# Improper integral An improper integral of the first kind. The integral may need to be defined on an unbounded domain. An improper Riemann integral of the second kind. The integral may fail to exist because of a vertical asymptote in the function.

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In calculus, an improper integral is the limit of a definite integral as an endpoint of the interval(s) of integration approaches either a specified real number or $\infty$ or $-\infty$ or, in some cases, as both endpoints approach limits. Such an integral is often written symbolically just like a standard definite integral, perhaps with infinity as a limit of integration.

Specifically, an improper integral is a limit of the form

$\lim _{b\to \infty }\int _{a}^{b}f(x)\,\mathrm {d} x,\qquad \lim _{a\to -\infty }\int _{a}^{b}f(x)\,\mathrm {d} x,$ or of the form

$\lim _{c\to b^{-}}\int _{a}^{c}f(x)\,\mathrm {d} x,\quad \lim _{c\to a^{+}}\int _{c}^{b}f(x)\,\mathrm {d} x,$ in which one takes a limit in one or the other (or sometimes both) endpoints Template:Harv. When a function is undefined at finitely many interior points of an interval, the improper integral over the interval is defined as the sum of the improper integrals over the intervals between these points.

By abuse of notation, improper integrals are often written symbolically just like standard definite integrals, perhaps with infinity among the limits of integration. When the definite integral exists (in the sense of either the Riemann integral or the more advanced Lebesgue integral), this ambiguity is resolved as both the proper and improper integral will coincide in value.

Often one is able to compute values for improper integrals, even when the function is not integrable in the conventional sense (as a Riemann integral, for instance) because of a singularity in the function, or poor behavior at infinity. Such integrals are often termed "properly improper", as they cannot be computed as a proper integral.

## Examples

The original definition of the Riemann integral does not apply to a function such as $1/{x^{2}}$ 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

$\int _{1}^{\infty }{\frac {1}{x^{2}}}\,\mathrm {d} x=\lim _{b\to \infty }\int _{1}^{b}{\frac {1}{x^{2}}}\,\mathrm {d} x=\lim _{b\to \infty }\left(-{\frac {1}{b}}+{\frac {1}{1}}\right)=1.$ The narrow definition of the Riemann integral also does not cover the function $1/{\sqrt {x}}$ on the interval [0, 1]. The problem here is that the integrand is unbounded in the domain of integration (the definition requires that both the domain of integration and the integrand be bounded). However, the improper integral does exist if understood as the limit

$\int _{0}^{1}{\frac {1}{\sqrt {x}}}\,\mathrm {d} x=\lim _{a\to 0^{+}}\int _{a}^{1}{\frac {1}{\sqrt {x}}}\,\mathrm {d} x=\lim _{a\to 0^{+}}(2-2{\sqrt {a}})=2.$ ## Convergence of the integral

An improper integral converges if the limit defining it exists. Thus for example one says that the improper integral

$\lim _{t\to \infty }\int _{a}^{t}f(x)\,\mathrm {d} x$ 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

$\lim _{b\to \infty }\int _{1}^{b}{\frac {1}{x}}\,\mathrm {d} x=\infty .$ However, other improper integrals may simply diverge in no particular direction, such as

$\lim _{b\to \infty }\int _{1}^{b}x\sin x\,\mathrm {d} x,$ which does not exist, even as an extended real number. This is called divergence by oscillation.

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

$\int _{-\infty }^{\infty }f(x)\,\mathrm {d} x$ can be defined by taking two separate limits; to wit

$\int _{-\infty }^{\infty }f(x)\,\mathrm {d} x=\lim _{a\to -\infty }\lim _{b\to \infty }\int _{a}^{b}f(x)\,\mathrm {d} x$ provided the double limit is finite. It can also be defined as a pair of distinct improper integrals of the first kind:

$\lim _{a\to -\infty }\int _{a}^{c}f(x)\,\mathrm {d} x+\lim _{b\to \infty }\int _{c}^{b}f(x)\,\mathrm {d} x$ where c is any convenient point at which to start the integration. This definition also applies when one of these integrals is infinite, or both if they have the same sign.

An example of an improper integrals where both endpoints are infinite is the Gaussian integral $\int _{-\infty }^{\infty }e^{-x^{2}}\,\mathrm {d} x={\sqrt {\pi }}$ . An example which evaluates to infinity is $\int _{-\infty }^{\infty }e^{x}\,\mathrm {d} x$ . But one cannot even define other integrals of this kind unambiguously, such as $\int _{-\infty }^{\infty }x\,\mathrm {d} x$ , since the double limit is infinite and the two-integral method

$\lim _{a\to -\infty }\int _{a}^{c}x\,\mathrm {d} x+\lim _{b\to \infty }\int _{c}^{b}x\,\mathrm {d} x$ yields $\infty -\infty$ . In this case, one can however define an improper integral in the sense of Cauchy principal value:

$\operatorname {p.v.} \int _{-\infty }^{\infty }x\,\mathrm {d} x=\lim _{b\to \infty }\int _{-b}^{b}x\,\mathrm {d} x=0.$ 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 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.

