Integral with e and log

Question

so I have the following integral $$\int_0^\infty e^{-\log^2x}dx$$ I tried the following substitution $\log x=t \, ; \log^2x =t^2$ and substituted got to $$\int_0^\infty e^{-t^2}e^{-t}dt$$ should I split this integral and use Gauss integral?or what method should I approach I get blocked after substituting.

Answer 1

Actually, the first step seems wrong. If $x=e^t \iff t = \ln x$ then $$ I = \int_0^\infty e^{-\ln^2 x}dx = \int_{t=-\infty}^{t=\infty}e^{-t^2}e^{-t}dt $$ and we can complete the square in the exponent, noting that $$ -t^2-t = 1/4-(t+1/2)^2 $$ and hence (using $u=t+1/2$) $$ I = e^{1/4} \int_{-\infty}^\infty e^{-(t+1/2)^2}dt = e^{1/4} \int_{-\infty}^\infty e^{-u^2}du = e^{1/4} \sqrt{\pi} = \sqrt[4]{\pi^2e} $$ using the usual trick with polar coordinates.