Integral of product of two error functions (erf)

In the course of my research I came across the following integral:

$$\int\nolimits_{-\infty}^{\infty}\operatorname{erf}(a+x)\operatorname{erf}(a-x)dx$$

where $$\operatorname{erf}(x)=\frac{2}{\sqrt{\pi}}\int_{0}^{x}e^{-t^2}dt$$ is the familiar error function.

Does anyone know if this is solvable? If so, a suggestion on how to do this would be very much appreciated. I can tolerate a solution in terms of $\operatorname{erf}$ functions of $a$. I have tried my usual sources in these matters (G&R, the tables on the Wolfram website, Abramowitz's Handbook of Mathematical Formulas, the web) but couldn't find anything of use.

Thank you!

$\operatorname{erf}(x)$ is an odd function, therefore, $$ \begin{align} \int_{-\infty}^\infty(\operatorname{erf}(a+x)+\operatorname{erf}(a-x))\;\mathrm{d}x &=\lim_{L\to\infty}\;\int_{-L}^L(\operatorname{erf}(x+a)-\operatorname{erf}(x-a))\;\mathrm{d}x\\ &=\lim_{L\to\infty}\;\int_{-L+a}^{L+a}\operatorname{erf}(x)\;\mathrm{d}x-\lim_{L\to\infty}\;\int_{-L-a}^{L-a}\operatorname{erf}(x)\;\mathrm{d}x\\ &=\lim_{L\to\infty}\;\int_{L-a}^{L+a}\operatorname{erf}(x)\;\mathrm{d}x-\lim_{L\to\infty}\;\int_{-L-a}^{-L+a}\operatorname{erf}(x)\;\mathrm{d}x\\ &=4a\tag{1} \end{align} $$ since $\lim\limits_{x\to\infty}\operatorname{erf}(x)=1$ and $\lim\limits_{x\to-\infty}\operatorname{erf}(x)=-1$.

Furthermore, $$ \begin{align} \int_{-\infty}^\infty\left(\operatorname{erf}(a+x)\operatorname{erf}(a-x)+1 \right)\;\mathrm{d}x &=\int_{-\infty}^\infty(\operatorname{erf}(a+x)+1)(\operatorname{erf}(a-x)+1)\;\mathrm{d}x\\ &-\int_{-\infty}^\infty(\operatorname{erf}(a+x)+\operatorname{erf}(a-x))\;\mathrm{d}x\tag{2} \end{align} $$ To evaluate $$ \begin{align} \int_{-\infty}^\infty(\operatorname{erf}(a+x)+1)(\operatorname{erf}(a-x)+1)\;\mathrm{d}x &=\frac{4}{\pi}\int_{-\infty}^\infty\int_{-\infty}^{a+x}\int_{-\infty}^{a-x}e^{-s^2-t^2}\;\mathrm{d}s\;\mathrm{d}t\;\mathrm{d}x \end{align} $$ note that $s\le a+x$ and $t\le a-x$; i.e. $s-a\le x\le a-t$ and $s+t\le2a$. Thus, $$ \begin{align} \frac{4}{\pi}\int_{-\infty}^\infty\int_{-\infty}^{a+x}\int_{-\infty}^{a-x}e^{-s^2-t^2}\;\mathrm{d}s\;\mathrm{d}t\;\mathrm{d}x &=\frac{4}{\pi}\int\int_{s+t\le2a}\int_{s-a}^{a-t}e^{-s^2-t^2}\;\mathrm{d}x\;\mathrm{d}s\;\mathrm{d}t\\ &=\frac{4}{\pi}\int_{-\infty}^\infty\int_{-\infty}^\infty(2a-s-t)_+\;e^{-s^2-t^2}\;\mathrm{d}s\;\mathrm{d}t \end{align} $$ Change variables: $u=(s+t)/\sqrt{2}$ and $v=(s-t)/\sqrt{2}$ so that $s=(u+v)/\sqrt{2}$ and $t=(u-v)/\sqrt{2}$: $$ \begin{align} \frac{4}{\pi}\int_{-\infty}^\infty\int_{-\infty}^\infty(2a-s-t)_+\;e^{-s^2-t^2}\;\mathrm{d}s\;\mathrm{d}t &=\frac{4}{\pi}\int_{-\infty}^\infty\int_{-\infty}^\infty(2a-\sqrt{2}u)_+\;e^{-u^2-v^2}\;\mathrm{d}u\;\mathrm{d}v\\ &=\frac{4}{\sqrt{\pi}}\int_{-\infty}^{\sqrt{2}a}(2a-\sqrt{2}u)\;e^{-u^2}\;\mathrm{d}u\\ &=4a(\operatorname{erf}(\sqrt{2}a)+1)-\frac{4}{\sqrt{\pi}}\int_{-\infty}^{\sqrt{2}a}\sqrt{2}u\;e^{-u^2}\;\mathrm{d}u\\ &=4a(\operatorname{erf}(\sqrt{2}a)+1)-\frac{2\sqrt{2}}{\sqrt{\pi}}\int_{-\infty}^{\sqrt{2}a}\;e^{-u^2}\;\mathrm{d}u^2\\ &=4a(\operatorname{erf}(\sqrt{2}a)+1)+\frac{2\sqrt{2}}{\sqrt{\pi}}\;e^{-2a^2} \end{align} $$ Therefore, $$ \int_{-\infty}^\infty\left(\operatorname{erf}(a+x)+1\right)\left(\operatorname{erf}(a-x)+1\right)\;\mathrm{d}x =4a\left(\operatorname{erf}(\sqrt{2}a)+1\right)+\frac{2\sqrt{2}}{\sqrt{\pi}}\;e^{-2a^2}\tag{3} $$ Thus, the convolution of $\operatorname{erf}(x)+1$ with itself is $2x(\operatorname{erf}(x/\sqrt{2})+1)+\frac{2\sqrt{2}}{\sqrt{\pi}}e^{-x^2/2}$.

Subtract $4a$ from $(3)$ using $(1)$ and $(2)$ to get $$ \int_{-\infty}^\infty\left(\operatorname{erf}(a+x)\operatorname{erf}(a-x)+1 \right)\;\mathrm{d}x =4a\operatorname{erf}(\sqrt{2}a)+\frac{2\sqrt{2}}{\sqrt{\pi}}\;e^{-2a^2}\tag{4} $$ My guess is you want either $(3)$ or $(4)$.