TITLE: A (likely) positivity property of the Lerch zeta-function
QUESTION [6 upvotes]: 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?

REPLY [2 votes]: 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.