Definitions

# Improper integral

In calculus, an improper integral is the limit of a definite integral as an endpoint of the interval of integration approaches either a specified real number or ∞ or −∞ or, in some cases, as both endpoints approach limits.

Specifically, an improper integral is a limit of the form

$lim_\left\{btoinfty\right\} int_a^bf\left(x\right), dx, qquad lim_\left\{ato -infty\right\} int_a^bf\left(x\right), dx,$
or of the form
$lim_\left\{cto b^-\right\} int_a^cf\left(x\right), dx,quad$
lim_{cto a^+} int_c^bf(x), dx, in which one takes a limit in one or the other (or sometimes both) endpoints . Improper integrals may also occur at an interior point of the domain of integration, or at multiple such points.

It is often necessary to use indefinite integrals in order to compute a value for integrals which may not exist in the conventional sense (as a Riemann integral, for instance) because of a singularity in the function, or an infinite endpoint of the domain of integration.

## Examples

The following integral does not exist as a Riemann integral
$int_1^infty frac\left\{1\right\}\left\{x^2\right\},dx$
because the domain of integration is unbounded. (The Riemann integral is only well-defined over a bounded domain.) However, it may be assigned a value as an improper integral by interpreting it instead as a limit
$lim_\left\{btoinfty\right\}int_1^bfrac\left\{1\right\}\left\{x^2\right\},dx=lim_\left\{btoinfty\right\}left\left[-frac\left\{1\right\}\left\{b\right\}+frac\left\{1\right\}\left\{1\right\}right\right] = 1$

The following integral also fails to exist as a Riemann integral:

$int_0^1 frac\left\{1\right\}\left\{sqrt\left\{x\right\}\right\},dx.$
Here the function is unbounded, and the Riemann integral is not well-defined for unbounded functions. However, if the integral is instead understood as the limit:
$lim_\left\{ato 0^+\right\}int_a^1frac\left\{1\right\}\left\{sqrt\left\{x\right\}\right\}, dx = lim_\left\{ato 0^+\right\}left\left[2sqrt\left\{1\right\}-2sqrt\left\{a\right\}right\right]=2,$
then the limit converges.

## Identities

$int_\left\{-infty\right\}^\left\{infty\right\} f\left(x\right) , dx = int_\left\{-infty\right\}^\left\{infty\right\} f\left(x - 1/x\right) , dx$

## Convergence of the integral

An improper integral converges if the limit defining it exists. Thus for example one says that the improper integral
$lim_\left\{ttoinfty\right\} int_a^t f\left(x\right), dx$
exists and is equal to L if the integrals under the limit exist for all sufficiently large t, and the value of the limit is equal to L.

It is also possible for an improper integral to diverge to infinity. In that case, one may assign the value of ∞ (or −∞) to the integral. For instance

$lim_\left\{btoinfty\right\}int_1^b frac\left\{1\right\}\left\{x\right\},dx = infty.$
However, other improper integrals may simply diverge in no particular direction, such as
$lim_\left\{btoinfty\right\}int_1^b xsin x, dx,$
which does not exist, even as an extended real number.

A limitation of the technique of improper integration is that the limit must be taken with respect to one endpoint at a time. Thus, for instance, an improper integral of the form

$int_\left\{-infty\right\}^infty f\left(x\right), dx$
is defined by taking two separate limits; to wit
$int_\left\{-infty\right\}^infty f\left(x\right), dx = lim_\left\{ato -infty\right\}lim_\left\{bto infty\right\} int_a^bf\left(x\right),dx,$
provided the double limit is finite. By the properties of the integral, this can also be written as a pair of distinct improper integrals of the first kind:
$lim_\left\{ato -infty\right\}int_a^cf\left(x\right), dx + lim_\left\{bto infty\right\} int_c^bf\left(x\right),dx$
where c is any convenient point at which to start the integration.

