Re: Help needed with integral.
 From: BRG <brg@xxxxxxxxxxx>
 Date: Thu, 17 Aug 2006 16:01:17 +0100
BRG wrote:
leo_morgan_@xxxxxxxxxxx wrote:
Hi!
When studying Fourier transforms I stumbled over an integral I cannot
solve.
(Integral from zero to infinity)[e^(v^2a^2/v^2)]dv, where a is a
constant.
Substitute t = v  a / v, which means that 2 v = t +/ sqrt(t^2+4a).
This shows that the integral is equal to
integral(e^(t^2)(1/+2*t/sqrt(t^2+4*a)) dt, inf, inf)/(2e^(2*a))
The first part of this integral evaluates to sqrt(pi) while the second
evaluates to zero over inf to inf because the integrand is odd. So we
obtain the answer sqrt(pi)/(2e^(2*a)).
The second term in the integrand above is a factor of 2 high but it
doesn't matter since its integral evaluates to zero anyway.
.
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