# Closed form for definite integral involving Erf and Gaussian?

## Question

Is there a closed form for integrals such as

$\int_{-\infty }^{\infty } e^{-y^2} \text{erf}(1-y) \, dy$

The integrant seems simple enough?

There is a table of different integrals involving the $\text{erf}$ function, where one can find an answer ($13$, p.$8$) to your question,
$$\int_{-\infty}^\infty e^{-y^2}\text{erf}(1-y)\:dy=\sqrt{\pi}\cdot \text{erf}\left(\frac1{\sqrt{2}}\right). \tag1$$
Proposition. One has, for any real number $b$,
$$\int_{-\infty}^\infty e^{-(y+b)^2}\text{erf}(y)\:dy=\sqrt{\pi}\cdot \text{erf}\left(\frac{b}{\sqrt{2}}\right). \tag2$$
Proof. One has \begin{align} \partial_b \left(\int_{-\infty}^\infty e^{-(y+b)^2}\text{erf}(y)\:dy \right)&=-2\int_{-\infty}^\infty (y+b)e^{-(y+b)^2}\text{erf}(y)\:dy \\\\&=\left[e^{-(y+b)^2}\text{erf}(y) \right]_{-\infty}^\infty-\frac{2}{\sqrt{\pi}}\int_{-\infty}^\infty e^{-(y+b)^2}e^{-y^2}dy \\\\&=0-\frac{2}{\sqrt{\pi}}\int_{-\infty}^\infty e^{-(y+b)^2}e^{-y^2}dy \\\\&=-\sqrt{2}\: e^{\large-\frac{b^2}{2}} \end{align} then one obtains $(2)$ by integrating the latter function. By putting $b=-1$ and making a change of variable one gets the desired integral.