It is sometimes possible to define improper integrals where both endpoints are infinite, such as the Gaussian integral $int_\left\{-infty\right\}^infty e^\left\{-x^2\right\},dx = sqrt\left\{pi\right\}$. But one cannot even define other integrals of this kind unambiguously, such as $int_\left\{-infty\right\}^infty x,dx$, since the double limit diverges:

$lim_\left\{ato -infty\right\}int_a^cx,dx+lim_\left\{btoinfty\right\}int_c^bx,dx$
In this case, one can however define an improper integral in the sense of Cauchy principal value:
$p.v.int_\left\{-infty\right\}^infty x,dx=lim_\left\{btoinfty\right\}int_\left\{-b\right\}^bx,dx = 0.$

The questions one must address in determining an improper integral are:

• Does the limit exist?
• Can the limit be computed?

The first question is an issue of mathematical analysis. The second one can be addressed by calculus techniques, but also in some cases by contour integration, Fourier transforms and other more advanced methods.

## Types of integrals

There is more than one theory of mathematical integration. From the point of view of calculus, the Riemann integral theory is usually assumed as the default theory. In using improper integrals, it can matter which integration theory is in play.

• For the Darboux integral, improper integration is necessary both for unbounded intervals (since one cannot divide the interval into finitely many subintervals of finite length) and for unbounded functions with finite integral (since, supposing it is unbounded above, then the upper integral will be infinite, but the lower integral will be finite).
• The Riemann integral, improper integration is also necessary for unbounded intervals and for unbounded functions, as with the Darboux integral.
• The Lebesgue integral deals differently with unbounded domains and unbounded functions, so that often an integral which only exists as an improper Riemann integral will exist as a (proper) Lebesgue integral, such as $int_1^infty frac\left\{1\right\}\left\{x^2\right\},dx$. On the other hand, there are also integrals that have an improper Riemann integral do not have a (proper) Lebesgue integral, such as $int_0^infty frac\left\{sin x\right\}\left\{x\right\},dx$. The Lebesgue theory does not see this as a deficiency: from the point of view of measure theory, $int_0^infty frac\left\{sin x\right\}\left\{x\right\},dx = infty - infty$ and cannot be defined satisfactorily. In some situations, however, it may be convenient to employ improper Lebesgue integrals as is the case, for instance, when defining the Cauchy principal value.
• For the Henstock-Kurzweil integral, improper integration is not necessary, and this is seen as a strength of the theory: it encompasses all Lebesgue integrable and improper Riemann integrable functions.

## Improper Riemann integrals and Lebesgue integrals

In some cases, the integral

$int_a^c f\left(x\right),dx,$

can be defined as an integral (a Lebesgue integral, for instance) without reference to the limit

$lim_\left\{bto c^-\right\}int_a^b f\left(x\right),dx,$

but cannot otherwise be conveniently computed. This often happens when the function f being integrated from a to c has a vertical asymptote at c, or if c = ∞ (see Figures 1 and 2). In such cases, the improper Riemann integral allows one to calculate the Lebesgue integral of the function. Specifically, the following theorem holds :

• If a function f is Riemann integrable on [a,b] for every b ≥ a, and the partial integrals

$int_a^b|f\left(x\right)|,dx$
are bounded as b → ∞, then the improper Riemann integrals
$int_a^infty f\left(x\right), dx,quadmbox\left\{and\right\} int_a^infty |f\left(x\right)|, dx$
both exist. Furthermore, f is Lebesgue integrable on [a, ∞), and its Lebesgue integral is equal to its improper Riemann integral.

For example, the integral

$int_0^inftyfrac\left\{dx\right\}\left\{1+x^2\right\}$
can be interpreted alternatively as the improper integral
$lim_\left\{btoinfty\right\}int_0^bfrac\left\{dx\right\}\left\{1+x^2\right\}=lim_\left\{btoinfty\right\}arctan\left\{b\right\}=frac\left\{pi\right\}\left\{2\right\},$
or it may be interpreted instead as a Lebesgue integral over the set (0, ∞). Since both of these kinds of integral agree, one is free to choose the first method to calculate the value of the integral, even if one ultimately wishes to regard it as a Lebesgue integral. Thus improper integrals are clearly useful tools for obtaining the actual values of integrals.

