If the term I X is meant to be imaginary I * Complex,the then the answer is not definitely true. Make the formal substitution
Together /@
Inactive[Integrate][
dx E^(-(x - I X)^2/(2 \[Sigma]^2))/(x^2/\[Sigma]^4 + 1),
{x, -\[Infinity], \[Infinity]}] /.
{x -> I X + \[Sigma] y , dx -> \[Sigma] dy } // FullSimplify
Inactive[Integrate][
(dy E^(-(y^2/2)) \[Sigma]^5)/
(\[Sigma]^4+(I X+y \[Sigma])^2), -oo< X+y \[Sigma] < oo]
ConditionalExpression[
(1/2)*dy* E^((X - \[Sigma]^2)^2/
(2*\[Sigma]^2))*\[Sigma]^2* (Pi*Erf[(X - \[Sigma]^2)/
(Sqrt[2]*\[Sigma])] - E^(2*X)*Pi*Erf[(X + \[Sigma]^2)/(Sqrt[2]*[Sigma])]
+ I*(Log[(I*\[Sigma])/(X - \[Sigma]^2)] - Log[(I*\[Sigma])/
(-X +\[Sigma]^2)] + E^(2*X)*(Log[-((I*\[Sigma])/
(X + \[Sigma]^2))] - Log[(I*\[Sigma])/(X + \[Sigma]^2)]))),
Re[X/\[Sigma]] != Re[\[Sigma]] && Re[X/\[Sigma] + \[Sigma]] != 0]
Since by inspection, the integrand has a singularity at the zeros of the denominator, the shifting of the path x -> I X + sigma y$x \rightarrow i X + \sigma y$, may change the integral. A simple check is the numricalnumerical integral over the interval of confidence for a Gaussian x by +-3 sigma
