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 June 3rd, 2013, 07:25 PM #1 Senior Member   Joined: Oct 2010 From: Changchun, China Posts: 492 Thanks: 14 Is integral just trick? , where June 3rd, 2013, 11:29 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, 11:52 PM #3 Senior Member   Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 522 Math Focus: Calculus/ODEs Re: Is integral just trick? I have moved this to the university level analysis forum.  June 4th, 2013, 12:04 AM #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 4th, 2013, 12:05 AM   #5
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Re: Is integral just trick?

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 Originally Posted by MarkFL I have moved this to the university level analysis forum. Thanks! Tags integral, trick ### simple trick for convolution theorem

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