If the term I X is meant to be imaginary I * Complex, 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 $x \rightarrow i X + \sigma y$, may change the integral. A simple check is the numerical integral over the interval of confidence for a Gaussian x by +-3 sigma