Dtjytyk

I used the result $$frac{2}{pi} mathrm{exp}(-z^{2}) intlimits_{0}^{infty} mathrm{exp}(-z^{2}x^{2}) frac{1}{x^{2}+1} mathrm{d}x = mathrm{erfc}(z)$$ to answer this MSE question. As I mentioned in the link, I obtained this result from the DLMF. I happened to find this solution after failing to evaluate the integral using a variety of substitutions. A solution would be appreciated.

Addendum

Expanding @Jack D'Aurizio's solution, we have

begin{align}
frac{2}{pi} mathrm{e}^{-z^{2}} intlimits_{0}^{infty} frac{mathrm{e}^{-z^{2}x^{2}}}{x^{2} + 1} mathrm{d}x &=
frac{2z}{pi} mathrm{e}^{-z^{2}} intlimits_{0}^{infty} frac{mathrm{e}^{-t^{2}}}{z^{2} + t^{2}} mathrm{d}t \
&= frac{z}{pi} mathrm{e}^{-z^{2}} intlimits_{-infty}^{infty} frac{mathrm{e}^{-t^{2}}}{z^{2} + t^{2}} mathrm{d}t
end{align}
we used the substitution $x=t/z$.

For the integral
begin{equation}
intlimits_{-infty}^{infty} frac{mathrm{e}^{-t^{2}}}{z^{2} + t^{2}} mathrm{d}t
end{equation}
we let $f(t) = mathrm{e}^{-t^{2}}$ and $g(t) = 1/(z^{2} + t^{2})$ and take Fourier transforms of each,
begin{equation}
mathrm{F}(s) = mathcal{F}[f(t)] = frac{mathrm{e}^{-s^{2}/4}}{sqrt{2}}
end{equation}
and
begin{equation}
mathrm{G}(s) = mathcal{F}[g(t)] = frac{1}{z}sqrt{frac{pi}{2}} mathrm{e}^{-z|s|}
end{equation}
then invoke Parseval's theorem
begin{equation}
intlimits_{-infty}^{infty} f(t)overline{g(t)} mathrm{d}t
= intlimits_{-infty}^{infty} mathrm{F}(s)overline{mathrm{G}(s)} mathrm{d}s
end{equation}
dropping constants, the integral becomes

begin{align}
intlimits_{-infty}^{infty} mathrm{e}^{-s^{2}/4} mathrm{e}^{-z|s|} mathrm{d}s
&= 2intlimits_{0}^{infty} mathrm{e}^{-s^{2}/4} mathrm{e}^{-z|s|} mathrm{d}s \
&= 2mathrm{e}^{z^{2}} intlimits_{0}^{infty} mathrm{e}^{-(s+2z)^{2}/4} mathrm{d}s \
&= 4mathrm{e}^{z^{2}} intlimits_{0}^{infty} mathrm{e}^{-y^{2}} mathrm{d}y \
&= 2sqrt{pi}mathrm{e}^{z^{2}} mathrm{erfc}(z)
end{align}
We completed the square in the exponent and used the substitution $y=z+s/2$.

Putting the pieces together yields our desired result
begin{align}
frac{2}{pi} mathrm{e}^{-z^{2}} intlimits_{0}^{infty} frac{mathrm{e}^{-z^{2}x^{2}}}{x^{2} + 1} mathrm{d}x &=
frac{z}{pi} mathrm{e}^{-z^{2}} intlimits_{-infty}^{infty} frac{mathrm{e}^{-t^{2}}}{z^{2} + t^{2}} mathrm{d}t \
&= frac{z}{pi} mathrm{e}^{-z^{2}} frac{1}{sqrt{2}} frac{1}{z} sqrt{frac{pi}{2}} 2sqrt{pi} mathrm{e}^{z^{2}} mathrm{erfc}(z) \
&= mathrm{erfc}(z)
end{align}

