## Definite integral of exp and error function

Hi,

I've been trying to evaluate the following integral

$$\int_{z}^{\infty}\exp\left(-y^{2}\right)\mathrm{erf}\left(b\left(y-c\right)\right)\,\mathrm{d}y$$

or equivalently

$$\int_{z}^{\infty}\exp\left(-y^{2}\right)\mathrm{erfc}\left(b\left(y-c\right)\right)\,\mathrm{d}y$$

$$\mathrm{erf}\left(x\right)=\frac{2}{\sqrt{\pi}} \int_{0}^{x}\exp\left(-u^{2}\right)\,\mathrm{d}u, \quad\quad \mathrm{erfc}\left(x\right)=1-\mathrm{erf}\left(x\right)=\frac{2}{\sqrt{\pi}} \int_{x}^{+\infty}\exp\left(-u^{2}\right)\,\mathrm{d}u$$

I guess I tried to employ all techniques I'm familiar with but with no result.
Can anyone help me with this one, please?
Thank you!
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 Recognitions: Science Advisor To the best of my knowledge, it can't be done analytically. I suggest you start with the erf representation and see if the two exponentials might be combined into one, so that you might have an erf for the integral.
 Thanks mathman for your reply. I guess I'm not able to deal with this integral. I have a question though. I'm not a mathematician nor a math student so I was wondering if anyone could explain to me why the integral $$\int_{-\infty}^{\infty}\exp\left(-y^{2}\right) \mathrm{erf}\left(b\left(y-c\right)\right)\,\mathrm{d}y=-\sqrt{\pi}\,\mathrm{erf}\left(\frac{bc}{\sqrt{1+b^{2}}}\right)$$ can be evaluated quite easily (using differentiation under integral sign method) and the integral from my original post seems to have no analytical solution? Thanks!

Recognitions:
 Hi JJacquelin! Your post helped me with showing that the constant of integration $$C=0$$ in a more general formula: $$\int_{-\infty}^{\infty}\exp\left(-b^{2}(x-c)^{2}\right)\mathrm{erf}\left(a\left(x-d\right)\right)\,\mathrm{d}x=\frac{\sqrt{\pi}}{b} \mathrm{erf}\left(\frac{ab\left(c-d\right)}{\sqrt{a{}^{2}+b{}^{2}}}\right),\quad b>0$$ Thank you!