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
- 1.1 Cautions
- 1.2 Questions
2 Types of integrals
3 Notation
4 Issues of definition
5 Issues of interpretation
6 Singularities
7 Infinite bounds of integration
8 Vertical asymptotes at bounds of integration
9 Cauchy principal value
10 References
11 External links

[edit] Formulation

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The Riemann integral is only be defined for a bounded function f (subject to certain other requirements) on a closed and bounded interval, [a,b], where a and b are real numbers. The Riemann integral is computed by taking the limit of finite sums of rectangles which approximate the signed area under the graph of a function. Thus expressions of the form $\int_1^\infty \frac{1}{x^2}\,dx$ or $\int_0^1 \frac{1}{\sqrt{x}}\,dx$ do not make sense as Riemann integrals:

In the first example, the interval $[1,\infty)$ is unbounded. 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.
In the second example, the function is unbounded: it tends to infinity at an endpoint of the domain: $\frac{1}{\sqrt{x}} \to +\infty$ as $x \to 0^+$ . In this case one must approximate the function by an infinitely high rectangle, which has infinite area.

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

$\int_a^b f(x)\,dx := \lim_{c\to b^-} \int_a^c f(x)\,dx \qquad\text{or}\qquad \int_a^b f(x)\,dx := \lim_{c\to a^+} \int_c^b f(x)\,dx,$

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 $f(x) \to \infty$ or $f(x) \to -\infty$ .

[edit] Cautions

Infinite value: Note that the value can be infinity, if the proper integrals are defined and diverge to infinity, for instance $\int_1^\infty \frac{1}{x}\,dx = \infty$ .

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 $\int_{-\infty}^\infty e^{-x^2}\,dx = \sqrt{\pi}$ , but one cannot even define $\int_{-\infty}^\infty x\,dx$ unambiguously, since the order in which you go to infinity matters:; $\begin{align} &\lim_{a \to \infty} \int_{-a}^a x\,dx &&= \lim_{a \to \infty} 0 &&= 0\\ &\lim_{a \to \infty} \int_{-a}^{a+1} x\,dx &&= \lim_{a \to \infty} a + \frac{1}{2} &&= +\infty\\ &\lim_{a \to \infty} \int_{-a}^{a-1} x\,dx &&= \lim_{a \to \infty} -a + \frac{1}{2} &&= -\infty \end{align}$

Further, if one takes the limit as first one endpoint goes to infinity, then the other, then one gets:

$\begin{align} &\lim_{a\to -\infty} \lim_{b\to \infty} \int_a^b x\,dx &&= \lim_{a\to -\infty} \infty = \infty\\ &\lim_{b\to \infty} \lim_{a\to -\infty} \int_a^b x\,dx &&= \lim_{b\to \infty} -\infty = -\infty \end{align}$

In this case, one can however define an improper integral in the sense of Cauchy principal value.

[edit] Questions

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.

[edit] 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).
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.

[edit] 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.

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

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

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

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

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

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

[edit] Issues of definition

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.

[edit] 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

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

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

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

[edit] 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.

[edit] Infinite bounds of integration

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.

[edit] 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

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

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:

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