June 3rd, 2013, 06:25 PM  #1 
Senior Member Joined: Oct 2010 From: Changchun, China Posts: 492 Thanks: 14  Is integral just trick? , where 
June 3rd, 2013, 10:29 PM  #2 
Senior Member Joined: Oct 2010 From: Changchun, China Posts: 492 Thanks: 14  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. 
June 3rd, 2013, 10:52 PM  #3 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs  Re: Is integral just trick?
I have moved this to the university level analysis forum. 
June 3rd, 2013, 11:04 PM  #4 
Senior Member Joined: Oct 2010 From: Changchun, China Posts: 492 Thanks: 14  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 . 
June 3rd, 2013, 11:05 PM  #5  
Senior Member Joined: Oct 2010 From: Changchun, China Posts: 492 Thanks: 14  Re: Is integral just trick? Quote:
 

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