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?

  • 1
    $\begingroup$ 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? $\endgroup$ – Gerry Myerson Dec 15 '12 at 5:42
  • $\begingroup$ @GerryMyerson I did check out the paper but it seems only to discuss the rate of convergence of the sum (pg. 242). $\endgroup$ – Argon Dec 15 '12 at 18:05

The closed form means an expression containing only elementary functions. For your case no such a form exists. For more informations read these links:






Some background are needed for your understanding and good luck with these referrences.


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