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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.
Formulation
The definite integral as commonly defined in calculus texts is only defined for a bounded function Failed to parse (Missing texvc executable; please see math/README to configure.): f
defined on a closed and bounded interval Failed to parse (Missing texvc executable; please see math/README to configure.): [a,b]
, where a and b are real numbers (not infinity, which is not a real number), for instance Failed to parse (Missing texvc executable; please see math/README to configure.): \int_0^1 x^2\,dx . (See Types of integral below for technicals.)
The key idea of integration is to compute areas as a limit of a finite sum of areas of rectangles.
Thus expressions of the form Failed to parse (Missing texvc executable; please see math/README to configure.): \int_1^\infty \frac{1}{x^2}\,dx
or Failed to parse (Missing texvc executable; please see math/README to configure.): \int_0^1 \frac{1}{\sqrt{x}}\,dx
do not make sense as definite integrals without some expanded definition, which role improper integrals fill.
This is because one computes an integral over an interval by approximating it by finitely many rectangles: subdivide the interval into finitely many pieces, each of finitely width and finite height, so the total area is defined, and then taking the limit as one subdivides the interval more and more finely.
This can break down in two ways:
- unbounded interval
- For example Failed to parse (Missing texvc executable; please see math/README to configure.): [1,\infty)
- in this case one cannot subdivide the interval into finitely many intervals of finite length: one must either use infinitely many rectangles (which yields an infinite sum), or use some rectangles which are infinitely large, hence having infinite area.
- unbounded function
- The function goes to infinity at some point, like Failed to parse (Missing texvc executable; please see math/README to configure.): \frac{1}{\sqrt{x}} \to +\infty
as Failed to parse (Missing texvc executable; please see math/README to configure.): x \to 0^+
. In this case one must approximate the function by an infinitely high rectangle, which has infinite area.[1]
Thus instead one defines the improper integral as being, not itself an integral in the above sense (the limit of the sum of areas of finitely many rectangles of finite size), but as a limit of integrals. It is thus a double limit: each proper integral is a limit, and the improper integral is a limit of these limits.
Formally, an improper integral is defined as
- Failed to parse (Missing texvc executable; please see math/README to configure.): \int_a^b f(x)\,dx := \lim_{c\to b^-} \int_a^c f(x)\,dx \qquad\text{or}\qquad \int_a^b f(x)\,dx := \lim_{c\to a^+} \int_c^b f(x)\,dx,
assuming that the proper definite integrals on the right hand side are defined, and that the limit of these integrals is itself defined.
One does this either if one of the endpoints is infinite, or if the function f "blows up" at one of the endpoints, meaning Failed to parse (Missing texvc executable; please see math/README to configure.): f(x) \to \infty
or Failed to parse (Missing texvc executable; please see math/README to configure.): f(x) \to -\infty
.
Cautions
- Infinite value
- Note that the value can be infinity, if the proper integrals are defined and diverge to infinity, for instance Failed to parse (Missing texvc executable; please see math/README to configure.): \int_1^\infty \frac{1}{x}\,dx = \infty
.
- Only one endpoint at a time
- One can only take the limit as one varies a single endpoint. One can sometimes define improper integrals where both endpoints are infinite, such as the Gaussian integral Failed to parse (Missing texvc executable; please see math/README to configure.): \int_{-\infty}^\infty e^{-x^2}\,dx = \sqrt{\pi}
, but one cannot even define Failed to parse (Missing texvc executable; please see math/README to configure.): \int_{-\infty}^\infty x\,dx
unambiguously, since the order in which you go to infinity matters:
- Failed to parse (Missing texvc executable; please see math/README to configure.): \begin{align} &\lim_{a \to \infty} \int_{-a}^a x\,dx &&= \lim_{a \to \infty} 0 &&= 0\\ &\lim_{a \to \infty} \int_{-a}^{a+1} x\,dx &&= \lim_{a \to \infty} a + \frac{1}{2} &&= +\infty\\ &\lim_{a \to \infty} \int_{-a}^{a-1} x\,dx &&= \lim_{a \to \infty} -a + \frac{1}{2} &&= -\infty \end{align}
Further, if one takes the limit as first one endpoint goes to infinity, then the other, then one gets:
- Failed to parse (Missing texvc executable; please see math/README to configure.): \begin{align} &\lim_{a\to -\infty} \lim_{b\to \infty} \int_a^b x\,dx &&= \lim_{a\to -\infty} \infty = \infty\\ &\lim_{b\to \infty} \lim_{a\to -\infty} \int_a^b x\,dx &&= \lim_{b\to \infty} -\infty = -\infty \end{align}
In this case, one can however define an improper integral in the sense of Cauchy principal value.
Questions
The basic questions of the theory are therefore:
The second part can be addressed by calculus techniques, but also in some cases by contour integration, Fourier transforms and other more advanced methods.
Types of integrals
Improper integrals are more or less useful for various notions of integration.
