I came across the following integral:
$$\int e^{-t^2}\cdot \text{erf}(a+b\cdot t)dt$$
Does anyone know whether it has a closed form? I have seen related solutions for the interval t=[0,inf] or for the case with a=0 with lower integration limit at t=0. A similar specific solution is also shown here http://alumnus.caltech.edu/~amir/bivariate.pdf but it lacks of generality. Thanks!
It definitely does.
I am going to call $I_{a,b}$ the integral you defined, and assume that the error function is defined as usual (like here for instance). The first thing to notice is that $I_{0,1}=0$ because the error function is odd. Let us consider $I$ as a function of $a$ and try to differentiate. Namely, set $$f(x) = \int \mathrm{e}^{-t^2}\cdot \mathrm{erf}(t+x)\,\mathrm{d} t$$ and let us compute $f'(x)$: $$f'(x) = \int \mathrm{e}^{-t^2}\cdot \frac{2}{\sqrt{\pi}}\mathrm{e}^{\mathrm{-(x+t)^2}}\,\mathrm{d}t = \frac{2}{\sqrt{\pi}}\int \mathrm{e}^{-2t^2-2xt-x^2}\,\mathrm{d}t=\sqrt{2}\mathrm{e}^{-\frac{x^2}{2}}.$$ where we used this useful lemma. Because $f(0)=0$, we have obtained $$I_{x,1}=f(x) = \int_0^x \sqrt{2}\mathrm{e}^{-\frac{t^2}{2}}\,\mathrm{d}t = \sqrt{\pi}\,\mathrm{erf}\left(\frac{x}{\sqrt{2}}\right).$$ Now, it should be clear that we can generalize this result into $$\int \mathrm{e}^{-\alpha t^2}\mathrm{erf}(t+x)\,\mathrm{d}t = \sqrt{\frac{\pi}{\alpha}}\mathrm{erf}\left(\sqrt{\frac{\alpha}{\alpha + 1}}x\right),$$ for any $\alpha > 0$. We are nearly done at this point, since the change of variable $s=bt$ yields $$I_{a,b}=\frac{1}{b}\int \mathrm{e}^{-\frac{s^2}{b^2}}\mathrm{erf}(a+s)\,\mathrm{d}s,$$ which we can compute. In the end, $$I_{a,b} = \sqrt{\pi} \, \mathrm{erf}\left(\frac{a}{\sqrt{1+b^2}}\right).$$