微積分計算~~高手入內

2009-12-03 5:24 am
cos(wx)*e^(-(ax)^2*pi) dx
x從0積到無窮

回答 (5)

2009-12-12 10:13 am
✔ 最佳答案
這題確實不簡單。
若w = 0,這只是屬於類似Gaussian integral。
但是w並不一定等於0,這就麻煩得多了。因為我在http://en.wikipedia.org/wiki/List_of_integrals及其附近都找不到你這類的improper integral。

2009-12-07 03:01:46 補充:
我亦想過這個improper integral是否可以化成以w為自變量的微分方程。因為我在http://tw.knowledge.yahoo.com/question/question?qid=1009100207053用laplace transform去解一條微分方程的時候無意中發現得出來的解與這題的improper integral十分類似。

不過你要留意的是,就算可以化成微分方程,該微分方程也要用其他方法可解才可行,否則也只是行不通。

2009-12-07 03:17:02 補充:
我亦試過取其indefinite integral用Wolfram Alpha計算(http://www.wolframalpha.com/input/?i=integrate%28cos%28wx%29e%5E%28-%28ax%29%5E2pi%29%29),但發現每一個答案代x = ∞都是很難化簡。

2009-12-07 03:26:47 補充:
因此,就要試下這「絕招」。就是在http://mathworld.wolfram.com/Erf.html內取第31號公式,然後盡量試下轉換該improper integral看看是否能夠轉換至這題。

2009-12-07 06:22:46 補充:
猛省,其實這題即是e^(-(ax)²π)的fourier cosine transform,在http://eqworld.ipmnet.ru/en/auxiliary/inttrans/FourCos3.pdf已經可以找到答案,但計算過程就欠奉了。

2009-12-12 02:13:32 補充:

圖片參考:http://i212.photobucket.com/albums/cc82/doraemonpaul/yahoo_knowledge/improperintegral/crazyimproperint0.jpg


圖片參考:http://i212.photobucket.com/albums/cc82/doraemonpaul/yahoo_knowledge/improperintegral/crazyimproperint1.jpg


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圖片參考:http://i212.photobucket.com/albums/cc82/doraemonpaul/yahoo_knowledge/improperintegral/crazyimproperint5.jpg


圖片參考:http://i212.photobucket.com/albums/cc82/doraemonpaul/yahoo_knowledge/improperintegral/crazyimproperint6.jpg


圖片參考:http://i212.photobucket.com/albums/cc82/doraemonpaul/yahoo_knowledge/improperintegral/crazyimproperint7.jpg


圖片參考:http://i212.photobucket.com/albums/cc82/doraemonpaul/yahoo_knowledge/improperintegral/crazyimproperint8.jpg


圖片參考:http://i212.photobucket.com/albums/cc82/doraemonpaul/yahoo_knowledge/improperintegral/crazyimproperint9.jpg

靈感來自http://tw.knowledge.yahoo.com/question/question?qid=1609120701127 + http://en.wikipedia.org/wiki/Gaussian_integral#By_Cartesian_coordinates + my wisdom of maths
2009-12-09 6:45 pm
the answer is 1/[2a*e^b], where b= w²/(4πa²)
2009-12-09 7:35 am
求定積分[x從0積到無窮]cos(wx)*e^(-(ax)^2*pi)dx:

1. 令{定積分[x從0積到無窮]cos(wx)*e^(-(ax)^2*pi)dx}=g(w);
2. 則g'(w) =定積分[x從0積到無窮]-x*sin(wx)*e^(-(ax)^2*pi)dx. Integration by parts yields g'(w)=-w/(2api)*g(w). i.e. g(w) satisfies the first order ode g'(w)+w/(2api)*g(w)=0, which has the general solution g(w)=C*e^(-w^2/4api).

3. The constant C (in 2) is acturally =g(0)=定積分[x從0積到無窮]e^(-(ax)^2*pi)dx. Letting x=z/(a*sqrt(pi)) we get g(0)=a*sqrt(pi)*[sqrt(pi)/2], where the famous integral: 定積分[z從0積到無窮]e^(-z^2) dz=sqrt(pi)/2 has been used. i.e. C=g(0)=a*pi/2.

4. Combining 1. 2. and 3. ==> 定積分[x從0積到無窮]cos(wx)*e^(-(ax)^2*pi)dx=(a*pi)/2*e^(-w^2/4api).
2009-12-03 10:29 am
^嗎?
次方的意思
x^2表
x的平方

2009-12-04 02:15:57 補充:
因為指數是
-pi乘ax的平方

2009-12-07 07:11:54 補充:
http://tw.knowledge.yahoo.com/question/question?qid=1609120701127
恩...因為都沒人回答
所以我換個方式問了再另外一篇(您也有回覆)
這表我有
但我好奇的是過程...
2009-12-03 8:39 am
那個"鈎"是什麼意思?

2009-12-03 04:20:06 補充:
那a後面的x幹嘛再加一個?
題目要不要再看一下?


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