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Improper integral

Concept in mathematical analysis

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Through Integral, Continuous function and Limit (mathematics)

In the simplest case of a real-valued function of a single variable integrated in the sense of Riemann (or Darboux) over a single interval, improper integrals may be in any of the following forms:

  1. ∫a∞f(x)dx
  2. ∫−∞bf(x)dx
  3. ∫−∞∞f(x)dx
  4. ∫abf(x)dx, where f(x) is undefined or discontinuous somewhere on [a,b]

The first three forms are improper because the integrals are taken over an unbounded interval. (They may be improper for other reasons, as well, as explained below.) Such an integral is sometimes described as being of the "first" type or kind if the integrand otherwise satisfies the assumptions of integration. Integrals in the fourth form that are improper because f(x) has a vertical asymptote somewhere on the interval [a,b] may be described as being of the "second" type or kind. Integrals that combine aspects of both types are sometimes described as being of the "third" type or kind.

In each case above, the improper integral must be rewritten using one or more limits, depending on what is causing the integral to be improper. For example, in case 1, if f(x) is continuous on the entire interval [a,∞), then

∫a∞f(x)dx=limb→∞∫abf(x)dx.

The limit on the right is taken to be the definition of the integral notation on the left.

If f(x) is only continuous on (a,∞) and not at a itself, then typically this is rewritten as

∫a∞f(x)dx=limt→a+∫tcf(x)dx+limb→∞∫cbf(x)dx,

for any choice of c>a. Here both limits must converge to a finite value for the improper integral to be said to converge. This requirement avoids the ambiguous case of adding positive and negative infinities (i.e., the "∞−∞" indeterminate form). Alternatively, an iterated limit could be used or a single limit based on the Cauchy principal value.

If f(x) is continuous on [a,d) and (d,∞), with a discontinuity of any kind at d, then

∫a∞f(x)dx=limt→d−∫atf(x)dx+limu→d+∫ucf(x)dx+limb→∞∫cbf(x)dx,

for any choice of c>d. The previous remarks about indeterminate forms, iterated limits, and the Cauchy principal value also apply here.

The function f(x) can have more discontinuities, in which case even more limits would be required (or a more complicated principal value expression).

Cases 2–4 are handled similarly. See the examples below.

Improper integrals can also be evaluated in the context of complex numbers, in higher dimensions, and in other theoretical frameworks such as Lebesgue integration or Henstock–Kurzweil integration. Integrals that are considered improper in one framework may not be in others.

Examples

The original definition of the Riemann integral does not apply to a function such as 1/x2 on the interval [1, ∞), because in this case the domain of integration is unbounded. However, the Riemann integral can often be extended by continuity, by defining the improper integral instead as a limit

∫1∞dxx2=limb→∞∫1bdxx2=limb→∞(−1b+11)=1.

The narrow definition of the Riemann integral also does not cover the function 1/x on the interval [0, 1]. The problem here is that the integrand is unbounded in the domain of integration. In other words, the definition of the Riemann integral requires that both the domain of integration and the integrand be bounded. However, the improper integral does exist if understood as the limit

∫01dxx=lima→0+∫a1dxx=lima→0+(2−2a)=2.

Sometimes integrals may have two singularities where they are improper. Consider, for example, the function 1/((x + 1)√x) integrated from 0 to ∞ (shown right). At the lower bound of the integration domain, as x goes to 0 the function goes to ∞, and the upper bound is itself ∞, though the function goes to 0. Thus this is a doubly improper integral. Integrated, say, from 1 to 3, an ordinary Riemann sum suffices to produce a result of π/6. To integrate from 1 to ∞, a Riemann sum is not possible. However, any finite upper bound, say t (with t > 1), gives a well-defined result, 2 arctan(√t) − π/2. This has a finite limit as t goes to infinity, namely π/2. Similarly, the integral from 1/3 to 1 allows a Riemann sum as well, coincidentally again producing π/6. Replacing 1/3 by an arbitrary positive value s (with s < 1) is equally safe, giving π/2 − 2 arctan(√s). This, too, has a finite limit as s goes to zero, namely π/2. Combining the limits of the two fragments, the result of this improper integral is

∫0∞dx(1+x)x=lims→0+∫s1dx(1+x)x+limt→∞∫1tdx(1+x)x=lims→0+(π2−2arctan⁡s)+limt→∞(2arctan⁡t−π2)=π2+(π−π2)=π.

This process does not guarantee success; a limit might fail to exist, or might be infinite. For example, over the bounded interval from 0 to 1 the integral of 1/x does not converge; and over the unbounded interval from 1 to ∞ the integral of 1/√x does not converge.

It might also happen that an integrand is unbounded near an interior point, in which case the integral must be split at that point. For the integral as a whole to converge, the limit integrals on both sides must exist and must be bounded. For example:

∫−11dxx23=lims→0−∫−1sdxx23+limt→0+∫t1dxx23=lims→0−3(1−−s3)+limt→0+3(1−t3)=3+3=6.

But the similar integral

∫−11dxx

cannot be assigned a value in this way, as the integrals above and below zero in the integral domain do not independently converge. (However, see Cauchy principal value.)

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