Vrftsjtryk

I am stuck with the integral:
$$int_{-infty}^{infty} frac{exp[-a(x-b)^2]}{1+cx^2} dx$$ where $a,c$ are real and $b$ is purely imaginary.

I tried to solve it by contour integration but the integral along the semicircular edge of the contour does not end up as 0. So the method doesnt work.

I dont have any idea about other ways to solve this. I am thinking about expanding the gaussian in terms of its Taylor series and integrating the respective terms. Still it is becoming complicated.

Does anyone have any idea about this?

Since $b$ is purely imaginary, this is a Fourier Transform.

Assuming $a>0$, $c>0$ ,and $Re(b) = 0$; make the substitution $- pi s = Im (b)$ or equivalently $-ipi s =b$ or $pi s = ib$

$$begin{align*}displaystyle & int_{-infty}^{infty} frac{expleft[-aleft(x-bright)^2right]}{1+cx^2} dx\
\
&= int_{-infty}^{infty} frac{expleft[-aleft(x+ipi sright)^2right]}{1+cx^2} dx\
\
&= int_{-infty}^{infty} frac{expleft[-ax^2+a(pi s)^2-2pi i axsright]}{cx^2+1} dx\
\
&= frac{a^2}{c}e^{a(pi s)^2}int_{-infty}^{infty} frac{e^{-frac{1}{a}(ax)^2}}{(ax)^2+frac{a^2}{c}}e^{-2pi i (ax)s} dx\
\
&= frac{a}{c}e^{a(pi s)^2}int_{-infty}^{infty} frac{e^{-frac{1}{a}y^2}}{y^2+frac{a^2}{c}}e^{-2pi iys} dy\
\
&= frac{a}{c}e^{a(pi s)^2}(2pi)^2frac{1}{2left(frac{2pi a}{sqrt{c}}right)}int_{-infty}^{infty} frac{2left(frac{2pi a}{sqrt{c}}right)}{(2pi y)^2+left(frac{2pi a}{sqrt{c}}right)^2}space e^{-pi^2left(frac{y}{pisqrt{a}}right)^2}space e^{-2pi iys} dy\
\
&= frac{pi}{sqrt{c}}e^{a(pi s)^2}mathscr{F}left{ frac{2left(frac{2pi a}{sqrt{c}}right)}{(2pi y)^2+left(frac{2pi a}{sqrt{c}}right)^2}space e^{-pileft(frac{y}{sqrt{pi a}}right)^2}right} \
\
&= frac{pi}{sqrt{c}}e^{a(pi s)^2}left[mathscr{F}left{ frac{2left(frac{2pi a}{sqrt{c}}right)}{(2pi y)^2+left(frac{2pi a}{sqrt{c}}right)^2}right} * mathscr{F}left{ e^{-pileft(frac{y}{sqrt{pi a}}right)^2}right} right]\
\
&= frac{pi}{sqrt{c}}e^{a(pi s)^2}left[e^{-frac{2 a}{sqrt{c}}|pi s|} * sqrt{pi a} e^{-aleft(pi sright)^2}right] \
\
&= pi sqrt{frac{pi a}{c}}e^{a(pi s)^2}int_{-infty}^{infty}e^{-frac{2 a}{sqrt{c}}|pi tau|} e^{-aleft(pi s -pi tauright)^2}space dtau \
\
&= pi sqrt{frac{pi a}{c}}e^{a(pi s)^2}left[int_{-infty}^{0}e^{frac{2 a}{sqrt{c}}pi tau} e^{-aleft(pi s -pi tauright)^2}space dtau +int_{0}^{infty}e^{-frac{2 a}{sqrt{c}}pi tau} e^{-aleft(pi s -pi tauright)^2}space dtau right]\
\
&= pi sqrt{frac{pi a}{c}}left(int_{-infty}^{0}expleft[-aleft([pitau]^2-2left[pi s+frac{1}{sqrt{c}}right]pi tauright)right]space dtau +int_{0}^{infty}expleft[-aleft([pitau]^2-2left[pi s-frac{1}{sqrt{c}}right]pi tauright)right]space dtau right)\
\
&= pi sqrt{frac{pi a}{c}}left(e^{aleft(pi s+frac{1}{sqrt{c}}right)^2}int_{-infty}^{0}expleft[-aleft(pitau-left[pi s+frac{1}{sqrt{c}}right]right)^2right]space dtau +e^{aleft(pi s-frac{1}{sqrt{c}}right)^2}int_{0}^{infty}expleft[-aleft(pitau-left[pi s-frac{1}{sqrt{c}}right]right)^2right]space dtau right)\
\
&= pi sqrt{frac{pi a}{c}}left(e^{aleft(pi s+frac{1}{sqrt{c}}right)^2}frac{1}{pisqrt{a}}int_{-infty}^{-sqrt{a}left(pi s+frac{1}{sqrt{c}}right)}e^{-u^2}space du +e^{aleft(pi s-frac{1}{sqrt{c}}right)^2}frac{1}{pisqrt{a}}int_{-sqrt{a}left(pi s-frac{1}{sqrt{c}}right)}^{infty}e^{-u^2}space duright)\
\
&= sqrt{frac{pi}{c}}left(e^{aleft(pi s+frac{1}{sqrt{c}}right)^2}int_{-infty}^{-sqrt{a}left(pi s+frac{1}{sqrt{c}}right)}e^{-u^2}space du +e^{aleft(pi s-frac{1}{sqrt{c}}right)^2}int_{-sqrt{a}left(pi s-frac{1}{sqrt{c}}right)}^{infty}e^{-u^2}space duright)\
\
&= frac{pi}{2}frac{1}{sqrt{c}}left(e^{aleft(pi s+frac{1}{sqrt{c}}right)^2}mathrm{erf}(u)biggr{|}_{-infty}^{-sqrt{a}left(pi s+frac{1}{sqrt{c}}right)} +e^{aleft(pi s-frac{1}{sqrt{c}}right)^2}mathrm{erf}(u)biggr{|}_{-sqrt{a}left(pi s-frac{1}{sqrt{c}}right)}^{infty}right)\
\
&= frac{pi}{2}frac{1}{sqrt{c}}left(e^{aleft(pi s+frac{1}{sqrt{c}}right)^2}left[mathrm{erf}left(-sqrt{a}left[pi s+frac{1}{sqrt{c}}right]right)+1right] +e^{aleft(pi s-frac{1}{sqrt{c}}right)^2}left[1 -mathrm{erf}left(-sqrt{a}left[pi s-frac{1}{sqrt{c}}right]right)right]right)\
\
&= frac{pi}{2}frac{1}{sqrt{c}}left(e^{left[-sqrt{a}left(ib+frac{1}{sqrt{c}}right)right]^2}left[mathrm{erf}left(-sqrt{a}left[ib+frac{1}{sqrt{c}}right]right)+1right] +e^{left[-sqrt{a}left(ib-frac{1}{sqrt{c}}right)right]^2}left[1 -mathrm{erf}left(-sqrt{a}left[ib-frac{1}{sqrt{c}}right]right)right]right)\
end{align*}$$