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I had attempted to evaluate

$$\int_2^\infty (\zeta(x)-1)\, dx \approx 0.605521788882$$

Upon writing out the zeta function as a sum, I got

$$\int_2^\infty \left(\frac{1}{2^x}+\frac{1}{3^x}+\cdots\right)\, dx = \sum_{n=2}^\infty \frac{1}{n^2\log n}$$

This sum is mentioned in the OEIS.

All my attempts to evaluate this sum have been fruitless. Does anyone know of a closed form, or perhaps, another interesting alternate form?

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I tried the inverse symbolic calculator, isc.carma.newcastle.edu.au/index --- standard search got me nothing, advanced search got some unexplained symbol (but no answer), so I expect there's nothing known and nothing simple possible. Did you check the Monthly paper linked at the OEIS? – Gerry Myerson Dec 15 '12 at 5:42
    
@GerryMyerson I did check out the paper but it seems only to discuss the rate of convergence of the sum (pg. 242). – Argon Dec 15 '12 at 18:05

Some time ago I derived a formula:

$$ \int\limits_{2}^\infty (\zeta(s)-1) ds=\frac{3}{2} - \int\limits_{0}^\infty \frac{H_x dx}{x^2(\mathrm{ln}^2 x+\pi^2)} $$ where $\mathrm{ln}(x) $ means natural logarithm and $H_x=\sum\limits_{n=1}^\infty \frac{1}{n}-\frac{1}{n+x}$

If it is true it will be a good start.

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