I want to show whether $I+J+\cdot\cdot\cdot$ converges or diverges and how to notate everything correctly.
$$ I=\int_0^1 \exp\bigg(\frac{1}{\ln(x)}\bigg)~dx$$
$$ G(x)=\int_0^x\exp\bigg(\frac{1}{\ln(x)}\bigg)~dx$$
$$ J=\int_0^1 G(x)~dx$$
$$ \cdot\cdot\cdot$$
So far I have $I=2K_1(2)$ where $K_1$ is a modified bessel function of the second kind, and $J\approx.2099$
Here's my desmos code that I've worked on for this problem: https://www.desmos.com/calculator/fj9pwsccii
By defining $$ f_n(x) = \int_{0}^{x}\int_{0}^{x_n}\ldots\int_{0}^{x_2}e^{\frac{1}{\log x_1}}\,dx_1\,dx_2\cdots dx_n $$ we have $$ f_n(x) = \int_{0<x_1 < x_2<\ldots<x_n<x}e^{\frac{1}{\log x_1}}\prod dx_k = \int_{0}^{x}\,\frac{1}{(n-1)!}(x-x_1)^{n-1}e^{\frac{1}{\log x_1}}dx_1$$ and by letting $x_1 = x v, dx_1 = x\,dv$ $$ f_n(x) = \frac{x^n}{(n-1)!}\int_{0}^{1}(1-v)^{n-1}e^{\frac{1}{\log x+\log v}}\,dv. $$ Since $\int_{0}^{1}v^m e^{\frac{1}{\log v}}\,dv = \frac{2}{\sqrt{m+1}}\,K_1\left(\frac{2}{\sqrt{m+1}}\right)$, by the binomial theorem
$$ f_n(1) = \frac{2}{(n-1)!}\sum_{k=0}^{n-1}\binom{n-1}{k}\frac{(-1)^k}{\sqrt{k+1}}\,K_1\left(\frac{2}{\sqrt{k+1}}\right). $$ It is interesting to point out that $$ \sum_{n\geq 1}f_n(1) = \int_{0}^{1}\sum_{n\geq 1}\frac{(1-v)^{n-1}}{(n-1)!}e^{\frac{1}{\log v}}\,dv=\int_{0}^{1}\exp\left(1-v+\frac{1}{\log v}\right)\,dv $$ is convergent like $f_1(1)$, since $e^{1-v}$ is bounded between two positive constants for $v\in[0,1]$.