So I know that $$\displaystyle \int_{0}^{\infty} \text{erf}(x) dx$$ does not converge so I am assuming that $$\displaystyle \int_{0}^{\infty} \frac{\text{erf}(x)}{x} dx$$ does not converge? Is there anyway to estimate these integrals?
By the way I arrived at that equation from the following:
$$\displaystyle \int_{0}^{p} \frac{\text{exp}(-r^2)r^2}{p} dr \approx \frac{\text{erf}(p)}{p}$$
so maybe my evaluation of that integral is wrong? Thanks for the help guys!
On
If $f$ is pdf of a standard normal ($f'=-xf$), integration by parts yields:
$$I(a,b)=\int_{a}^{b} r^2f(r)\,dr = \Phi(b)-\Phi(a)+af(a)-bf(b)$$
For $a=0$ and $b$ large we get asymptotics: $$I(0,p)\approx\frac 1 2-(b+\frac 1 b)f(b)\approx \frac 1 2-bf(b)$$ You can translate the above into your integral.
Since: $$ \text{erf}(x)=\int_0^x\frac{2}{\sqrt{\pi}}\exp\left(-y^2\right)\mathrm{d}y $$ Thus, since all the functions are non-negative, we can change the order of integration - take care of the change of the integration limits (due to Fubini): $$ \displaystyle \int_{0}^{\infty} \frac{\text{erf}(x)}{x}\mathrm{d}x=\int_0^\infty\int_{0}^x\frac{2}{\sqrt{\pi}x}e^{-y^2}\mathrm{d}y\mathrm{d}x=\int_{0}^\infty\int_{y}^\infty \frac{2}{\sqrt{\pi}x}e^{-y^2}\mathrm{d}x\mathrm{d}y $$ And the inner integral does not converge for any $y$.