What is $\int_0^{\infty} e^{\frac{-(\ln(x)-\mu)^2}{2 \sigma^2}}\, dx$

Question

Is there a closed form solution for this?

$$\int_0^{\infty} \exp\left({\frac{-(\ln(x)-\mu)^2}{2 \sigma^2}}\right)\, dx$$

Answer 1

Of course. By setting $x=e^t$ and completing the square we have:

$$ \int_{0}^{+\infty}\exp\left(-\frac{(\log x-\mu)^2}{2\sigma^2}\right)\,dx=\int_{-\infty}^{+\infty}\exp\left(t-\frac{(t-\mu)^2}{2\sigma^2}\right)\,dt=\color{red}{\sigma\,\sqrt{2\pi}\, e^{\mu+\frac{\sigma^2}{2}}}. $$

Answer 2

$$I=\int_0^\infty e^{\frac{-(\ln(x)-\mu)^2}{2\sigma^2}}dx$$ for simplicity I am going to let $a=2\sigma^2$ and $x=e^t$ $$I=\int_{-\infty}^\infty e^{\frac{-(t-\mu)^2}{a}}e^tdt=\int_{-\infty}^\infty e^{\frac{-t^2+2\mu t-\mu^2+at}{a}}dt=\int_{-\infty}^\infty e^{-\left(t-\frac{2\mu+a}{2}\right)^2+\left(\mu a+\frac{a^2}{4}\right)}dt$$$$ =e^{\mu a+\frac{a^2}{4}}\int_{-\infty}^\infty e^{-\left(t-\frac{2\mu+a}{2}\right)^2}dt$$ now by letting $$u=t-\frac{2\mu+a}{2}$$ you can easily solve this