A (likely) positivity property of the Lerch zeta-function

Question

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?

Answer 1

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.

Answer 2

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.