Distribution of a random variable that is a cumulative distribution

Question

Define $$x = \int_{-\infty}^{\lambda} e^{-t^2/2} dt$$ and suppose $$\lambda \text{ follows } Normal(0, 1)$$ What is the distribution of $x$?

I've been trying for hours. Any hint would be appreciated.

Answer 1

$$X = \sqrt{2\pi}\, \Phi(\lambda) = \sqrt{2\pi} \Pr[Z \le \lambda],$$ where $Z$ is standard normal, so $0 < X < 1$. Then $$\Pr[X \le x] = \Pr[ \sqrt{2\pi} \, \Phi(\lambda) \le x] = \Pr[\lambda \le \Phi^{-1}(x/\sqrt{2\pi})].$$ But $\lambda$ is itself standard normal, so this is simply $$\Pr[X \le x] = \Phi(\Phi^{-1}(x/\sqrt{2\pi})) = x/\sqrt{2\pi},$$ and it immediately follows that $X$ is uniform on $(0,\sqrt{2\pi})$.