6
$\begingroup$

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?

pics

$\endgroup$
0

2 Answers 2

7
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.