How do I even??

The substitution $e^{\alpha u} = e^{\alpha x} - 1$ gives $e^{\alpha u}\,du = e^{\alpha x}\,dx$ , so that $dx = \tfrac{e^{\alpha u}}{1 +e^{\alpha u}}\,du$ , and hence $f(\alpha) \; = \; \int_{-\infty}^\infty e^{-u^2} \frac{e^{\alpha u}}{1 + e^{\alpha u}}\,du \; = \; \int_{-\infty}^\infty e^{-v^2} \frac{1}{1 + e^{\alpha v}}\,dv$ using the substitution $u = -v$ . But then $f(\alpha) \; = \; \tfrac12\int_{-\infty}^\infty e^{-u^2}\,du \; =\; \boxed{\tfrac12\sqrt{\pi}}$ for any $\alpha > 0$ .