Consider the function
$$ f(x)=\lim_{k \to \infty}\bigg(\int_0^x \sum_{n=1}^k e^{\frac{\log n}{\log r}}~dr \bigg)\bigg( \int_0^1 \sum_{n=1}^k e^{\frac{\log n}{\log r}}~dr \bigg)^{-1} $$
I want to find a closed form for
$$\sum_{n=1}^\infty f\bigg( e^{-\sqrt{\log n}} \bigg). $$
I simplified this to
$$\sum_{n=1}^\infty f\bigg( e^{-\sqrt{\log n}} \bigg)=\sum_{n=1}^\infty \bigg( \lim_{k \to \infty}\bigg(\int_0^{e^{-\sqrt{\log n}}} \sum_{n=1}^k e^{\frac{\log n}{\log r}}~dr \bigg)\bigg( \int_0^1 \sum_{n=1}^k e^{\frac{\log n}{\log r}}~dr \bigg)^{-1}\bigg)$$
After simplyfying more I arrived at:
$$ = 1/2+1/4\bigg(\lim_{k\to\infty}\bigg( \sum_{n=1}^k e^{-2\sqrt{\log n}} \bigg)\bigg(\sum_{n=1}^k \sqrt{\log n}~K_1\big(2\sqrt{{\log n}} \big) \bigg)^{-1}\bigg). $$
Is there a closed form for the limit in parentheses?
$K_1$ is the modified bessel function.
But I'm unable to make progress on simplifying the last limit.
I did think to look at the asymptotic expansion for large $x$ here:
$$K_{\nu}(x)\sim \sqrt{\frac{\pi}{2x}}e^{-x}\Big(1+\frac{4\nu^2-1}{8x}+\frac{(4\nu^2-1)(4\nu^2-9)}{2!(8x)^2}+\ldots\Big).$$
Then some calculations based on the first 2 terms of the expansion seem to suggest $$\sum_{n=1}^\infty f\bigg( e^{-\sqrt{\log n}} \bigg)=1/2$$
because we basically have
$$ 1/2+C/4\bigg( \lim_{k\to \infty} \frac{\sum_{n=1}^k e^{-2\sqrt{\log n}}}{\sum_{n=1}^k(\log n)^{1/4}e^{-2\sqrt{\log n}}} \bigg) $$
