Re: Help needed with integral.

BRG wrote:
leo_morgan_@xxxxxxxxxxx wrote:

When studying Fourier transforms I stumbled over an integral I cannot

(Integral from zero to infinity)[e^(-v^2-a^2/v^2)]dv, where a is a

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.