Can this integral be evaluated with Gamma function?

Question

The integral is $\int_{0}^\infty e^{-\frac{(lnx)^2}2} \,dx $

If I substitute $x = e^z$, and further $\frac{z^2}2 = t$ then it is not in the form of Gamma function.

Answer 1

Substitute $x=e^z$ $$\int_{0}^\infty e^{-\frac{(lnx)^2}2} \,dx=\int_{-\infty}^\infty e^{-\frac{z^2}2} e^z\,dz=\int_{-\infty}^\infty e^{-\frac12(z^2-2z)}\,dz$$ Complete the square: $z^2-2z=(z-1)^2-1$. $$ \int_{-\infty}^\infty e^{-\frac12(z^2-2z)}\,dz = \int_{-\infty}^\infty e^{-\frac{(z-1)^2}2}\,e^{\frac12}\,dz=e^{\frac12} 2\int_1^\infty e^{-\frac{(z-1)^2}2}\,dz$$. Now you can substitute $t=\frac{(z-1)^2}2$ and use gamma-function, if it is required.

Answer 2

You may notice that $$\int_{0}^{+\infty}e^{-\log^2(x)/2}\,dx \stackrel{x\mapsto e^z}{=} \int_{\mathbb{R}}\exp\left(\frac{2z-z^2}{2}\right)\,dz \stackrel{z\mapsto t+1}{=}\sqrt{e}\int_{\mathbb{R}}e^{-t^2/2}\,dt=\color{red}{\sqrt{2\pi e}}.$$