# What is the indefinite integral of the error function times a gaussian?

Wikipedia lists

$$\int\limits_{-\infty}^{\infty} \operatorname{erf}(ax+b) \frac{1}{\sqrt{2\pi\sigma^2}} \exp\left( \frac{-(x-\mu)^2}{2 \sigma^2}\right) \, dx \\ = \operatorname{erf}\left(\frac{a\mu+b}{\sqrt{1+2a^2\sigma^2}}\right)$$

How is this obtained? Is there an indefinite solution to this integral?

• Maybe you can exchange the order of integration to obtain the result? Jan 25 '20 at 21:02
• This is $\int_{-\infty}^{\infty}(2\Phi((ax+b)\sqrt{2})-1)\Phi'(\frac{x-\mu}{\sigma})dx$. Can you complete the proof? Jan 25 '20 at 21:11

There's an elegant solution based on @fGDu94's hint using probability theory. (I'm sure I've used this argument in a previous question of which this one might be a duplicate, but I can't find it.) Let's first restate the claimed result with $$\Phi$$ instead of $$\operatorname{erf}$$ as$$\int_{\Bbb R}(2\Phi((ax+b)\sqrt{2})-1)\frac{d}{dx}\Phi\left(\frac{x-\mu}{\sigma}\right)dx=2\Phi\Bigg((a\mu+b)\sqrt{\frac{2}{1+2a^2\sigma^2}}\Bigg)-1,$$or equivalently as$$\int_{\Bbb R}\Phi((ax+b)\sqrt{2})\frac{d}{dx}\Phi\left(\frac{x-\mu}{\sigma}\right)dx=\Phi\Bigg((a\mu+b)\sqrt{\frac{2}{1+2a^2\sigma^2}}\Bigg).$$The left-hand side is the mean of $$\Phi(Y)$$ with $$Y:=(aX+b)\sqrt{2}$$ for $$X\sim N(\mu,\,\sigma^2)$$ so that $$Y\sim N((a\mu+b)\sqrt{2},\,2a^2\sigma^2)$$. But in terms of $$Z\sim N(0,\,1)$$ independent of $$Y$$ so that $$Y-Z\sim N((a\mu+b)\sqrt{2},\,1+2a^2\sigma^2)$$,$$\Phi(Y)=P(Z\le Y)=P(Y-Z\ge 0)=\Phi\Bigg((a\mu+b)\sqrt{\frac{2}{1+2a^2\sigma^2}}\Bigg).$$