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June 3rd, 2013, 07:25 PM   #1
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Is integral just trick?

, where
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June 3rd, 2013, 11:29 PM   #2
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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.
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June 3rd, 2013, 11:52 PM   #3
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Re: Is integral just trick?

I have moved this to the university level analysis forum.
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June 4th, 2013, 12:04 AM   #4
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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 .
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June 4th, 2013, 12:05 AM   #5
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Re: Is integral just trick?

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