Convergence of $\int_1^\infty e^{-\ln^2(x)}dx$.

Question

I`m interested in the convergence of the integral : $$\int_1^\infty e^{-\ln^2(x)}dx$$ I've tried using algebraic identities and some substitutions which lead me no where. Some examples to what I tried : $$\int_1^\infty e^{-\ln^2(x)}dx=\int_0^\infty e^{-t^2}e^{t}dt$$ and $$\int_1^\infty e^{-\ln^2(x)}dx=\int_1^\infty e^{-\ln(x)\ln(x)}dx=\int_1^\infty \frac 1 x^{\ln(x)} dx.$$

I also tried to use Cauchy convergence test and failed to succeed. Can anyone give me a hint?

Answer 1

Following your first approach, by letting $t=\ln(x)$ then $x=e^t$, $dx=e^t dt$ and we have that $$\int_1^\infty e^{-\ln^2(x)}dx=\int_0^\infty e^{-t^2}e^{t}dt\leq \int_0^\infty e^{-t+1}dt=[-e^{-t+1}]_0^\infty=e$$ because $-t^2+t\leq -t+1$.

Answer 2

We have that for $x>e^2 \implies \ln x >2$

$$e^{-\ln^2(x)}=(e^{-\ln(x)})^{\ln x}=\left(\frac1x\right)^{\ln x}<\frac1{x^2}$$

Answer 3

FYI, this integral can be evaluated exactly as $$\frac{e^{1/4}\sqrt\pi}2\left(\text{erf}\left(\frac12\right)+1\right)$$