“improper integrals”的概念、定义、翻译、参考文献-科学参考

improper integrals

Mathematics

Riemann integration deals only with bounded functions on bounded intervals. But other integrals may be considered as improper integrals. The first kind is one in which the interval of integration is infinite as, for example, in

$\int_{a}^{\infty} f (x) d x .$

This integral exists as an improper integral and has value l, if the the integral from a to X tends to a limit l as X → ∞. For example,

$\int_{1}^{X} \frac{1}{x^{2}} d x = 1 - \frac{1}{X}$

and, as X → ∞, the right‐hand side tends to 1. So, as an improper Riemann integral

$\int_{1}^{\infty} \frac{1}{x^{2}} d x = 1 .$

Similar definitions can be given for improper integrals from −∞ to a and from −∞ to ∞.

A second kind of improper integral is one in which the function becomes infinite at some point. For example, in

$\int_{0}^{1} \frac{1}{\sqrt{x}} d x .$

The function $1 / \sqrt{x}$ is not bounded on the interval (0, 1], and so is not Riemann integrable. However, the function is bounded on the interval [δ‎, 1], where 0<δ‎<1, and

$\int_{δ}^{1} \frac{1}{\sqrt{x}} d x = 2 - 2 \sqrt{δ} .$

As δ‎ → 0, the right‐hand side tends to the limit 2. So the integral above, from 0 to 1, exists as an improper integral with value 2.

单词	improper integrals
释义	improper integrals Mathematics Riemann integration deals only with bounded functions on bounded intervals. But other integrals may be considered as improper integrals. The first kind is one in which the interval of integration is infinite as, for example, in $\int_{a}^{\infty} f (x) d x .$ This integral exists as an improper integral and has value l, if the the integral from a to X tends to a limit l as X → ∞. For example, $\int_{1}^{X} \frac{1}{x^{2}} d x = 1 - \frac{1}{X}$ and, as X → ∞, the right‐hand side tends to 1. So, as an improper Riemann integral $\int_{1}^{\infty} \frac{1}{x^{2}} d x = 1 .$ Similar definitions can be given for improper integrals from −∞ to a and from −∞ to ∞. A second kind of improper integral is one in which the function becomes infinite at some point. For example, in $\int_{0}^{1} \frac{1}{\sqrt{x}} d x .$ The function $1 / \sqrt{x}$ is not bounded on the interval (0, 1], and so is not Riemann integrable. However, the function is bounded on the interval [δ‎, 1], where 0<δ‎<1, and $\int_{δ}^{1} \frac{1}{\sqrt{x}} d x = 2 - 2 \sqrt{δ} .$ As δ‎ → 0, the right‐hand side tends to the limit 2. So the integral above, from 0 to 1, exists as an improper integral with value 2.
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