and y = 1/x

Why an endless region can have finite area

Compare how the two curves in the graph add area as bb doubles. From bb to 2b2b, the curve 1x\tfrac{1}{x} adds ln⁡2b−ln⁡b=ln⁡2\ln 2b - \ln b = \ln 2: the same amount every time, forever. The curve 1x2\tfrac{1}{x^2} adds 1b−12b=12b\tfrac{1}{b} - \tfrac{1}{2b} = \tfrac{1}{2b}: half as much with each doubling. Amounts that keep halving add up to a finite total; equal amounts do not. Whether an endless region has finite area depends on how fast its height falls, not on how long the region is.

The same test, run for every power, gives the benchmark integrals:

∫1∞dxxp converges if p>1, to 1p−1, and diverges if p≤1\int_1^{\infty} \frac{dx}{x^p} \text{ converges if } p > 1 \text{, to } \frac{1}{p - 1}\text{, and diverges if } p \le 1

Integrands that blow up

An integrand can also be unbounded at an endpoint. Then the limit approaches that endpoint from inside the interval:

∫01dxx=lim⁡a→0+∫a1x−1/2 dx=lim⁡a→0+(2−2a)=2\int_0^1 \frac{dx}{\sqrt{x}} = \lim_{a \to 0^+} \int_a^1 x^{-1/2}\,dx = \lim_{a \to 0^+} \left(2 - 2\sqrt{a}\right) = 2

Near 00 the benchmark flips: ∫01dxxp\int_0^1 \tfrac{dx}{x^p} converges if p<1p < 1 and diverges if p≥1p \ge 1. If the integrand is unbounded inside the interval, split there and treat each piece as improper.

Worked examples

Common mistakes

Practice problems

  1. Evaluate ∫1∞dxx3\displaystyle\int_1^{\infty} \frac{dx}{x^3}.

    Answer

    12\tfrac{1}{2}

    Full solution

    ∫1bx−3 dx=12−12b2→12\int_1^b x^{-3}\,dx = \tfrac{1}{2} - \tfrac{1}{2b^2} \to \tfrac{1}{2}. It is a pp-integral with p=3p = 3: 13−1\tfrac{1}{3 - 1}.

  2. Evaluate ∫−∞∞dx1+x2\displaystyle\int_{-\infty}^{\infty} \frac{dx}{1 + x^2}.

    Answer

    π\pi

    Full solution

    Split at 00. ∫0bdx1+x2=arctan⁡b→π2\int_0^b \tfrac{dx}{1 + x^2} = \arctan b \to \tfrac{\pi}{2}, and by symmetry the left half is also π2\tfrac{\pi}{2}.

  3. Does ∫1∞dxx\displaystyle\int_1^{\infty} \frac{dx}{\sqrt{x}} converge?

    Answer

    No, it diverges.

    Full solution

    It is a pp-integral with p=12≤1p = \tfrac{1}{2} \le 1. Directly: ∫1bx−1/2 dx=2b−2\int_1^b x^{-1/2}\,dx = 2\sqrt{b} - 2, which grows without bound.

  4. Evaluate ∫0∞e−2x dx\displaystyle\int_0^{\infty} e^{-2x}\,dx.

    Answer

    12\tfrac{1}{2}

    Full solution

    ∫0be−2x dx=12(1−e−2b)→12\int_0^b e^{-2x}\,dx = \tfrac{1}{2}\left(1 - e^{-2b}\right) \to \tfrac{1}{2}.

  5. Evaluate ∫0∞dx1+x2\displaystyle\int_0^{\infty} \frac{dx}{1 + x^2}.

    Answer

    π2\tfrac{\pi}{2}

    Full solution

    lim⁡b→∞arctan⁡b=π2\lim_{b \to \infty} \arctan b = \tfrac{\pi}{2}.

  6. Evaluate ∫01dxx1/3\displaystyle\int_0^1 \frac{dx}{x^{1/3}}.

    Answer

    32\tfrac{3}{2}

    Full solution

    ∫a1x−1/3 dx=32(1−a2/3)→32\int_a^1 x^{-1/3}\,dx = \tfrac{3}{2}\left(1 - a^{2/3}\right) \to \tfrac{3}{2} as a→0+a \to 0^+.

  7. Does ∫01dxx\displaystyle\int_0^1 \frac{dx}{x} converge?

    Answer

    No, it diverges.

    Full solution

    ∫a1dxx=−ln⁡a\int_a^1 \tfrac{dx}{x} = -\ln a, which grows without bound as a→0+a \to 0^+.

  8. Evaluate ∫0∞xe−x dx\displaystyle\int_0^{\infty} x e^{-x}\,dx.

    Answer

    11

    Full solution

    By parts, ∫0bxe−x dx=[−xe−x−e−x]0b=1−be−b−e−b\int_0^b x e^{-x}\,dx = \big[-x e^{-x} - e^{-x}\big]_0^b = 1 - b e^{-b} - e^{-b}. By L’Hôpital’s rule, be−b=beb→0b e^{-b} = \tfrac{b}{e^b} \to 0, so the limit is 11.

  9. Does ∫03dx(x−1)2\displaystyle\int_0^3 \frac{dx}{(x - 1)^2} converge?

    Answer

    No, it diverges.

    Full solution

    The integrand is unbounded at x=1x = 1, inside the interval. Split there: ∫01dx(x−1)2=lim⁡c→1−(11−c−1)\int_0^1 \tfrac{dx}{(x - 1)^2} = \lim_{c \to 1^-}\left(\tfrac{1}{1 - c} - 1\right), which grows without bound.

  10. A student writes ∫−21dxx2=[−1x]−21=−1−12=−32\displaystyle\int_{-2}^{1} \frac{dx}{x^2} = \left[-\frac{1}{x}\right]_{-2}^{1} = -1 - \frac{1}{2} = -\frac{3}{2}. What went wrong?

    Hint

    Where is 1x2\tfrac{1}{x^2} undefined?

    Answer

    The integrand is unbounded at x=0x = 0, inside the interval. The integral diverges.

    Full solution

    The Fundamental Theorem needs a continuous integrand on the whole interval. Splitting at 00, ∫01dxx2\int_0^1 \tfrac{dx}{x^2} diverges, so the integral does too.

    A negative answer for a positive integrand was the warning sign.

Frequently asked questions

What makes an integral improper?

An infinite limit of integration, such as ∫ from 1 to ∞, or an integrand that becomes unbounded somewhere on the interval, such as 1/√x near 0.

What does it mean for an improper integral to converge?

The limit that defines it exists as a finite number. That number is the value of the integral.

For which p does ∫ from 1 to ∞ of 1/xᵖ converge?

For p > 1, with value 1/(p − 1). For p ≤ 1 it diverges; p = 1 gives ln b, which grows without bound.

If f(x) goes to 0, does ∫ from 1 to ∞ of f converge?

Not necessarily. 1/x goes to 0, yet its integral from 1 to ∞ diverges. The height has to fall fast enough.

Why is ∫ from −1 to 1 of 1/x² not equal to −2?

The integrand is unbounded at x = 0, inside the interval, so the Fundamental Theorem does not apply. Split at 0: each piece diverges, so the integral diverges.

What to learn next

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