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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?

Mathematica graphics

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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.

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