Improper integral

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

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.

Figure 1

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.

Figure 2

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.

1 Infinite bounds of integration
2 Vertical asymptotes at bounds of integration
3 Cauchy principal value
4 External links

[edit] Infinite bounds of integration

The most basic of improper integrals are integrals 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 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 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. See Cauchy principle value below.

[edit] Vertical asymptotes at bounds of integration

Consider

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

This integral involves a function with a vertical asymptote at x = 0.

One can evaluate this integral by evaluating from b (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 only if the limit converges. Here is an example of an integral which does not converge:

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

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:

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