- 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).
- For the Riemann integral, improper integration is necessary for unbounded intervals (as for Darboux) but is not necessary for unbounded functions (since the tags ensure that the Riemann sums are always finite).
- For the Lebesgue integral, deals differently with unbounded domains and unbounded functions, and one does not distinguish an "improper Lebesgue integral": some integrals that require an improper Riemann integral have a Lebesgue integral (such as Failed to parse (Missing texvc executable; please see math/README to configure.): \int_1^\infty \frac{1}{x^2}\,dx
), while other integrals that have an improper Riemann integral do not have a Lebesgue integral, such as Failed to parse (Missing texvc executable; please see math/README to configure.): \int_0^\infty \sin x\,dx . The Lebesgue theory does not see this as a deficiency: from the point of view of measure theory, Failed to parse (Missing texvc executable; please see math/README to configure.): \int_0^\infty \sin x\,dx = \infty - \infty
and cannot be defined satisfactorily.
- 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.
Notation
It is common to use notation that is reminiscent of a typical integral, however, each symbol of these symbols stand for an improper integral.
- Failed to parse (Missing texvc executable; please see math/README to configure.): \int_a^\infty f(x)\,dx\, := \lim_{t\to \infty}\int_a^t f(x)\,dx\,
- Failed to parse (Missing texvc executable; please see math/README to configure.): \int_{-\infty} ^ {b} f(x)\,dx\,:= \lim_{t\to -\infty}\int_t^b f(x)\,dx\,
- Failed to parse (Missing texvc executable; please see math/README to configure.): \int_{-\infty}^{\infty} f(x)\,dx\, := \lim_{t\to -\infty}\int_t^a f(x)\,dx\,+\lim_{t\to \infty} \int_a^t f(x)\,dx\,
- Failed to parse (Missing texvc executable; please see math/README to configure.): \int_a^b f(x)\,dx\, := \lim_{t\to b^-}\int_a^t f(x)\,dx, \,\mbox{where} \,\, \lim_{x\to b^-} |f(x)| = \infty
- Failed to parse (Missing texvc executable; please see math/README to configure.): \int_a^b f(x)\,dx\, := \lim_{t\to a^+}\int_t^b f(x)\,dx, \,\mbox{where} \,\, \lim_{x\to a^+} |f(x)| = \infty
- Failed to parse (Missing texvc executable; please see math/README to configure.): \int_a^b f(x)\,dx\,:=\lim_{t\to c^-} \int_a^t f(x)\,dx\, + \lim_{t\to c^+ } \int_t^b f(x)\,dx, \,\mbox{where} \,\, \lim_{x\to c} |f(x)| = \infty
Issues of definition
In some cases, the integral
- Failed to parse (Missing texvc executable; please see math/README to configure.): \int_a^c f(x)\,dx\,
can be defined without reference to the limit
- Failed to parse (Missing texvc executable; please see math/README to configure.): \lim_{b\to 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 some cases, 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.
Issues of interpretation
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 other words, in calculus discussions of what expressions with the integral sign means). In studying improper integrals, it can matter which integration theory is in play.
The integral
- Failed to parse (Missing texvc executable; please see math/README to configure.): \int_0^\infty\frac{dx}{1+x^2}
can be interpreted as
- Failed to parse (Missing texvc executable; please see math/README to configure.): \lim_{b\to\infty}\int_0^b\frac{dx}{1+x^2}=\lim_{b\to\infty}\arctan{b}=\frac{\pi}{2},
but from the point of view of mathematical analysis it is not necessary to interpret it that way, since it may be interpreted instead as a Lebesgue integral over the set (0, ∞). On the other hand, the use of the limit of definite integrals over finite ranges is clearly useful, if only as a way to calculate actual values.
In contrast,
- Failed to parse (Missing texvc executable; please see math/README to configure.): \int_0^\infty\frac{\sin(x)}{x}\,dx
cannot be interpreted as a Lebesgue integral, since
- Failed to parse (Missing texvc executable; please see math/README to configure.): \int_0^\infty\left|\frac{\sin(x)}{x}\right|\,dx=\infty.
This is therefore a "properly" improper integral, whose value is given by
- Failed to parse (Missing texvc executable; please see math/README to configure.): \int_0^\infty\frac{\sin(x)}{x}\,dx=\lim_{b\rightarrow\infty}\int_0^b\frac{\sin(x)}{x}\,dx=\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.
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.
Infinite bounds of integration
The most basic of improper integrals are integrals with an unbounded interval of integration such as:
- Failed to parse (Missing texvc executable; please see math/README to configure.): \int_0^\infty {dx \over x^2+1}.
