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?

 A: There is a table of different integrals involving the $\text{erf}$ function, where one can find an answer ($13$, p.$8$) to your question, 

$$
\int_{-\infty}^\infty e^{-y^2}\text{erf}(1-y)\:dy=\sqrt{\pi}\cdot \text{erf}\left(\frac1{\sqrt{2}}\right). \tag1
$$

Let's find a way to obtain the given closed form.
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.
