Improper integral

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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.

1 Formulation
2 Notation
3 Issues of definition
4 Issues of interpretation
5 Singularities
6 Infinite bounds of integration
7 Vertical asymptotes at bounds of integration
8 Cauchy principal value
9 External links

More precisely, an improper integral is not a kind of integral, but means an expression of the form

$\lim_{b\to c}\int_a^b f(x)\,dx$

where c is either ±∞, or a real number such that |f(x)| → ∞ as x → c^-.

$\lim_{b\to a}\int_b^c f(x)\,dx$

where a is either ±∞, or a real number such that |f(x)| → ∞ as x → a⁺.

The basic questions of the theory are therefore:

Does the limit exist (an issue of mathematical analysis)?
Can the limit be computed?

The second part can be addressed by calculus techniques, but also in some cases by contour integration, Fourier transforms and other more advanced methods.

It is common to use notation that is reminiscent of a typical integral, however, each symbol of these symbols stand for an improper integral.

$\int_a^\infty f(x)\,dx\, := \lim_{t\to \infty}\int_a^t f(x)\,dx\,$

$\int_{-\infty} ^ {b} f(x)\,dx\,:= \lim_{t\to -\infty}\int_t^b f(x)\,dx\,$

$\int_{-\infty}^{\infty} f(x)\,dx\, := \lim_{t\to -\infty}\int_t^a f(x)\,dx\,+\lim_{t\to \infty} \int_a^t f(x)\,dx\,$

$\int_a^b f(x)\,dx\, := \lim_{t\to b^-}\int_a^t f(x)\,dx, \,\mbox{where} \,\, \lim_{x\to b^-} |f(x)| = \infty$

$\int_a^b f(x)\,dx\, := \lim_{t\to a^+}\int_t^b f(x)\,dx, \,\mbox{where} \,\, \lim_{x\to a^+} |f(x)| = \infty$

$\int_a^b f(x)\,dx\,:=\lim_{t\to c^-} \int_a^t f(x)\,dx\, + \lim_{t\to c^+ } \int_t^b f(x)\,dx, \,\mbox{where} \,\, \lim_{x\to c} |f(x)| = \infty$

Figure 1.

Figure 2

In some cases, the integral

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

can be defined without reference to the limit

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

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.

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

$\int_0^\infty\frac{dx}{1+x^2}$

can be interpreted as

$\lim_{b\to\infty}\int_0^b\frac{dx}{1+x^2}=\lim_{b\to\infty}\arctan{b}=\frac{\pi}{2},$

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,

$\int_0^\infty\frac{\sin(x)}{x}\,dx$

cannot be interpreted as a Lebesgue integral, since

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

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

$\int_0^\infty\frac{\sin(x)}{x}\,dx=\lim_{b\rightarrow\infty}\int_0^b\frac{\sin(x)}{x}\,dx=\frac{\pi}{2}.$

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.

The most basic of improper integrals are integrals with an unbounded interval of integration such as:

$\int_0^\infty {dx \over x^2+1}.$

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

$\lim_{b\rightarrow\infty}\int_0^b\frac{dx}{1+x^2}=\lim_{b\rightarrow\infty}\arctan b-\arctan 0=\pi/2-0=\pi/2.$

The improper integral converges if and only if the limit converges. Here is an example of an integral which does not converge:

$\int_1^\infty {dx \over x} = \lim_{b\rightarrow\infty}\int_1^b\frac{dx}{x}=\lim_{b\rightarrow\infty}\ln b=\infty$

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:

$\int_{-\infty}^{+\infty} f(x) \,dx = \int_{-\infty}^a f(x) \,dx + \int_a^{+\infty} f(x) \,dx$

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.

Some improper integral involve a function with a vertical asymptote, such as the following with an asymptote at x = 0

$\int_0^1 \frac{dx}{x^{2/3}}.$

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 $3 x 1 / 3,$ so the integral can be evaluated as

$\lim_{b\rightarrow 0^+}\int_b^1\frac{dx}{x^{2/3}}=3 \cdot 1^{1/3}-\lim_{b\rightarrow 0^+}3 b^{1/3}=3-0=3.$

The improper integral converges if and only if the limit converges. Here is an example of an integral which does not converge:

$\int_0^1 {dx \over x} = \lim_{b\rightarrow 0^+}\int_b^1\frac{dx}{x}=0-\lim_{b\rightarrow 0^+}\ln b=\infty$

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:

$\int_a^c f(x) \,dx = \int_a^b f(x) \,dx + \int_b^c f(x) \,dx$

where b is the location of a vertical asymptote.

Main article: Cauchy principal value

Consider the difference in values of two limits:

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

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

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

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

Similarly, we have

$\lim_{a\rightarrow\infty}\int_{-a}^a\frac{2x\,dx}{x^2+1}=0,$

but

$\lim_{a\rightarrow\infty}\int_{-2a}^a\frac{2x\,dx}{x^2+1}=-\ln 4.$

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

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

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.

Numerical Methods to Solve Improper Integrals at Holistic Numerical Methods Institute

Retrieved from "http://en.wikipedia.org/wiki/Improper_integral"

Category: Mathematical analysis

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