In other cases, however, the integral from a to c is not even defined, because the integrals of the positive and negative parts of f(xdx from a to c are both infinite, but nonetheless the limit may exist. Such cases are "properly improper" integrals, i.e. their values cannot be defined except as such limits. For example,

$int_0^inftyfrac\left\{sin\left(x\right)\right\}\left\{x\right\},dx$

cannot be interpreted as a Lebesgue integral, since

$int_0^inftyleft|frac\left\{sin\left(x\right)\right\}\left\{x\right\}right|,dx=infty.$

This is therefore a "properly" improper integral, whose value is given by

$int_0^inftyfrac\left\{sin\left(x\right)\right\}\left\{x\right\},dx=lim_\left\{brightarrowinfty\right\}int_0^bfrac\left\{sin\left(x\right)\right\}\left\{x\right\},dx=frac\left\{pi\right\}\left\{2\right\}.$

## Singularities

One can speak of the singularities of an improper integral, meaning those points of the extended real number line at which limits are used.

Such an integral is often written symbolically just like a standard definite integral, perhaps with infinity as a limit of integration. But that conceals the limiting process. By using the more advanced Lebesgue integral, rather than the Riemann integral, one can in some cases bypass this requirement, but if one simply wants to evaluate the limit to a definite answer, that technical fix may not necessarily help. It is more or less essential in the theoretical treatment for the Fourier transform, with pervasive use of integrals over the whole real line.

## Cauchy principal value

Consider the difference in values of two limits:

$lim_\left\{arightarrow 0+\right\}left\left(int_\left\{-1\right\}^\left\{-a\right\}frac\left\{dx\right\}\left\{x\right\}+int_a^1frac\left\{dx\right\}\left\{x\right\}right\right)=0,$

$lim_\left\{arightarrow 0+\right\}left\left(int_\left\{-1\right\}^\left\{-a\right\}frac\left\{dx\right\}\left\{x\right\}+int_\left\{2a\right\}^1frac\left\{dx\right\}\left\{x\right\}right\right)=-ln 2.$

The former is the Cauchy principal value of the otherwise ill-defined expression

$int_\left\{-1\right\}^1frac\left\{dx\right\}\left\{x\right\}\left\{ \right\}$
left(mbox{which} mbox{gives} -infty+inftyright).

Similarly, we have

$lim_\left\{arightarrowinfty\right\}int_\left\{-a\right\}^afrac\left\{2x,dx\right\}\left\{x^2+1\right\}=0,$

but

$lim_\left\{arightarrowinfty\right\}int_\left\{-2a\right\}^afrac\left\{2x,dx\right\}\left\{x^2+1\right\}=-ln 4.$

The former is the principal value of the otherwise ill-defined expression

$int_\left\{-infty\right\}^inftyfrac\left\{2x,dx\right\}\left\{x^2+1\right\}\left\{ \right\}$
left(mbox{which} mbox{gives} -infty+inftyright).

All of the above limits are cases of the indeterminate form ∞ − ∞.

These pathologies do not affect "Lebesgue-integrable" functions, that is, functions the integrals of whose absolute values are finite.

## Summability

An indefinite integral may diverge in the sense that the limit defining it may not exist. In this case, there are more sophisticated definitions of the limit which can produce a convergent value for the improper integral. These are called summability methods.

One summability method, popular in Fourier analysis, is that of Cesàro summation. The integral

$int_0^infty f\left(x\right),dx$

is Cesàro summable (C, α) if

$lim_\left\{lambdatoinfty\right\}int_0^lambdaleft\left(1-frac\left\{x\right\}\left\{lambda\right\}right\right)^alpha f\left(x\right), dx$

exists and is finite . The value of this limit, should it exist, is the (C, α) sum of the integral.

An integral is (C, 0) summability precisely when it exists as an improper integral. However, there are integrals which are (C, α) summable for α > 0 which fail to converge as improper integrals (in the sense of Riemann or Lebesgue). One example is the integral

$int_0^inftysin x, dx$

which fails to exist as an improper integral, but is (C,α) summable for every α > 0, with value 1. This is an integral version of Grandi's series.

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