Improper integral

An improper integral is a term from the mathematical branch of analysis . With the help of this integral term it is possible to integrate functions which have individual singularities or whose domain is unlimited and which therefore cannot be integrated in the actual sense .

The improper integral can be understood as an extension of the Riemann integral , the Lebesgue integral or other integration terms. Often, however, it is considered in connection with the Riemann integral, since in particular the (actual) Lebesgue integral can already integrate many functions that are only improperly Riemann integrable.

definition

There are two reasons why improper integrals are considered. On the one hand, you want to integrate functions across unlimited areas, for example from to . This is simply not possible with the Riemann integral. Improper integrals that solve this problem are called improper integrals of the first kind. It is also of interest to integrate functions that have a singularity on the edge of their domain . Inauthentic integrals that make this possible are called improper integrals of the second kind. It is possible that improper integrals are improperly of the first kind on one limit and improperly of the second kind on the other. However, it is irrelevant for the definition of the improper integral what kind the integral is. ${\displaystyle -\infty }$ ${\displaystyle \infty }$

Integration area with a critical limit

Be and a function . So the improper integral in the case of convergence is defined by ${\displaystyle -\infty <a<b\leq \infty }$ ${\displaystyle f\colon {[a,b[}\to \mathbb {R} }$

${\displaystyle \int _{a}^{b}f(x)\mathrm {d} x:=\lim _{\beta \nearrow b}\int _{a}^{\beta }f(x)\mathrm {d} x\,.}$

The improper integral for and is defined analogously . ${\displaystyle -\infty \leq a<b<\infty }$ ${\displaystyle f\colon {]a,b]}\to \mathbb {R} }$

Integration area with two critical limits

Be and a function . So the improper integral in the case of convergence is defined by ${\displaystyle -\infty \leq a<b\leq \infty }$ ${\displaystyle f\colon {]a,b[}\to \mathbb {R} }$

${\displaystyle \int _{a}^{b}f(x)\mathrm {d} x:=\int _{a}^{c}f(x)\mathrm {d} x+\int _{c}^{b}f(x)\mathrm {d} x\,,}$

where applies and the two right integrals are improper integrals with a critical limit. That means written out ${\displaystyle a<c<b}$

${\displaystyle \int _{a}^{b}f(x)\mathrm {d} x:=\lim _{\alpha \searrow a}\int _{\alpha }^{c}f(x)\mathrm {d} x+\lim _{\beta \nearrow b}\int _{c}^{\beta }f(x)\mathrm {d} x\,.}$

The convergence and the value of the integral does not depend on the choice of . ${\displaystyle c}$

Examples

Two broken rational functions

If an antiderivative is known, the integral at the neighboring point can be evaluated as in the actual case and then the limit value for can be calculated. One example is the integral ${\displaystyle \beta }$ ${\displaystyle \beta \nearrow b}$

${\displaystyle \int _{0}^{1}{\frac {\mathrm {d} x}{\sqrt {x}}}=2{\sqrt {x}}{\Big |}_{0}^{1}=2\,,}$

in which the integrand at has a singularity and therefore does not exist as a (actual) Riemann integral. If one understands the integral as an improper Riemann integral of the second kind, then the following applies ${\displaystyle x=0}$

${\displaystyle \lim _{\alpha \searrow 0}\int _{\alpha }^{1}{\frac {1}{\sqrt {x}}}\,\mathrm {d} x=\lim _{\alpha \searrow 0}\left[2{\sqrt {1}}-2{\sqrt {\alpha }}\right]=2\,.}$

The integral

${\displaystyle \int _{1}^{\infty }{\frac {1}{x^{2}}}\,\mathrm {d} x}$

has an unlimited domain and is therefore an improper integral of the first kind. It holds

${\displaystyle \lim _{\beta \to \infty }\int _{1}^{\beta }{\frac {1}{x^{2}}}\,\mathrm {d} x=\lim _{\beta \to \infty }\left[-{\frac {1}{\beta }}+{\frac {1}{1}}\right]=1\,.}$

Gaussian error integral

→ Main article : Error integral

The Gaussian error integral

${\displaystyle \int _{-\infty }^{\infty }e^{-{\frac {1}{2}}x^{2}}\,\mathrm {d} x={\sqrt {2\pi }}}$

is an improper Riemann integral of the first kind. In the sense of Lebesgue's integration theory , the integral also exists in the actual sense.

Relationship between proper and improper Riemann and Lebesgue integrals

• Every Riemann-integrable function can also be Lebesgue-integrable.
• Thus every improperly Riemann-integrable function can also improperly Lebesgue-integrable.
• There are functions that are improperly Riemann-integrable, but not Lebesgue-integrable, consider the integral, for example

${\displaystyle \int _{1}^{\infty }{\frac {\sin x}{x}}\,\mathrm {d} x.}$

(It does not exist in the Lebesgue sense, since for every Lebesgue-integrable function its absolute value can also be Lebesgue-integrable, which is associated with useful properties of the function spaces defined by the Lebesgue integral , which are therefore lost in the improper Lebesgue integral).

• On the other hand, there are functions that can be Lebesgue-integrable, but not improperly Riemann-integrable. For example, consider the Dirichlet function on a limited interval.

Web links

Commons : Improper integral  - collection of images, videos and audio files

• Christoph Bock: Elements of Analysis (PDF; 1.2 MB) Section 8.33

Individual evidence

1. ↑ ^{a b} Konrad Königsberger : Analysis 1 . Springer-Verlag, Berlin et al., 2004, ISBN 3-540-41282-4 , p. 218.