As stated above, this need not be defined as an improper integral, since it can be construed as a Lebesgue integral instead. Nonetheless, for purposes of actually computing this integral, it is more convenient to treat it as an improper integral, i.e., to evaluate it when the upper bound of integration is finite and then take the limit as that bound approaches ∞. The antiderivative of the function being integrated is arctan x. The integral is
- Failed to parse (Missing texvc executable; please see math/README to configure.): \lim_{b\rightarrow\infty}\int_0^b\frac{dx}{1+x^2}=\lim_{b\rightarrow\infty}\arctan b-\arctan 0=\pi/2-0=\pi/2.
The improper integral converges if and only if the limit converges. Here is an example of an integral which does not converge:
- Failed to parse (Missing texvc executable; please see math/README to configure.): \int_1^\infty {dx \over x} = \lim_{b\rightarrow\infty}\int_1^b\frac{dx}{x}=\lim_{b\rightarrow\infty}\ln b=\infty
Sometimes both bounds will be infinite. In such a case it can be broken up into the sum of two improper integrals, one on each half:
- Failed to parse (Missing texvc executable; please see math/README to configure.): \int_{-\infty}^{+\infty} f(x) \,dx = \int_{-\infty}^a f(x) \,dx + \int_a^{+\infty} f(x) \,dx
where a is an arbitrary finite number.
In this case the improper integral converges if and only if both integrals converge. If one integral diverges to positive infinity, and the other diverges to negative infinity, then the integral is indeterminate, and you can get different answers depending on how the two limits for the two integrals are related. The Cauchy principal value addresses this issue.
Vertical asymptotes at bounds of integration
Some improper integral involve a function with a vertical asymptote, such as the following with an asymptote at x = 0
- Failed to parse (Missing texvc executable; please see math/README to configure.): \int_0^1 \frac{dx}{x^{2/3}}.
One can evaluate this integral by evaluating from b (a number greater than 0) to 1, and then take the limit as b approaches 0 from the right (since the interval we are integrating over is to the right of 0). One should note that the antiderivative of the above function is Failed to parse (Missing texvc executable; please see math/README to configure.): 3 x^{1/3},
so the integral can be evaluated as
- Failed to parse (Missing texvc executable; please see math/README to configure.): \lim_{b\rightarrow 0^+}\int_b^1\frac{dx}{x^{2/3}}=3 \cdot 1^{1/3}-\lim_{b\rightarrow 0^+}3 b^{1/3}=3-0=3.
The improper integral converges if and only if the limit converges. Here is an example of an integral which does not converge:
- Failed to parse (Missing texvc executable; please see math/README to configure.): \int_0^1 {dx \over x} = \lim_{b\rightarrow 0^+}\int_b^1\frac{dx}{x}=0-\lim_{b\rightarrow 0^+}\ln b=\infty
Sometimes you integrate over an interval that crosses a vertical asymptote. In such a case, you can break the integral up into the sum of two improper integrals, one on each side:
- Failed to parse (Missing texvc executable; please see math/README to configure.): \int_a^c f(x) \,dx = \int_a^b f(x) \,dx + \int_b^c f(x) \,dx
where b is the location of a vertical asymptote.
In this case the improper integral converges if and only if both integrals converge. If one integral diverges to positive infinity, and the other diverges to negative infinity, then the integral is indeterminate, and you can get different answers depending on how the two limits for the two integrals are related. The Cauchy principal value addresses this issue.
Cauchy principal value
-
Consider the difference in values of two limits:
- Failed to parse (Missing texvc executable; please see math/README to configure.): \lim_{a\rightarrow 0+}\left(\int_{-1}^{-a}\frac{dx}{x}+\int_a^1\frac{dx}{x}\right)=0,
- Failed to parse (Missing texvc executable; please see math/README to configure.): \lim_{a\rightarrow 0+}\left(\int_{-1}^{-a}\frac{dx}{x}+\int_{2a}^1\frac{dx}{x}\right)=-\ln 2.
The former is the Cauchy principal value of the otherwise ill-defined expression
- Failed to parse (Missing texvc executable; please see math/README to configure.): \int_{-1}^1\frac{dx}{x}{\ } \left(\mbox{which}\ \mbox{gives}\ -\infty+\infty\right).
Similarly, we have
- Failed to parse (Missing texvc executable; please see math/README to configure.): \lim_{a\rightarrow\infty}\int_{-a}^a\frac{2x\,dx}{x^2+1}=0,
but
- Failed to parse (Missing texvc executable; please see math/README to configure.): \lim_{a\rightarrow\infty}\int_{-2a}^a\frac{2x\,dx}{x^2+1}=-\ln 4.
The former is the principal value of the otherwise ill-defined expression
- Failed to parse (Missing texvc executable; please see math/README to configure.): \int_{-\infty}^\infty\frac{2x\,dx}{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.
External links
ca:Integral impròpia da:Uegentligt integral ro:Integrală improprie es:Integral impropia fr:Intégrale impropre it:Integrale improprio he:אינטגרל לא אמיתי nl:Oneigenlijke integraal ja:広義積分 pl:Całka niewłaściwa ru:Несобственный интеграл
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