微積分計算~~高手入內
cos(wx)*e^(-(ax)^2*pi) dx
x從0積到無窮
回答 (5)
the answer is 1/[2a*e^b], where b= w²/(4πa²)
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).
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?
收錄日期: 2021-05-02 10:05:55
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20091202000016KK07694
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