If I calculate the value of $\int\limits_{3}^\infty \frac{1}{x\ln(x)(\ln(\ln(x))^2)}dx$ by wolfram alpha it says $\approx 10.663$. However, If I do it by hand via a substitution and integration by parts I get a different value:
\begin{align*}&\int\limits_{3}^\beta\frac{1}{x\ln(x)(\ln(\ln(x))^2)}dx=\int\limits_{e^3}^{e^\beta}\frac{1}{y(\ln(y)^2)}dy=\frac{\ln(y)}{\ln(y)^2}\Big|_{e^3}^{e^\beta}+2\int\limits_{e^3}^{e^\beta}\frac{\ln(y)}{y(\ln(y)^3)}dy=\frac{1}{\ln(y)}\Big|_{e^3}^{e^\beta}+2\int\limits_{e^3}^{e^\beta}\frac{1}{y(\ln(y)^2)}dy\\&\implies -\int\limits_{e^3}^{e^\beta}\frac{1}{y(\ln(y)^2)}dy=\frac{1}{\beta}-\frac{1}{3}.\end{align*}If we let $\beta\to\infty$ we get $\int\limits_{3}^\infty \frac{1}{x\ln(x)(\ln(\ln(x))^2)}dx=\frac{1}{3}$. Where is my mistake?