Quod Erat Demonstrandum

2010/01/04

廣義積分定義

Filed under: Pure Mathematics — johnmayhk @ 2:57 下午
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聖誕假前某天,中七同學大師問了幾個問題,包括

\int_{-\infty}^{\infty} f(x)dx

何義?

我當時答

\displaystyle \lim_{a \rightarrow \infty}\int_{-a}^{a} f(x)dx

大師質疑:「上下極限為何一定要設定為 a 和 -a?」

是,我也不知道。維基一下,看到

\int_{-\infty}^{\infty} f(x)dx = \displaystyle \lim_{a \rightarrow -\infty} \displaystyle \lim_{b \rightarrow \infty}\int_{a}^{b} f(x)dx

較一般吧。

誠如當時大師舉了

\int_{-\infty}^{\infty} xdx

用上下限為 \pm a 的定義,得

\int_{-\infty}^{\infty} xdx = \displaystyle \lim_{a \rightarrow \infty}\int_{-a}^{a} xdx = 0(或以更一般的 odd function 取代 x 也可)

但若考慮(比如說)

\displaystyle \lim_{a \rightarrow \infty}\int_{-a}^{2a} xdx = +\infty

\int_{-\infty}^{\infty} xdx 是什麼?

另外,在證明

時,其中一個方法也是從 \displaystyle \lim_{a \rightarrow \infty}\int_{-a}^{a} e^{-x^2}dx 入手。

究竟,此等廣義積分(improper integrals)之定義為何?有待高手指示了。

* Don’t worry, the topic of improper integrals is out-of-syllabus now.

3 則迴響 »

  1. You are using a limit of a Riemann integral to define an indefinite integral, that’s why you see the problem. If you define the indefinite integral as a Lebesgue integral, the problem is solved because there is no need to have the limit. You are just doing integration on the set of all real numbers. In deed, to make the Lebesgue integral equal to Riemann integral, you need the upper limit and lower limit to be +a and -a respectively before doing the limit. Otherwise, you are weighting the positive part of the real line and the negative part of the real line different.

    迴響 由 Alienz Thame — 2010/01/05 @ 1:19 上午 | 回覆

    • Thank you Alienz Thame!

      That means, to match the Lebesgue and Riemann, we MUST take +a and -a to ensure even weighting of the positive and negative parts on the real line.

      迴響 由 johnmayhk — 2010/01/06 @ 5:21 下午 | 回覆

      • I am not sure if setting +a to -a is a necessary condition but surely it is sufficient. In mathematics, dealing with infinity is always not intuitive!

        迴響 由 Alienz Thame — 2010/01/07 @ 5:30 上午


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