Is integral just trick? , where 
Re: Is integral just trick? A simple trick: Let . Note that , so we have: , with notations: (the Fourier transform of f(x)) and ('conjugate' Fourier transform of f(x)). Since , it is easy to get that: . Thus, we have Now the problem is: how can we get the Fourier transform ? The answer is convolution theorem! Let and , we get (h(x), g(x) and f(x) are all absolutely integrable). The convolution theorem claims that: , where and denotes the Fourier transform of and , respectively. . (Fourier transform of a Gaussian function is still Gaussian) . By applying the convolution theorem, we have , with notations and . Thus, the integral equals: . Moreover, if , the integral will be degenerated to the form: , which shall be equal to . Of course, we can use Feynman's method to solve this 'degenerated' form and achieve the same conclusion. 
Re: Is integral just trick? I have moved this to the university level analysis forum. :D 
Re: Is integral just trick? By the way, If we consider the convolution operator , and have: . If is even and integrable absolutely, we can verify that the cosine functions will meet the operator equation: , This conclusion indicates that is the 'eigen' function of this operator. The corresponding eigenvalue is . 
Re: Is integral just trick? Quote:

All times are GMT 8. The time now is 05:56 AM. 
Copyright © 2019 My Math Forum. All rights reserved.