1
$\begingroup$

Is there a closed form for the constant given by:

$$\sum_{n=2}^\infty \frac{Ei(-(n-1)\log(2))}{n}$$

(Where $Ei$ is the exponential integral)?

Could we generalize it for:

$$I(k)=\sum_{n=2}^\infty \frac{Ei(-(n-1)\log(k))}{n}$$

?

My try: As it is given that $k$ will be a positive integer, I have already proved that these series converge at least for every $k>1$. To obtain a closed form, I have tried to substitute the exponential integral by both its main definition and its series expansion, with no success. Mathematica does not give any result either. Any help?

$\endgroup$
  • $\begingroup$ much better (+1). i will give it a shot tomorrow $\endgroup$ – tired Feb 13 '17 at 21:19
  • $\begingroup$ @tired Thank you very much! $\endgroup$ – user3141592 Feb 13 '17 at 21:29
  • $\begingroup$ i had no luck with this one two...i get some interesting integral representations which are close to something which can be solved in closed form. But not close enough i fear $\endgroup$ – tired Feb 18 '17 at 12:11
  • $\begingroup$ @tired Yes, playing with it I have been able to leave it in terms of improper integrals, but had no luck finding a closed form for them $\endgroup$ – user3141592 Feb 18 '17 at 12:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.