I have to compute following integrals analytically: $$I=\int_0^{A} e^{-\frac{x^2}{K}}e^{-\frac{(A-x)^2}{K}}\frac{{\rm erf}(\lambda x)}{x}dx$$ and: $$J=\int_B^{\infty} e^{-\gamma x}e^{-\frac{x^2}{K}}\frac{{\rm erf}(\lambda x)}{x}dx$$ where $A,B,K,\lambda$ and $\gamma$ are constants such that: $(1/\lambda^2)\ll A^2\ll K$, $B^2\ll K$ and $\lambda,\gamma = O(1)$, and $\rm erf$ is the error function.
Is there any way to evaluate those integrals ( maybe to some good approximations)?
An honest attempt that didn't lead to anything useful so far.
Let's reduce the number of parameters:
$$I=\int_0^{A} e^{-\frac{x^2}{K}}e^{-\frac{(A-x)^2}{K}}\frac{{\rm erf}(\lambda x)}{x}dx$$
$$x=A y, \qquad \frac{A^2}{K} = \alpha^2 \ll 1, \qquad \lambda A = \beta \gg 1$$
$$I(\alpha,\beta)=\int_0^1 e^{-2\alpha^2 (y^2+y+1/2)} \frac{{\rm erf}(\beta y)}{y}dy$$
First the note on approximations. Since $\alpha$ is small and the upper limit is $1$, we can always expand the exponential as a series. However, the resulting integrals will be complicated.
On the other hand, the error function has a nice asymptotic expansion for large arguments and a nice series for small arguments. The problem here, $\beta y$ can be both small for $y<1\beta$ and large for $y>1/\beta$, so either we need to separate the integral into two terms or use some other method.
Let's try Feynman's trick:
$$2\alpha^2 (y^2+y+1/2)+\beta^2 y^2=(2 \alpha^2+\beta^2) y^2+2 \alpha^2 y+\alpha^2 = \\ = (2 \alpha^2+\beta^2) \left(y^2+2 \frac{\alpha^2}{2 \alpha^2+\beta^2} y +\frac{\alpha^4}{(2 \alpha^2+\beta^2)^2}\right)-\frac{\alpha^4}{2 \alpha^2+\beta^2}+\alpha^2 = \\ = (2 \alpha^2+\beta^2) \left(y+\frac{\alpha^2}{2 \alpha^2+\beta^2}\right)^2+\frac{\alpha^2 (\alpha^2+\beta^2)}{2 \alpha^2+\beta^2} $$
Let's rename the parameters:
$$p^2=2 \alpha^2+\beta^2 \gg 1 \\ q=\frac{\alpha^2}{2 \alpha^2+\beta^2} \ll 1 \\ r^2=\frac{\alpha^2 (\alpha^2+\beta^2)}{2 \alpha^2+\beta^2} \ll 1$$
Now we have:
$$Y(p,q,r)=\frac{2}{\sqrt{\pi}} e^{-r^2} \int_0^1 e^{-p^2 (y+q)^2}dy$$
We also can see from the original expression that:
$$I(\alpha,0)=0$$
So we need to find the following integral:
$$I(\alpha,\beta)= \int_0^{\beta} \frac{ 1}{ \sqrt{2 \alpha^2 +t^2}} \exp \left(-\alpha^2\frac{\alpha^2+t^2}{2 \alpha^2+t^2} \right) \times \\ \times \left(\operatorname{erf} \left(\frac{3\alpha^2+t^2}{\sqrt{2 \alpha^2+t^2}} \right)-\operatorname{erf} \left(\frac{\alpha^2}{\sqrt{2 \alpha^2+t^2}} \right) \right) d t$$
So I have made the integral more complicated, not less. Nice.
Note that $pq \ll 1$, but $p(1+q) \gg 1$, so again, not very nice for approximations.
However, we can simplifty a little by substitution:
$$t= \sqrt{2} \alpha u$$
$$I(\alpha,\beta)= \int_0^{\beta/(\sqrt{2} \alpha)} \frac{ 1}{ \sqrt{1 +u^2}} \exp \left(-\frac{\alpha^2}{2}\frac{1+2u^2}{1+u^2} \right) \times \\ \times \left(\operatorname{erf} \left(\frac{\alpha}{\sqrt{2}} \frac{3+2u^2}{\sqrt{1+u^2}} \right)-\operatorname{erf} \left(\frac{\alpha}{\sqrt{2}} \frac{1}{\sqrt{1+u^2}} \right) \right) d u$$
Note that this is correct numerically (gives the same values as the original integral), I checked. No mistakes so far.
Now let's make a substitution:
$$u=\sinh v, \qquad \gamma = \operatorname{arcsinh} \frac{\beta}{\sqrt{2} \alpha}, \qquad \delta =\frac{\alpha}{\sqrt{2}} $$
$$I(\alpha,\beta)= \int_0^\gamma \exp \left(-\delta^2 \frac{\cosh 2 v}{\cosh^2 v} \right) \left(\operatorname{erf} \left( \delta \frac{2+\cosh 2 v}{\cosh v} \right)-\operatorname{erf} \left(\frac{\delta}{\cosh v} \right) \right) d v$$
This looks fun, but there's no point in trying further substitutions. I guess some other methods are needed.