The problem is to show that $\Re L(b/2,1/2,p+1)>0$ for all real $b\ne0$ and all real $p>-1$, where $$L(\lambda,c,s):=\sum_{k=0}^\infty\frac{\exp(2\pi i\lambda k)}{(k+c)^s}$$ is the Lerch zeta-function; see e.g. Wikipedia. Below are pictures of the sets $\{\big(b,p,\Re L(b/2,1/2,p+1)/2^p\big)\colon\,0.03<b<5,-0.9<p<15\}$ and $\{\big(b,p,\mathrm{sgn}\,\Re L(b/2,1/2,p+1)\big)\colon\,0.03<b<10,-0.9<p<15\}$,
which suggest that the statement is true. This problem arises as part of the problem stated at Is the integral always nonzero?



2 Answers 2


From Prudnikov,Brychkov,Marichev Integral and Series, Vol.1, sect. 5.4.3, formula 1 we derive that $$ \sum_{k=0}^\infty \frac{\cos(ak)}{(k+1/2)^{p+1}}= \frac{1}{\Gamma(s)}\int_0^{\infty}\frac{t^p e^{-t/2}(1-\cos(2\pi a)e^{-t})} {1-2\cos(2\pi a) e^{-t}+e^{-2t}}\,dt. $$

As everything under the integral's sigh is positive so the series is positive too.

  • $\begingroup$ Thank you Sergei. On the left-hand side, did you mean $2\pi a$ in place of $a$ and, on the right-hand side, did you mean $p+1$ in place of $s$? $\endgroup$ May 8, 2015 at 14:09
  • $\begingroup$ yes, and $s=p+1$ was your choice. $\endgroup$
    – Sergei
    May 8, 2015 at 16:08

Yes Sergei, looking back I see I asked a pretty strange question -- sorry. Indeed, writing $$\frac{e^{iak}}{(k+c)^{p+1}}=\frac1{\Gamma(p+1)}\,\int_0^\infty t^p e^{iak-(k+c)t}\,dt $$ for real $a$, real $p>-1$, $k=0,1,\dots$, and real $c>0$ and then writing $$\Re\sum_{k=0}^\infty e^{iak-(k+c)t}=\Re\frac{e^{-ct}}{1-e^{ia-t}} =\frac{e^{-ct}(1-e^{-t}\cos a)}{|1-e^{ia-t}|^2}>0 $$ for real $a$ and real $t>0$, one has $$\Re\sum_{k=0}^\infty\frac{e^{iak}}{(k+c)^{p+1}} =\frac1{\Gamma(p+1)}\,\int_0^\infty t^p \frac{e^{-ct}(1-e^{-t}\cos a)}{|1-e^{ia-t}|^2}\,dt>0 $$ for real $a\ne0$ and real $c>0$. (The interchange of the summation and integration, for $a\ne0$, is justified by dominated convergence.)

Essentially, what the above proof shows is that $\Re\sum_{k=0}^\infty e^{iak}g(k)>0$ for any nonzero function $g$ such that $g$ is completely monotone on $[0,\infty)$ and $g(\infty-)=0$ -- which is of course obvious.

  • $\begingroup$ Added a comment concerning an arbitrary completely monotone function in place of $1/(\cdot+c)^{p+1}$. $\endgroup$ May 8, 2015 at 16:26
  • $\begingroup$ I think your question was not a silly one. On the other hand, your post above is not an answer to your own question (rather, it is a comment), so it should be included in your original post, as part of the question or as a remark following the question. Only answers (or partial answers) should go in the answer fields. $\endgroup$
    – GH from MO
    May 8, 2015 at 16:39
  • $\begingroup$ GH: Thank you for the comment. My latest post details Sergei's answer (by providing a proof of the identity he cited) and complements it (by the remark about completely monotone functions). So, I don't see why that post belongs with the question rather than with the answers. $\endgroup$ May 8, 2015 at 18:07
  • $\begingroup$ OK, fair enough. I admit I did not read your post carefully. $\endgroup$
    – GH from MO
    May 8, 2015 at 18:27
  • $\begingroup$ But looking back, since Sergei's idea resolved the original question, it should be accepted, no? (so that the question attains "answered" status) $\endgroup$
    – Suvrit
    May 8, 2015 at 20:31

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