$$\int^{\infty}_0 \frac{\log t+\Gamma (0,t) + \gamma}{e^t-1}~dt=\sum_{k = 2}^{\infty} \frac{(-1)^k \zeta(k)}{k-1}=1.2577468869$$
Here $\gamma$ -Euler constant, $\Gamma (0,t)$ - incomplete gamma function:
$$\Gamma (0,t)=\int_t^{\infty} \frac{e^{-p}}{p} dp=\cfrac{\exp(-t)}{t+1-\cfrac{1}{t+3-\cfrac{4}{t+5-\cfrac{9}{t+7-\cdots}}}}$$
What is so special about this integral? None of the terms converge on their own, according to WolframAlpha:
$$\int^{\infty}_0 \frac{\log t}{e^t-1} ~dt \to \infty$$
$$\int^{\infty}_0 \frac{\Gamma(0,t)}{e^t-1} ~dt \to \infty$$
$$\int^{\infty}_0 \frac{\gamma}{e^t-1} ~dt \to \infty$$
Moreover, no two terms together converge (again, according to WolframAlpha).
This integral is equal to another integral and several infinite series:
$$\int^{\infty}_0 \frac{\log t+\Gamma (0,t) + \gamma}{e^t-1}~dt=$$
$$=\frac{\pi^2}{4}-1-4\int_{0}^{\infty} \frac{\arctan x}{1+x^{2}} \frac{dx}{e^{\pi x}+1}=\sum_{k=1}^{\infty} \frac{\ln (1+\frac{1}{k})}{k}=\sum_{k = 2}^{\infty} \frac{(-1)^k \zeta(k)}{k-1}$$
See this answer and this answer for clarification.
And now proof. See this question for the proof of the following integral:
$$\int^1_0 \frac{\log^n x}{1-x}dx=(-1)^n~ n!~ \zeta(n+1)$$
Now we change the variable under the integral:
$$\log x=-t$$
$$\int^{\infty}_0 \frac{(-1)^n t^n \exp(-t)}{1-\exp(-t)}dt=(-1)^n~ n!~ \zeta(n+1)$$
Now we divide both sides by $n!~n$ and sum from $n=1$ to $\infty$:
$$\sum_{n=1}^{\infty} \frac{(-1)^n t^n }{n!n}=-\log t-\Gamma (0,t) - \gamma$$
See WolframAlpha.
Finally, we can write:
$$-\int^{\infty}_0 \frac{\exp(-t)}{1-\exp(-t)} (\log t+\Gamma (0,t) + \gamma)~dt=\sum_{n = 1}^{\infty} \frac{(-1)^n \zeta(n+1)}{n}$$
$$\int^{\infty}_0 \frac{\log t+\Gamma (0,t) + \gamma}{e^t-1}~dt=\sum_{n = 2}^{\infty} \frac{(-1)^n \zeta(n)}{n-1}$$
The rest of the equalitites, including the numerical value of the integral follows from this one.
Is it a known integral? Have you seen it before? Some reference of other proofs would be nice, since I don't know if anyone tried to prove it the same way
