When I type

1 in the box lim x to, and

zeta(x)-1/(x^x-1) in the box Function,

of this online calculator (Wolfram Alpha) one has as output $$\lim_{x\to 1}\left(\zeta(x)-\frac{1}{x^x-1}\right)=1+\gamma,$$ where $\zeta(x)$ is the Riemann zeta function, and $\gamma$ is the Euler- Mascheroni constant.

I know that $$\lim_{x\to 1^{+}}\left(\zeta(x)-\frac{1}{x-1}\right)=\gamma,$$ and understand the proof.

Question. What about $$\lim_{x\to 1}\left(\zeta(x)-\frac{1}{x^x-1}\right)=1+\gamma?$$ Canyou give a proof of previous limit when $x\to 1^{+}$? Thanks in advance.

  • 1
    $\begingroup$ What about looking at the difference between $1/(x-1)$ and $1/(x^x-1)$. Just Taylor expand around $x=1$, and you will find the extra $1$. $\endgroup$ – mickep Jan 21 '16 at 14:23
  • $\begingroup$ Thanks @mickep, I will try your hint, but you or other user feel free to provide an answer. $\endgroup$ – user243301 Jan 21 '16 at 14:25

I do not know if this answers your question; so, please forgive me if I am off-topic.

Built around $x=1$, we have $$\zeta(x)=\frac{1}{x-1}+\gamma -\gamma _1 (x-1)+O\left((x-1)^2\right)$$ where appears the Stieltjes constant. On the other hand, starting from $$x^x=1+(x-1)+(x-1)^2+\frac{1}{2} (x-1)^3+O\left((x-1)^4\right) $$ $$\frac{1}{x^x-1}=\frac{1}{x-1}-1+\frac{x-1}{2}+O\left((x-1)^2\right)$$ So, $$\zeta(x)-\frac{1}{x^x-1}=(1+\gamma )-\left(\gamma _1+\frac{1}{2}\right) (x-1)+O\left((x-1)^2\right)$$ and then the result.


Making the problem more general, it is quite simple to show that $$\zeta(x)-\frac{1}{x^{x^n}-1}=(n+\gamma )- \left(\gamma _1+\frac{n^2}{2}\right)(x-1)+O\left((x-1)^2\right)$$

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  • $\begingroup$ (+1) I think you are on topic, and this is more or less what I was suggesting in my comment. $\endgroup$ – mickep Jan 21 '16 at 14:27
  • $\begingroup$ @mickep. Thank you and this is exactly what you suggested while I was typing. Cheers. $\endgroup$ – Claude Leibovici Jan 21 '16 at 14:28
  • $\begingroup$ I will read your proof with more detail, but I accept your answer, very thanks much @ClaudeLeibovici $\endgroup$ – user243301 Jan 21 '16 at 14:30
  • $\begingroup$ @JuanLG. You are very welcome ! What I did matches almost exactly what mickep was suggesting in comments. $\endgroup$ – Claude Leibovici Jan 21 '16 at 14:33
  • $\begingroup$ Are your mathematics and your results, mickep and Claude Leibovici. Thanks $\endgroup$ – user243301 Jan 21 '16 at 14:34

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