<s>[INST] Integral involving nested Log I've been trying to solve this integral for a few days.
$$\int_0^{\infty}\left(\frac{1}{n}\left(t+n\right)\ln\left(\frac{t+n}{t}\right)-\ln\left(\frac{1}{n}\left(t+n\right)\ln\left(\frac{t+n}{t}\right)\right)-1\right)dt$$
For $n\gt0$.
I'm able to solve most of the integral until I got stuck trying to solve
$$\int\log\left(\log\left(\frac{t+n}{t}\right)\right)dt$$
Edit:
We first see that with a substitution we take $n$ out of the problem. Thus the integral we want to solve has a value of $0.38033\dots$  @Yuriy S has helped find an alternate form for integral.  I want to contribute another alternate form that can be derived from Yuriy's form.
$$I_1=-\frac{1}{4}+\int_{0}^{\infty}\left(-\frac{e^s-1}{2e^s}+\ln\left(e^s-1\right)-\ln\left(s\right)\right)\frac{e^s}{\left(e^s-1\right)^2}ds$$
Another Update:
I discovered that
$$\begin{align}
I_1+\frac14&=-\int_x^\infty\frac{1}{t(e^t-1)}dt-\left(-\frac1x-\frac{\ln{x}}{2}+\sum_{n=2}^\infty\frac{B_n}{n!(n-1)}x^{n-1}\right)
\\&=\sum_{n=1}^\infty \text{Ei}(-xn)-\left(-\frac1x-\frac{\ln{x}}{2}+\sum_{n=2}^\infty\frac{B_n}{n!(n-1)}x^{n-1}\right)
\end{align}$$ for $0\lt x\lt 2\pi$.  Here the non-integral part on the rhs is the series expansion of the integral part at $x=0$.
 [/INST] We can actually find a term by term expansion of the $f(s)=s+(e^s-1)\left(\ln(e^s-1)-\ln s-1\right)$ used in YuriyS's answer. If we rearrange $f(s)$, we get $$f(x) = (x+1-e^x) + (e^x-1)(\ln(e^x-1)-\ln(x))$$
If we look at $x+1-e^x$, this has a known Taylor series (which converges for all real) of $$-\sum_{n=2}^\infty \frac{x^n}{n!}$$
We also know that $$e^x-1 = \sum_{n=1}^\infty \frac{x^n}{n!}$$ which again converges for all real.
$\ln(e^x-1)-\ln(x)$ is a bit more tricky. If we differentiate it, we get $$\frac{e^x}{e^x-1}-\frac{1}{x} = 1+\frac{1}{e^x-1}-\frac{1}{x}$$ Here we can use the fact that $\frac{x}{e^x-1} = \sum_{n=0}^\infty \frac{B_n}{n!} x^n$ where $B_n$ are the Bernoulli numbers. If we divide by $x$ and add $1-\frac{1}{x}$, we get $$1+\frac{1}{e^x-1}-\frac{1}{x} = \frac{1}{2}+\sum_{n=2}^{\infty}\frac{B_n}{n!} x^{n-1}$$ Integrating, we then have that $$\ln(e^x-1)-\ln(x) = \frac{x}{2} + \sum_{n=2}^\infty\frac{B_n}{n! \cdot n}x^n$$
We now have that $$f(x) = -\sum_{n=2}^\infty \frac{x^n}{n!} + \sum_{n=1}^\infty \frac{x^n}{n!} \cdot \left(\frac{x}{2} + \sum_{n=2}^\infty\frac{B_n}{n! \cdot n}x^n\right) = -\sum_{n=2}^\infty \frac{x^n}{n!} + \frac{x}{2} \sum_{n=1}^\infty \frac{x^n}{n!}  + \sum_{n=1}^\infty \frac{x^n}{n!} \cdot \sum_{m=2}^\infty\frac{B_m}{m! \cdot m}x^m$$
$$f(x) = -\sum_{n=2}^\infty \frac{x^n}{n!} +  \sum_{n=2}^\infty \frac{x^{n}}{2(n-1)!}  + \sum_{n=3}^\infty \cdot \sum_{m=2}^{n-1}\frac{B_m}{(n-m)!m! \cdot m}x^n$$
Finally, we get the closed form for $a_n$ in $f(x) = \sum_{n=3}^\infty a_n x^n$ as $$a_n = \frac{n-2}{2(n!)}+\sum_{m=2}^{n-1}\frac{B_m}{(n-m)!m! \cdot m} = \sum_{m=2}^{n-1}\left(\frac{B_m}{(n-m)!m! \cdot m}+\frac{1}{2(n!)}\right)$$ 
Using the same $I_1 = \int_0^\infty f(s) \frac{e^s ds}{(e^s-1)^3}$ as YuriyS, we now want to find $$I_1 = \sum_{n=3}^\infty a_n \frac{n!}{2} (\zeta(n-1)-\zeta(n)) = \sum_{n=2}^\infty \left(a_{n+1}\frac{(n+1)!}{2}-a_{n}\frac{n!}{2}\right)\zeta(n)$$ I am not exactly sure what to do from here, but at least it is in the form of an infinite series instead of an integral.
Edit: As YuriyS mentioned in the comments, $a_n n!$ can be stated neatly as $b_n=\frac{n-2}{2}+\sum_{m=2}^{n-1}\frac{B_m \binom{n}{m}}{m}$. This means that $$I_1 = \sum_{n=3}^\infty \frac{b_n}{2} (\zeta(n-1)-\zeta(n)) = \sum_{n=2}^\infty \left(\frac{b_{n+1}}{2}-\frac{b_n}{2}\right)\zeta(n)$$
Edit 2: Unfortunately, these series will diverge, as mentioned in the comments, making it impossible for use in calculating $I_1$.
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