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

Furthermore,
$$
\begin{align}
\int_{-\infty}^\infty\left(\operatorname{erf}(a+x)\operatorname{erf}(a-x)+1 \right)\;\mathrm{d}x
&amp;=\int_{-\infty}^\infty(\operatorname{erf}(a+x)+1)(\operatorname{erf}(a-x)+1)\;\mathrm{d}x\\
&amp;-\int_{-\infty}^\infty(\operatorname{erf}(a+x)+\operatorname{erf}(a-x))\;\mathrm{d}x\tag{2}
\end{align}
$$
To evaluate
$$
\begin{align}
\int_{-\infty}^\infty(\operatorname{erf}(a+x)+1)(\operatorname{erf}(a-x)+1)\;\mathrm{d}x
&amp;=\frac{4}{\pi}\int_{-\infty}^\infty\int_{-\infty}^{a+x}\int_{-\infty}^{a-x}e^{-s^2-t^2}\;\mathrm{d}s\;\mathrm{d}t\;\mathrm{d}x
\end{align}
$$
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,
$$
\begin{align}
\frac{4}{\pi}\int_{-\infty}^\infty\int_{-\infty}^{a+x}\int_{-\infty}^{a-x}e^{-s^2-t^2}\;\mathrm{d}s\;\mathrm{d}t\;\mathrm{d}x
&amp;=\frac{4}{\pi}\int\int_{s+t\le2a}\int_{s-a}^{a-t}e^{-s^2-t^2}\;\mathrm{d}x\;\mathrm{d}s\;\mathrm{d}t\\
&amp;=\frac{4}{\pi}\int_{-\infty}^\infty\int_{-\infty}^\infty(2a-s-t)_+\;e^{-s^2-t^2}\;\mathrm{d}s\;\mathrm{d}t
\end{align}
$$
Change variables: $u=(s+t)/\sqrt{2}$ and $v=(s-t)/\sqrt{2}$ so that $s=(u+v)/\sqrt{2}$ and $t=(u-v)/\sqrt{2}$:
$$
\begin{align}
\frac{4}{\pi}\int_{-\infty}^\infty\int_{-\infty}^\infty(2a-s-t)_+\;e^{-s^2-t^2}\;\mathrm{d}s\;\mathrm{d}t
&amp;=\frac{4}{\pi}\int_{-\infty}^\infty\int_{-\infty}^\infty(2a-\sqrt{2}u)_+\;e^{-u^2-v^2}\;\mathrm{d}u\;\mathrm{d}v\\
&amp;=\frac{4}{\sqrt{\pi}}\int_{-\infty}^{\sqrt{2}a}(2a-\sqrt{2}u)\;e^{-u^2}\;\mathrm{d}u\\
&amp;=4a(\operatorname{erf}(\sqrt{2}a)+1)-\frac{4}{\sqrt{\pi}}\int_{-\infty}^{\sqrt{2}a}\sqrt{2}u\;e^{-u^2}\;\mathrm{d}u\\
&amp;=4a(\operatorname{erf}(\sqrt{2}a)+1)-\frac{2\sqrt{2}}{\sqrt{\pi}}\int_{-\infty}^{\sqrt{2}a}\;e^{-u^2}\;\mathrm{d}u^2\\
&amp;=4a(\operatorname{erf}(\sqrt{2}a)+1)+\frac{2\sqrt{2}}{\sqrt{\pi}}\;e^{-2a^2}
\end{align}
$$
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 $(4)$.
$\int_{-\infty}^\infty 4/\pi \int_0^{a+x} e^{-t^{2}} dx $ . $ \int_0^{a-x} e^{-t^{2}} $ dx $= \int_{-\infty}^\infty 4/\pi [-(1/2t) e^{-t^{2}}]_0^{a+x} [-(1/2t) e^{-t^{2}}]_0^{a-x}$ = $ \int_{-\infty}^\infty (4/\pi) ((1/2a) e^{-a^{2}})^{2} - ((1/2x) e^{-x^{2}})^{2} $ = $[-((1/2x) e^{-x^{2}})^{2}]_{-\infty}^\infty$ = 
$\int_{-\infty}^\infty -((1/2x) e^{-x^{2}})^{2}$=
$\int_{-\infty}^\infty -((1/4x^{2}) e^{-2x^{2}})] $ = $ [-(1/4x) e^{-2x^{2}}]_{-\infty}^\infty - \int_{-\infty}^\infty e^{-2x^{2}}$ = 
$[-(1/4x) e^{-2x^{2}}$+$(1/4x) e^{-2x^{2}}]_{-\infty}^\infty$ =0