$f(x)=\sum\limits_{k = 2 }^ \infty e^{-kx} \ln(k) $
$\int\limits_0^{\infty}\int\limits_x^{\infty}\, f(\gamma)\, d\gamma dx=\sum\limits_{k = 2 }^ \infty \frac{1}{k^2} \ln(k) $
$\int\limits_0^{\infty}\int\limits_x^{\infty} f(\gamma)\, d\gamma dx=\sum\limits_{k=2}^ \infty\ln(k^{\frac{1}{k^2}})=\ln(\prod\limits_{k=2}^{\infty}k^{\frac{1}{k^2}}) $
$\prod\limits_{k=2}^{ \infty }k^{\frac{1}{k^2}}=\prod\limits_{k=2}^{ \infty } \sqrt[k^2]{k}=e^{\int\limits_0^{\infty}\int\limits_x^{\infty} f(\gamma) \,d\gamma dx}$
$f(x)=\sum\limits_{k = 2 }^ \infty e^{-kx} \ln(k) $
$f(x)=\sum\limits_{k = 1 }^ \infty e^{-(k+1)x} \ln(k+1) $
$f(x)=e^{-x}\sum\limits_{k = 1 }^ \infty e^{-kx} \ln(k+1) $
$f(x)=e^{-x}\sum\limits_{n = 1 }^ \infty \frac{(-1)^{n+1}}{n} \sum\limits_{k = 1 }^ \infty k^n e^{-kx}$
We know that
$\sum\limits_{k = 1 }^ \infty e^{-kx}= \frac{1}{e^{x}-1} $
$\sum\limits_{k = 1 }^ \infty k^n e^{-kx}= (-1)^n\frac{d^n}{dx^n}(\frac{1}{e^{x}-1}) $
$f(x)=e^{-x}\sum\limits_{n = 1 }^ \infty \frac{(-1)^{n+1}}{n} \sum\limits_{k = 1 }^ \infty k^n e^{-kx} = e^{-x}\sum\limits_{n = 1 }^ \infty \frac{(-1)^{n+1}}{n} (-1)^n\frac{d^n}{dx^n}(\frac{1}{e^{x}-1})$
$f(x)=-e^{-x}\sum\limits_{n = 1 }^ \infty \frac{1}{n} \frac{d^n}{dx^n}(\frac{1}{e^x-1})$
$\int\limits_0^{\infty}\int\limits_x^{\infty} f(\gamma) \,d\gamma dx= -\int\limits_0^{\infty}\int\limits_x^{\infty} e^{-\gamma}\sum\limits_{n = 1 }^ \infty \frac{1}{n} \frac{d^n}{d\gamma^n}(\frac{1}{e^{\gamma}-1})\, d\gamma dx$
I have lost my way after that.
Is it possible to find a closed form in my way? or I need to follow a different way. I need your mathematical sense. Thanks a lot for answers and advice.