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Let $$F(a)=int_{-infty}^infty frac{e^{-x^2}}{x^2+a^2} dx, quad a>0.$$ Is it possible to relate $F(a)$ to some known (special) functions?

Parameterize this integral by adding a second parameter, $t$:
$$I(t):= int_{-infty}^infty frac{e^{-(x^2+a^2)t}}{x^2+a^2}dx$$
Differentiating with respect to $t$, we have
$$I'(t)=-int_{-infty}^infty e^{-(x^2+a^2)t}dx=-sqrt{frac{pi}{t}} e^{-a^2 t}$$
This shows us that
$$begin{align}
I(t)
&=I(0)-int_0^t sqrt{frac{pi}{x}}e^{-a^2 x}dx\
&=frac{pi}{a}-2int_0^{sqrt{t}} sqrt{pi}e^{-a^2 x^2}dx\
&=frac{pi}{a}-frac{pi text{erf}(asqrt{t})}{a}\
&=frac{pi text{erfc}(asqrt{t})}{a}\
end{align}$$
Which gives us the desired value of your integral:
$$int_{-infty}^infty frac{e^{-x^2}}{x^2+a^2}dx=frac{pi e^{a^2}text{erfc}(a)}{a}$$
Delicious!

As an alternative we can use the following property of the Fourier transform:$$ int_{-infty}^{+infty}f(t)g(t),dt = int_{-infty}^{+infty}(mathscr{F} f)(s)(mathscr{F}^{-1}g)(s),ds$$ Simplifying we have: $$int_{-infty}^infty e^{-x^2}frac{1}{x^2+a^2 } dx=2int_{0}^inftyleft(frac{e^{-x^2/4}}{sqrt 2}right)left(frac{sqrt{pi/2} e^{-ax} }{a}right)dx=$$$$=frac{sqrt{pi}}{a}int_0^infty e^{-(x^2/4+ax) } dx=frac{sqrt{pi}}{a}sqrt{pi} e^{a^2} text{erf}left(a+frac{x} {2} right)bigg|_0^infty=$$$$=frac{pi e^{a^2} } {a}left(1-text{erf}(a)right)=color{blue}{frac{ pi e^{a^2} } {a} text
{erfc}(a)}$$