4

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?

Robert Z
  • 145,942
Sar
  • 907
  • 3
    If you don't have to actually compute it, then you're basically done with the first one. –  Jul 23 '18 at 15:49
  • 2
    $t^2-t=(t-\frac{1}{2})^2-\frac{1}{4}$. The integral becomes $e^{-\frac{1}{4}}\int_{-\frac{1}{2}}^{\infty}e^{-u^2}du$. – herb steinberg Jul 23 '18 at 15:55

3 Answers3

9

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 $-t^2+t\leq -t+1$.

Robert Z
  • 145,942
7

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

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

user
  • 154,566
0

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

Szeto
  • 11,159