I am trying to evaluate
$$I = \int_1^{\infty } \left(\frac{\operatorname{erf}\left(a -b\log (x)\right)}{2 x^2}-\frac{\operatorname{erf}\left(a + b\log (x)\right)}{2 x}\right) \, dx$$
Let $\log (x) = v$ gives $x = \mathrm{e}^v$ and $dx = \mathrm{e}^v dv $
$$I = \frac{2}{\sqrt{\pi}}\int_0^{\infty } \mathrm{e}^{-v}\int_0^{a - bv} \mathrm{e}^{-u^2} du dv - \frac{2}{\sqrt{\pi}}\int_0^{\infty } \int_0^{a + bv} \mathrm{e}^{-u^2} du dv$$
I am not sure how to proceed from here though. Any help would be greatly appreciated.
Answer
Let me consider the two integrals $$I = \int \mathrm{e}^{-v}\int_0^{a - bv} \mathrm{e}^{-u^2} du\,dv$$ $$J= \int \int_0^{a + bv} \mathrm{e}^{-u^2} du\,dv$$ First $$\int_0^{a - bv} \mathrm{e}^{-u^2} du =\frac{\sqrt{\pi } }{2} \text{erf}(a-b v)$$ which makes $$I=\frac{ \sqrt{\pi }}{2} \int e^{-v} \text{erf}(a-b v)\,dv$$ This one can be integrated by parts and, after completing the square, we get $$I=\frac{ \sqrt{\pi }}{2} \left(e^{\frac{1-4 a b}{4 b^2}} \text{erf}\left(a-b
v-\frac{1}{2 b}\right)-e^{-v} \text{erf}(a-b v)\right)$$ In a similar manner $$\int_0^{a + bv} \mathrm{e}^{-u^2} du=\frac{\sqrt{\pi }}{2} \text{erf}(a+b v)$$ which makes $$J=\frac{\sqrt{\pi }}{2}\int \text{erf}(a+b v)\,dv$$ Again, an obvious change of variable and integration by parts lead to $$J=\frac{\sqrt{\pi }}{2} \left(\left(\frac{a}{b}+v\right) \text{erf}(a+b
v)+\frac{e^{-(a+b v)^2}}{\sqrt{\pi } b}\right)$$ There is no problem for the lower bound $(v=0)$; for the upper bound, consider $v=t$ and I suppose that you need to look at the asymptotics for the $\text{erf}(z)$ function when $z\to \pm \infty$ (have a look here). For infinitely large values of $z$ $$\text{erf}(z)=1-\frac{1}{\sqrt{\pi }}e^{-z^2} \left(\frac{1}{ z}-\frac{1}{2
z^3}+O\left(\frac{1}{z^4}\right)\right)$$
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