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Improper integral

It is recommended that the reader be familiar with antiderivatives, integrals, and limitss.

In calculus, an improper integral is defined as the limit of a definite integral, as an endpoint, or both endpoints, of the interval approaches either a specified real number or ∞ or −∞. The integral

can be interpreted as

but 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

cannot be interpreted as a Lebesgue integral, since

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

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.

Table of contents
1 Infinite bounds of integration
2 Vertical asymptotes at bounds of integration
3 Cauchy principal values

Infinite bounds of integration

The most basic of improper integrals are integrals such as: ∫0 dx / (x2 + 1). Such an integral can be evaluted by noting the antiderivative: arctan x. The integral is

Vertical asymptotes at bounds of integration

Consider ∫01 dx / x(2/3). This integral involves a function with a vertical asymptote at x = 0.

One can evaluate this integral by evaluating from b to 1, and then take the limit as b approaches 0. One should note that the antiderivative of the above function is (3/2)(x(2/3)); which can be evaluated by direct substition: (3/2)(1 − b(2/3)). The limit as b approaches 0 equals: (3/2) − 0 = 3/2 = 1.5.

Cauchy principal values

Consider the difference in values of two limits:

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

Similarly, we have


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

These pathologies do not afflict "Lebesgue-integrable" functions, that is, functions the integrals of whose absolute values are finite.