# Improper integral

 Topics in calculus Differentiation Integration

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, or both endpoints, of the interval approaches either a specified real number or ∞ or −∞.

If the function f being integrated from a to c has a discontinuity at c, especially in the form of a vertical asymptote, or if c = ∞, then there may be no more convenient way to compute the integral

[itex]\int_a^c f(x)\,dx\,[itex]

than by finding the limit

[itex]\lim_{b\to c-}\int_a^b f(x)\,dx.\,[itex]

In some cases, the integral from a to c is not even defined, because the integrals of the positive and negative parts of f(xdx 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.

The integral

[itex]\int_0^\infty\frac{dx}{1+x^2}[itex]

can be interpreted as

[itex]\lim_{b\rightarrow\infty}\int_0^b\frac{dx}{1+x^2}.[itex]

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,

[itex]\int_0^\infty\frac{\sin(x)}{x}\,dx[itex]

cannot be interpreted as a Lebesgue integral, since

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

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

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

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.

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## Infinite bounds of integration

The most basic of improper integrals are integrals such as:

[itex]\int_0^\infty {dx \over x^2+1}.[itex]

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

[itex]\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.[itex]

## Vertical asymptotes at bounds of integration

Consider

[itex]\int_0^1 \frac{dx}{x^{2/3}}.[itex]

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)(x1/3);

which can be evaluated by direct substitution to give the value

3 × (1 − b1/3).

The limit as b → 0 is 3 − 0 = 3

## Cauchy principal values

Consider the difference in values of two limits:

[itex]\lim_{a\rightarrow 0+}\left(\int_{-1}^{-a}\frac{dx}{x}+\int_a^1\frac{dx}{x}\right)=0,[itex]
[itex]\lim_{a\rightarrow 0+}\left(\int_{-1}^{-a}\frac{dx}{x}+\int_{2a}^1\frac{dx}{x}\right)=-\ln 2.[itex]

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

[itex]\int_{-1}^1\frac{dx}{x}{\ }

\left(\mbox{which}\ \mbox{gives}\ -\infty+\infty\right).[itex]

Similarly, we have

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

but

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

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

[itex]\int_{-\infty}^\infty\frac{2x\,dx}{x^2+1}{\ }

\left(\mbox{which}\ \mbox{gives}\ -\infty+\infty\right).[itex]

All of the above limits are cases of the indeterminate form ∞ − ∞.

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

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