Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
    
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

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

http://www.frm.utn.edu.ar/analisisdsys/MATERIAL/Funcion_Gamma.pdf

http://en.wikipedia.org/wiki/Hölder%27s_theorem

http://en.wikipedia.org/wiki/Gamma_function#19th-20th_centuries:_characterizing_the_gamma_function

http://divizio.perso.math.cnrs.fr/PREPRINTS/16-JourneeAnnuelleSMF/difftransc.pdf

http://www.tandfonline.com/doi/abs/10.1080/17476930903394788?journalCode=gcov20

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

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.