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 stainburg June 3rd, 2013 06:25 PM

Is integral just trick?

, where

 stainburg June 3rd, 2013 10:29 PM

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.

 MarkFL June 3rd, 2013 10:52 PM

Re: Is integral just trick?

I have moved this to the university level analysis forum. :D

 stainburg June 3rd, 2013 11:04 PM

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 .

 stainburg June 3rd, 2013 11:05 PM

Re: Is integral just trick?

Quote:
 Originally Posted by MarkFL I have moved this to the university level analysis forum. :D
Thanks!

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