Quod Erat Demonstrandum

2015/11/02

一個冇咩用嘅rule

Filed under: NSS,Pure Mathematics — johnmayhk @ 1:53 下午
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積分沒有甚麼 product rule,除非特殊情況,比如:

u=u(x)v=v(x) 二次可導,若 \frac{d^2u}{dx^2}=au\frac{d^2v}{dx^2}=bv 其中 a,b 為常數;則

\int uvdx=\frac{1}{b-a}(u\frac{dv}{dx}-v\frac{du}{dx})+C

先唔證明,用之。

e.g.1

\int e^x\sin xdx=?

u=e^x \Rightarrow \frac{d^2u}{dx^2}=e^x  故 a=1

v=\sin x \Rightarrow \frac{d^2v}{dx^2}=-\sin x  故 b=-1

於是

\int e^x\sin xdx=\frac{1}{-1-1}(e^x\cos x-\sin xe^x)+C=\frac{1}{2}e^x(\sin x-\cos x)+C

e.g.2

\int e^{3x}\cos 4xdx=?

u=e^{3x} \Rightarrow \frac{d^2u}{dx^2}=9e^{3x}  故 a=9

v=\cos 4x \Rightarrow \frac{d^2v}{dx^2}=-16\cos 4x  故 b=-16

於是

\int e^{3x}\cos 4xdx

=\frac{1}{-16-6}(e^{3x}(-4\sin 4x)-\cos 4x(3e^{3x}))+C=\frac{1}{25}e^{3x}(4\sin 4x+3\cos 4x)+C

看到此,同學可能感覺到,除了 e^x, \sin x\cos x,似乎沒有其他函數可以應用這個所謂 product rule。對嗎?

好了,都要證明一下這個近乎冇用的 rule:

\int uvdx

=\frac{1}{b}\int u\frac{d^2v}{dx^2}dx

=\frac{1}{b}\int ud(\frac{dv}{dx})

=\frac{1}{b}(u\frac{dv}{dx}-\int \frac{dv}{dx}du)

=\frac{1}{b}(u\frac{dv}{dx}-\int \frac{dv}{dx}\frac{du}{dx}dx)

=\frac{1}{b}(u\frac{dv}{dx}-\int \frac{du}{dx}dv)

=\frac{1}{b}(u\frac{dv}{dx}-v\frac{du}{dx}+\int vd(\frac{du}{dx}))

=\frac{1}{b}(u\frac{dv}{dx}-v\frac{du}{dx}+\int v(\frac{d^2u}{dx^2})dx)

=\frac{1}{b}(u\frac{dv}{dx}-v\frac{du}{dx}+a\int uvdx)

移項,重組,得

\int uvdx=\frac{1}{b-a}(u\frac{dv}{dx}-v\frac{du}{dx})+C

當然,表列法有較大的延伸,見:

https://johnmayhk.wordpress.com/2012/02/07/integration-by-parts/

1 則迴響 »

  1. The cow is so funny

    迴響 由 Ken — 2015/11/02 @ 5:52 下午 | 回覆


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