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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

- $lim\_\{btoinfty\}\; int\_a^bf(x),\; dx,\; qquad\; lim\_\{ato\; -infty\}\; int\_a^bf(x),\; dx,$

- $lim\_\{cto\; b^-\}\; int\_a^cf(x),\; dx,quad$

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.

- $int\_1^infty\; frac\{1\}\{x^2\},dx$

- $lim\_\{btoinfty\}int\_1^bfrac\{1\}\{x^2\},dx=lim\_\{btoinfty\}left[-frac\{1\}\{b\}+frac\{1\}\{1\}right]\; =\; 1$

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

- $int\_0^1\; frac\{1\}\{sqrt\{x\}\},dx.$

- $lim\_\{ato\; 0^+\}int\_a^1frac\{1\}\{sqrt\{x\}\},\; dx\; =\; lim\_\{ato\; 0^+\}left[2sqrt\{1\}-2sqrt\{a\}right]=2,$

- $int\_\{-infty\}^\{infty\}\; f(x)\; ,\; dx\; =\; int\_\{-infty\}^\{infty\}\; f(x\; -\; 1/x)\; ,\; dx$

- $lim\_\{ttoinfty\}\; int\_a^t\; f(x),\; dx$

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

- $lim\_\{btoinfty\}int\_1^b\; frac\{1\}\{x\},dx\; =\; infty.$

- $lim\_\{btoinfty\}int\_1^b\; xsin\; x,\; dx,$

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

- $int\_\{-infty\}^infty\; f(x),\; dx$

- $int\_\{-infty\}^infty\; f(x),\; dx\; =\; lim\_\{ato\; -infty\}lim\_\{bto\; infty\}\; int\_a^bf(x),dx,$

- $lim\_\{ato\; -infty\}int\_a^cf(x),\; dx\; +\; lim\_\{bto\; infty\}\; int\_c^bf(x),dx$

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

- $lim\_\{ato\; -infty\}int\_a^cx,dx+lim\_\{btoinfty\}int\_c^bx,dx$

- $p.v.int\_\{-infty\}^infty\; x,dx=lim\_\{btoinfty\}int\_\{-b\}^bx,dx\; =\; 0.$

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.

- 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 $int\_1^infty\; frac\{1\}\{x^2\},dx$. On the other hand, there are also integrals that have an improper Riemann integral do not have a (proper) Lebesgue integral, such as $int\_0^infty\; frac\{sin\; x\}\{x\},dx$. The Lebesgue theory does not see this as a deficiency: from the point of view of measure theory, $int\_0^infty\; frac\{sin\; x\}\{x\},dx\; =\; infty\; -\; infty$ 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.

In some cases, the integral

- $int\_a^c\; f(x),dx,$

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

- $lim\_\{bto\; c^-\}int\_a^b\; f(x),dx,$

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

- $int\_a^b|f(x)|,dx$

- are bounded as b → ∞, then the improper Riemann integrals

- $int\_a^infty\; f(x),\; dx,quadmbox\{and\}\; int\_a^infty\; |f(x)|,\; dx$

- both exist. Furthermore, f is Lebesgue integrable on [a, ∞), and its Lebesgue integral is equal to its improper Riemann integral.

For example, the integral

- $int\_0^inftyfrac\{dx\}\{1+x^2\}$

- $lim\_\{btoinfty\}int\_0^bfrac\{dx\}\{1+x^2\}=lim\_\{btoinfty\}arctan\{b\}=frac\{pi\}\{2\},$

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(x) dx 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,

- $int\_0^inftyfrac\{sin(x)\}\{x\},dx$

cannot be interpreted as a Lebesgue integral, since

- $int\_0^inftyleft|frac\{sin(x)\}\{x\}right|,dx=infty.$

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

- $int\_0^inftyfrac\{sin(x)\}\{x\},dx=lim\_\{brightarrowinfty\}int\_0^bfrac\{sin(x)\}\{x\},dx=frac\{pi\}\{2\}.$

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.

- $lim\_\{arightarrow\; 0+\}left(int\_\{-1\}^\{-a\}frac\{dx\}\{x\}+int\_a^1frac\{dx\}\{x\}right)=0,$

- $lim\_\{arightarrow\; 0+\}left(int\_\{-1\}^\{-a\}frac\{dx\}\{x\}+int\_\{2a\}^1frac\{dx\}\{x\}right)=-ln\; 2.$

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

- $int\_\{-1\}^1frac\{dx\}\{x\}\{\; \}$

Similarly, we have

- $lim\_\{arightarrowinfty\}int\_\{-a\}^afrac\{2x,dx\}\{x^2+1\}=0,$

but

- $lim\_\{arightarrowinfty\}int\_\{-2a\}^afrac\{2x,dx\}\{x^2+1\}=-ln\; 4.$

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

- $int\_\{-infty\}^inftyfrac\{2x,dx\}\{x^2+1\}\{\; \}$

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.

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

- $int\_0^infty\; f(x),dx$

is Cesàro summable (C, α) if

- $lim\_\{lambdatoinfty\}int\_0^lambdaleft(1-frac\{x\}\{lambda\}right)^alpha\; f(x),\; dx$

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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- Numerical Methods to Solve Improper Integrals at Holistic Numerical Methods Institute

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Last updated on Monday September 15, 2008 at 23:07:48 PDT (GMT -0700)

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Last updated on Monday September 15, 2008 at 23:07:48 PDT (GMT -0700)

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