# 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^2-a^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.

.

**Follow-Ups**:**Re: Help needed with integral.***From:*raymond manzoni

**References**:**Re: Help needed with integral.***From:*BRG

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