Integral with complex parameter

Question

Can I obtain an analytic expression for the following integral?

$$I = \int_{-\infty}^{\infty} d\nu \, \frac{e^{-\frac{(\nu -iX)^2}{2\sigma^2}}}{1+ \frac{\nu^2}{\sigma^4}}$$

Here $\sigma>0$, and $X \in \mathbb{C}$.

I tried using the below command but it returned the same expression. Any help will be great in this regard.

Integrate[E^(-(ν - I X)^2/(2 σ^2))/(ν^2/σ^4 + 1), {ν,-∞, ∞}, Assumptions -> σ > 0]

Answer 1

Looks like:

$$\int_{-\infty }^{\infty } \frac{e^{-\frac{(\nu -i X)^2}{2 \sigma ^2}}}{\frac{\nu ^2}{\sigma ^4}+1} \, d\nu =-\frac{1}{2} e^{\frac{\left(X-\sigma ^2\right)^2}{2 \sigma ^2}} \pi \sigma ^2 \left(-2+\text{erfc}\left(\frac{X-\sigma ^2}{\sqrt{2} \sigma }\right)-e^{2 X} \text{erfc}\left(\frac{X+\sigma ^2}{\sqrt{2} \sigma }\right)\right)$$

 N[With[{X = 3 + 4 I, \[Sigma] = 3}, -(1/2) E^((X - \[Sigma]^2)^2/(2 \[Sigma]^2)) \[Pi] \[Sigma]^2 (-2 + 
 Erfc[(X - \[Sigma]^2)/(Sqrt[2] \[Sigma])] - 
 E^(2 X) Erfc[(X + \[Sigma]^2)/(Sqrt[2] \[Sigma])])], 30]

 (*6.29144555383952196652073437413 + 1.23139881428285232437943972824 I*)

 With[{X = 3 + 4 I, \[Sigma] = 3}, NIntegrate[
 E^(-(\[Nu] - I X)^2/(2 \[Sigma]^2))/(\[Nu]^2/\[Sigma]^4 + 
 1), {\[Nu], -Infinity, Infinity}, WorkingPrecision -> 31]]

 (*6.291445553839521966520734374127 + 1.231398814282852324379439728238 I*)

I use FourierTransform to solve this integral:

 FourierTransform[(Exp[-(x^2/(2 \[Omega]^2)) + X^2/(
 2 \[Omega]^2)]*\[Omega]^4)/(\[Omega]^4 + x^2), x, W, 
 Assumptions -> {\[Omega] > 0}] /. W ->  X/\[Omega]^2
 (* ? *)

I waited half an hour and got no response, so I used Maple 2023 and got a response in a fraction of a second.

Answer 2

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