I would appreciate it if you could help proving the following identity which are two forms of complimentary error function
$$\frac{2}{\sqrt{\pi}}\int_z^\infty e^{-t^2} dt=\frac{2}{\pi}e^{-z^2}\int_0^\infty \frac{e^{-z^2t^2}}{t^2+1}dt$$
Thanks in advance
Let
$$F(a) = \int_0^{\infty} dt \frac{e^{-a (t^2+1)}}{t^2+1}$$
$$\frac{d}{da} F(a) = -\int_0^{\infty} dt \, e^{-a (t^2+1)} = -\frac12 e^{-a}\sqrt{\frac{\pi}{a}}$$
$$F(0) = \int_0^{\infty} \frac{dt}{t^2+1} = \frac{\pi}{2}$$
$$F(a) = \frac{\pi}{2}\underbrace{-\frac{\sqrt{\pi}}{2} \int_0^a da' \, a'^{-1/2} \, e^{-a'}}_{a'=v^2} = \frac{\pi}{2} -\sqrt{\pi} \int_0^{\sqrt{a}} dv \, e^{-v^2} = \frac{\pi}{2} \text{erfc}(\sqrt{a}) $$
Therefore
$$\text{erfc}(z) = \frac{2}{\pi} \int_0^{\infty} dt \frac{e^{-z^2 (t^2+1)}}{t^2+1}$$