With the substitution $x=frac{t}{z}$, the integral on the left becomes

$$I=frac{2}{pi z e^{z^2}}int_{0}^{+infty}frac{e^{-t^2}}{1+frac{t^2}{z^2}},dt = frac{1}{pi z e^{z^2}}int_{-infty}^{+infty}frac{e^{-t^2}}{1+frac{t^2}{z^2}},dt $$
and we may switch to Fourier transforms. Since
$$mathcal{F}(e^{-t^2}) = frac{1}{sqrt{2}}e^{-s^2/4},qquad mathcal{F}left(frac{1}{1+frac{t^2}{z^2}}right)=zsqrt{frac{pi}{2}} e^{-z|s|}$$
$I$ boils down to an integral of the form $int_{0}^{+infty}expleft(-(s-xi)^2right),ds$ that is straightforward to convert in a expression involving the (complementary) error function.

As an alternative, you may use differentiation under the integral sign to prove that both sides of your equation fulfill the same differential equation with the same initial constraints, then invoke the uniqueness part of the Cauchy-Lipschitz theorem:
$$ frac{d}{dz} LHS = -frac{2}{pi}int_{0}^{+infty}2z e^{-z^2 (x^2+1)},dx,qquad frac{d}{dz}RHS = -frac{2}{sqrt{pi}}e^{-z^2}.$$
We have $frac{d}{dz}(LHS-RHS)=0$, and $(LHS-RHS)(0)=1$.

An interesting consequence is the following (tight) approximation for the $text{erfc}$ function:

$$text{erfc}(z)=frac{2e^{-z^2}}{pi}int_{0}^{+infty}frac{e^{-z^2 x^2}}{x^2+1},dxleq frac{2e^{-z^2}}{pi}int_{0}^{+infty}frac{dx}{(x^2+1)(x^2 z^2+1)}=frac{1}{(1+z)e^{z^2}}.$$

Assuming $z>0$,

$$ begin{align}int_{0}^{infty} frac{e^{-z^{2}x^{2}}}{1+x^{2}} , dx &= int_{0}^{infty}e^{-z^{2}x^{2}} int_{0}^{infty}e^{-t(1+x^{2})} , dt , dx \ &= int_{0}^{infty} e^{-t} int_{0}^{infty}e^{-(z^{2}+t)x^{2}} , dx , dt tag{1}\ &= frac{sqrt{pi}}{2}int_{0}^{infty} frac{e^{-t}}{sqrt{z^{2}+t}} , dt tag{2}\ &= frac{sqrt{pi}}{2} , e^{z^{2}}int_{z^{2}}^{infty}frac{e^{-u}}{sqrt{u}} , du \ &= sqrt{pi} , e^{z^{2}} int_{z}^{infty} e^{-w^{2}} , dw \ &= frac{pi}{2} , e^{z^{2}}operatorname{erfc}(z) end{align}$$

$(1)$ Tonelli's theorem

$(2)$ $int_{0}^{infty} e^{-ax^{2}} , dx = frac{sqrt{pi}}{2} frac{1}{sqrt{a}}$ for $a>0$

$(3)$ Let $u = z^{2}+t$.

$(4)$ Let $w=sqrt{u}$.

begin{eqnarray}
&&intlimits_0^infty frac{e^{-z^2 x^2}}{1+x^2} dx=\
&&intlimits_0^infty frac{e^{-frac{1}{2} (sqrt{2}z)^2 x^2}}{1+x^2} dx=\
&&2 pi T(sqrt{2} z, infty) e^{frac{1}{2} (sqrt{2}z)^2}\
&&2 pi intlimits_{sqrt{2} z}^infty frac{e^{-1/2 xi^2}}{sqrt{2 pi}} frac{1}{2} underbrace{erf(frac{infty cdot xi}{sqrt{2}})}_{1} dxi e^{frac{1}{2} (sqrt{2}z)^2}=\
&&frac{pi}{2} erfc(z) e^{z^2}
end{eqnarray}
where $T(h,a)$ is the Owen's T function https://en.wikipedia.org/wiki/Owen%27s_T_function .