Is there a closed form solution for this definite integral? I'm trying to derive the cumulative distribution function for a statistical distribution in the lognormal family.
$$\int_0^{p} \exp\left({\frac{-(\ln(x)-\mu)^2}{2 \sigma^2}}\right)\, dx$$
where $$0<p<\infty$$
Let $$I=\int_{0}^{p} \exp[-(\ln x- a)^2/(2b^2)] ~dx.$$ Let $x=e^t$, then $$I=\int_{0}^{\ln p} \exp[-(t-a)^2/(2b^2)+t]~ dt= \int_{0}^{\ln p} \exp[-(t^2-2at-2b^2 t+a^2)/(2b^2)] dt=$$ $$ I=\exp[((a+b^2)^2-a^2)/(2b^2)] \int_{0}^{p} \exp[-(t-(a+b^2))^2/(2b^2)].$$ Let $(t-a-b^2)(/b\sqrt{2})=u$, then $$\Rightarrow I= \exp[((a+b^2)^2-a^2)/(2 b^2)] \int_{-(a+b^2)/(b \sqrt{2})}^{(\ln p-a-b^2)/(b \sqrt{2})} ~~~e^{-u^2} b \sqrt{2}~ du.$$ So $$I=\sqrt{2\pi}~b~\exp[((a+b^2)^2-a^2)/(2b^2)]~~ ~[Erf[(\ln p-a-b^2)/(b\sqrt{2})]-Erf[~(a+b^2)/(b \sqrt{2})]~].$$