## Improper Riemann integrals and Lebesgue integrals

In some cases, the integral

$\int _{a}^{c}f(x)\,\mathrm {d} x\,$ can be defined as an integral (a Lebesgue integral, for instance) without reference to the limit

$\lim _{b\to c^{-}}\int _{a}^{b}f(x)\,\mathrm {d} x\,$ 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 Template:Harv:

• If a function f is Riemann integrable on [a,b] for every b ≥ a, and the partial integrals
$\int _{a}^{b}|f(x)|\,\mathrm {d} x$ are bounded as b → ∞, then the improper Riemann integrals
$\int _{a}^{\infty }f(x)\,\mathrm {d} x,\quad {\mbox{and}}\ \int _{a}^{\infty }|f(x)|\,\mathrm {d} x$ both exist. Furthermore, f is Lebesgue integrable on [a, ∞), and its Lebesgue integral is equal to its improper Riemann integral.

For example, the integral

$\int _{0}^{\infty }{\frac {\mathrm {d} x}{1+x^{2}}}$ can be interpreted alternatively as the improper integral

$\lim _{b\to \infty }\int _{0}^{b}{\frac {\mathrm {d} x}{1+x^{2}}}=\lim _{b\to \infty }\arctan {b}={\frac {\pi }{2}},$ 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, a Lebesgue integral between finite endpoints may not even be defined, because the integrals of the positive and negative parts of f are both infinite, but the improper Riemann integral may still exist. Such cases are "properly improper" integrals, i.e. their values cannot be defined except as such limits. For example,

$\int _{0}^{\infty }{\frac {\sin(x)}{x}}\,\mathrm {d} x$ cannot be interpreted as a Lebesgue integral, since

$\int _{0}^{\infty }\left|{\frac {\sin(x)}{x}}\right|\,\mathrm {d} x=\infty .$ But $f(x)=\sin(x)/x$ is nevertheless integrable between any two finite endpoints, and its integral between 0 and ∞ is usually understood as the limit of the integral:

$\int _{0}^{\infty }{\frac {\sin(x)}{x}}\,\mathrm {d} x=\lim _{b\rightarrow \infty }\int _{0}^{b}{\frac {\sin(x)}{x}}\,\mathrm {d} x={\frac {\pi }{2}}.$ ## 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.

## Cauchy principal value

{{#invoke:main|main}} Consider the difference in values of two limits:

$\lim _{a\rightarrow 0+}\left(\int _{-1}^{-a}{\frac {\mathrm {d} x}{x}}+\int _{a}^{1}{\frac {\mathrm {d} x}{x}}\right)=0,$ $\lim _{a\rightarrow 0+}\left(\int _{-1}^{-a}{\frac {\mathrm {d} x}{x}}+\int _{2a}^{1}{\frac {\mathrm {d} x}{x}}\right)=-\ln 2.$ The former is the Cauchy principal value of the otherwise ill-defined expression

$\int _{-1}^{1}{\frac {\mathrm {d} x}{x}}{\ }\left({\mbox{which}}\ {\mbox{gives}}\ -\infty +\infty \right).$ Similarly, we have

$\lim _{a\rightarrow \infty }\int _{-a}^{a}{\frac {2x\,\mathrm {d} x}{x^{2}+1}}=0,$ but

$\lim _{a\rightarrow \infty }\int _{-2a}^{a}{\frac {2x\,\mathrm {d} x}{x^{2}+1}}=-\ln 4.$ The former is the principal value of the otherwise ill-defined expression

$\int _{-\infty }^{\infty }{\frac {2x\,\mathrm {d} x}{x^{2}+1}}{\ }\left({\mbox{which}}\ {\mbox{gives}}\ -\infty +\infty \right).$ 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

$\int _{0}^{\infty }f(x)\,\mathrm {d} x$ is Cesàro summable (C, α) if

$\lim _{\lambda \to \infty }\int _{0}^{\lambda }\left(1-{\frac {x}{\lambda }}\right)^{\alpha }f(x)\,\mathrm {d} x$ exists and is finite Template:Harv. The value of this limit, should it exist, is the (C, α) sum of the integral.

An integral is (C, 0) summable 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

$\int _{0}^{\infty }\sin x\,\mathrm {d} x$ which fails to exist as an improper integral, but is (C,α) summable for every α > 0. This is an integral version of Grandi's series.

## Bibliography

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