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As we know, the classical linear double Euler sums is defined by $${S_{p,q}} = \sum\limits_{n = 1}^\infty {\frac{{{\zeta _n}\left( p \right)}}{{{n^q}}}} \;$$ where $p, q\ (q \ge 2)$ are positive integer, $${\zeta _n}\left( p \right) = \sum\limits_{k = 1}^n {\frac{1}{{{k^p}}}}. $$ Philippe Flajolet and Bruno Salvy prove that for an odd weight $ m = p+q$, the linear Euler sums $S_{ p,q} $are reducible to Riemann zeta value. I have a question, if $p,\ q\ (q>1)$ are real number, then the fractional Euler sums $S_{ p,q} $ can be evaluated? sunch as $${S_{\frac{1}{2},\frac{3}{2}}} = \sum\limits_{n = 1}^\infty {\frac{{{\zeta _n}\left( {\frac{1}{2}} \right)}}{{{n^{\frac{3}{2}}}}}} = ?.$$

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  • Hmm, but $\frac 12$ and $\frac 32$ are not really real numbers, which is what I read in (and expect from) your question? – Gottfried Helms Nov 20 '15 at 21:32
  • A more familiar notation for $~\zeta_n(p)~$ is $~H_n^{(p)},~$ which can be written as $~\zeta(p)~-~\displaystyle\sum_{k=1}^\infty\frac1{(n+k)^p}.~$ For the special case $p=1$ we have $~\zeta_n(1)~=~H_n~=~n\cdot\displaystyle\sum_{k=1}^\infty\frac1{k~(n+k)}~=~\int_0^1\frac{~1-x^n}{1-x~}~dx.~$ All of this is spelled out rather explicitly in the Wikipedia article on harmonic numbers, which can be found quite easily by way of a simple Google search. – Lucian Apr 07 '17 at 15:21
  • In general, $$S_{p,~q} ~=~ \zeta(p)\zeta(q)-\frac1{\Gamma(p)}\int_{0}^{1}\frac{\log^{p-1}\big(1/x\big)\cdot\text{Li}_{q}(x)}{1-x}~dx.$$ – Lucian Apr 07 '17 at 17:47

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