✔ 最佳答案
By definition F(w)=F{ e^-( a。x^2 ) } = integral[x=-infinity to infinity] {e^(iwx)e^(-ax^2)} dx = integral[x=-infinity to infinity]{e^(-ax^2+iwx)} dx .
the exponent of the integrand: (-ax^2+iwx)=-a[x-(iw/(2a)]^2+[-w^2/(4a^2)]
thus F(w)= e^[-w^2/(4a^2)] * integral[x=-infinity to infinity]{e^(-a[x-(iw/(2a)]^2} dx. Using the famous integral integral[x=-infinity to infinity]{e^(-x^2} dx =sqrt(pi) , we have F(w)=e^[-w^2/(4a^2)] *sqrt(pi/a).
2010-03-25 02:36:08 補充:
更正(-ax^2+iwx)=-a[x-(iw/(2a)]^2+[-w^2/(4a^2)] --->(-ax^2+iwx)=-a[x-(iw/(2a)]^2+[-w^2/(4a)]
故 F(w)=e^[-w^2/(4a^2)] *sqrt(pi/a) 應為F(w)=e^[-w^2/(4a)] *sqrt(pi/a)
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