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 January 23rd, 2013, 05:01 AM #1 Newbie   Joined: Jan 2013 Posts: 9 Thanks: 0 integral (fourier transform) integral (from -00 to 00) e^(-ix^2)*e^(-ixb)dx = ? where b > 0
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.

 January 24th, 2013, 01:35 AM #3 Newbie   Joined: Jan 2013 Posts: 9 Thanks: 0 Re: integral (fourier transform) It indeed has an i in it and that's what makes it harder.
 January 24th, 2013, 03:04 PM #4 Global Moderator   Joined: May 2007 Posts: 6,704 Thanks: 669 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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