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January 23rd, 2013, 05:01 AM   #1
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integral (fourier transform)

integral (from -00 to 00) e^(-ix^2)*e^(-ixb)dx = ? where b > 0
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January 23rd, 2013, 01:20 PM   #2
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Re: integral (fourier transform)

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Originally Posted by strammer
integral (from -00 to 00) e^(-ix^2)*e^(-ixb)dx = ? where b > 0
It looks like the first exponent should not have an i in it. In that case you are looking at the Fourier transform of a normal distribution, with some constants.
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January 24th, 2013, 01:35 AM   #3
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Re: integral (fourier transform)

It indeed has an i in it and that's what makes it harder.
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January 24th, 2013, 03:04 PM   #4
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Re: integral (fourier transform)

In that case the integral doesn't exist, since the integrand has magnitude = 1 for the entire domain of integration.
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