Monday Math 133

Find the limit $\lim_{s\to1}\left(1-x^{1-s}\right)\zeta(s)$ as a function of positive, real x; where ζ(s) is the Riemann zeta function.

The Riemann zeta function has a pole at s=1, while the term in the brackets is zero at s=1. Recall from here that the Dirichlet eta function η(s), which is defined for all complex numbers s, is related to the Riemann zeta function by
$\eta(s)=(1-2^{1-s})\zeta(s)$ ,
and thus
$\zeta(s)=\frac{\eta(s)}{1-2^{1-s}}$ .
Therefore,
$\lim_{s\to1}\left(1-x^{1-s}\right)\zeta(s)=\lim_{s\to1}\frac{1-x^{1-s}}{1-2^{1-s}}\eta(s)$ ,
and since η(1)=ln2, we then see
$\lim_{s\to1}\left(1-x^{1-s}\right)\zeta(s)=(\ln2)\lim_{s\to1}\frac{1-x^{1-s}}{1-2^{1-s}}$ .
This last limit has indeterminate form 0/0, and so is amenable to L’Hôpital’s rule:
$\begin{eqnarray}\lim_{s\to1}\frac{1-x^{1-s}}{1-2^{1-s}}&=&\lim_{s\to1}\frac{\frac{d}{ds}\left(1-x^{1-s}\right)}{\frac{d}{ds}\left(1-2^{1-s}\right)}\\&=&\lim_{s\to1}\frac{x^{1-s}\ln{x}}{2^{1-s}\ln{2}}\\&=&\frac{\ln{x}}{\ln2}\end{eqnarray}$ ,
and so
$\lim_{s\to1}\left(1-x^{1-s}\right)\zeta(s)=(\ln2)\frac{\ln{x}}{\ln2}=\ln{x}$ .

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Tags: Dirichlet Eta Function, L'Hôpital's Rule, Math, Monday Math, Riemann Zeta Function

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Monday Math 133

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