What about $\lim_{x\to 1}\left(\zeta(x)-\frac{1}{x^x-1}\right)=1+\gamma$?

Question

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.

Answer 1

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.

Edit

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