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I have recently stumbled upon a curious approach to smoothing of the partial sums of Ordinary Dirichlet Series, and which I would like to share with others. Theorem 11.18 of Apostol's "Introduction to Analytic Number Theory" teaches how to express said partial sums (more precisely: partial sums - 1/2 the last term) by means of Perron's formula (here we are considering the example $f(n) = (-1)^{n-1}$, i.e. the Dirichlet series for the $ \eta(s)$ function, $ s=\sigma+it$):

$ \sum_{n\leq x} ^{*}\frac{(-1)^{n-1}}{n^{s}} \ = \ \frac{1}{2\pi i} \ \int_{c-\infty i} ^{c+\infty i} \eta (s + z) \ \frac{x^{z}}{z} \ \,dz $

provided that $ \sigma > \sigma_a - c$, then $c>0$ and $x>0$ can be arbitrary.

The integral term is necessarily a discontinuous function of $x$:

  • when $x=N$ (integer) $\Rightarrow$ the above integral yields $\sum_{1} ^{N}\frac{(-1)^{n-1}}{n^{s}} - \frac{(-1)^{N-1}}{2N^{s}}$
  • when $N<x<N+1$ $\Rightarrow$ the above integral yields $\sum_{1} ^{N}\frac{(-1)^{n-1}}{n^{s}}$
  • when $x=N+1$ $\Rightarrow$ the above integral yields $\sum_{1} ^{N+1}\frac{(-1)^{n-1}}{n^{s}} - \frac{(-1)^{N}}{2(N+1)^{s}}$

However, would it be possible to turn it into a continuous function of $x$ by progressively "bending", as a function of $x$, the integration path away from the vertical line $z=c$ ? We could for example try to to "bend" said vertical path of integration by means of a function whose contribution to such "bending" would vanish when $x = integer + \frac{1}{2}$.
As the purpose of such a function is to act as a bridge builder (pontifex in latin) in between consecutive partial sums, we could use the name Pontifex Path Bending Function, or PPBF in short. Many different PPBFs might be able to do the job, and the following is just a simple example:

$P(t,x) = H \ e^{-b t^2} cos[\pi x]^2 $, $H$ and $b$ positive constants,

which we can then use to bend the integration path from $z=c+it$ to $z=c+P(t,x)+it$

I have currently no idea about whether or not there might exist peculiar PPBF choices that, combined with some of the known alternative expressions for $\eta(s)$, would allow to simplify the above integral into a more approachable analytical expression, but I was instead able to numerically verify the PPBF trick by means of MATHEMATICA's NIntegrate.

The following figure illustrates how the NIntegrate evaluation of the "PPBF corrected" Perron Integral formula (black dots) compares with the partial sum values computed by simply adding up the corresponding number of terms (red dots, not very visible but they are located at the vertices):
enter image description here
The following two figures zoom around the two vertices:
enter image description here
enter image description here

I shall however remark that the above two interesting "revolving" patterns (around the corresponding partial sum values) still needs to be verified with respect to the possibility that they might be somehow affected by artefacts resulting from intrisic numerical computation errors. Perhaps, someone more familiar with the inner workings of NIntegrate may suggest a better choice for the various NIntegrate parameters. In order to get a rough feeling about said numerical errors, the simulation here below illustrates the case $P(t,x) = 0$ (i.e. no "path bending"), whereby the spread of the black dots around the corresponding vertex is related to said numerical error (ideally they should all overlap on top of the vertex itself, except for the two dots corresponding to integer $x$). Overall, we can say that such errors appear sufficiently small to affect only in a negligible way the conclusion that the PPBF trick is indeed effective in turning the Perron Formula's Integral into a continuous function of $x$.
enter image description here
Finally, here are the MATHEMATICA lines for the generation of the above plots:

s= 0.5+60*I ; c=0.51; K=23 ; J=50; b=0.01; H=1;
sm=20; X1=26.5; X2=27;
sm is the number of samples we wish to evaluate for X1 <= x <= X2
p[t_, i_]:=Exp[-b*t^2]HCos[Pi*i/sm]^2 ;
Dp[t_, i_]:=-2*btExp[-b*t^2]HCos[Pi*i/sm]^2 ;
Monitor[result0 = Table[{NIntegrate[1/(2*Pi)*(Dp[t,i]Im[ Zeta[s+c+p[t, i]+tI](1-2^(1-(s+c+p[t, i]+tI)))(i/sm)^(c+p[t, i]+tI)/(c+p[t,i]+tI)]+Re[ Zeta[s+c+p[t, i]+tI](1-2^(1-(s+c+p[t, i]+tI)))(i/sm)^(c+p[t, i]+tI)/(c+p[t, i]+t*I)]), {t,-2200,2200}, Method->{"GlobalAdaptive","MaxErrorIncreases"->10000}, MaxRecursion->50], NIntegrate[-1/(2*Pi)*(Dp[t,i]Re[ Zeta[s+c+p[t, i]+tI](1-2^(1-(s+c+p[t, i]+tI)))(i/sm)^(c+p[t, i]+tI)/(c+p[t, i]+tI)]-Im[ Zeta[s+c+p[t, i]+tI](1-2^(1-(s+c+p[t, i]+tI)))(i/sm)^(c+p[t, i]+tI)/(c+p[t, i]+t*I)]), {t,-2200,2200}, Method->{"GlobalAdaptive","MaxErrorIncreases"->10000}, MaxRecursion->50]}, {i, X1*sm, X2*sm}], i];
NIntegrate::slwcon: Numerical integration converging too slowly; suspect one of the following: singularity, value of the integration is 0, highly oscillatory integrand, or WorkingPrecision too small. >>
NIntegrate::slwcon: Numerical integration converging too slowly; suspect one of the following: singularity, value of the integration is 0, highly oscillatory integrand, or WorkingPrecision too small. >>
NIntegrate::slwcon: Numerical integration converging too slowly; suspect one of the following: singularity, value of the integration is 0, highly oscillatory integrand, or WorkingPrecision too small. >>
General::stop: Further output of NIntegrate::slwcon will be suppressed during this calculation. >>
plotIntegrals=ListPlot[result0, PlotStyle->RGBColor[0,0,0], PlotStyle->{PointSize[0.05]}];
plotIntSegm=ListLinePlot[result0, PlotStyle->RGBColor[0,0,0]];
a=0; t=60 ; d=0.51; K=22 ; J=50;
Sk = Sum[(-1)^(n-1)/n^(0.5+a+tI), {n, 1, K}];
V1 = Table[ {Re[Sk+Sum[(-1)^(n-1)/n^(0.5+a+t
I), {n, K+1,K+1+i}]], Im[Sk+Sum[(-1)^(n-1)/n^(0.5+a+t*I), {n, K+1,K+1+i}]] }, {i, 1, J}];
plotSums=ListPlot[V1, PlotStyle->RGBColor[1,0,0], PlotStyle->{PointSize1}];
segments=ListLinePlot[V1, PlotStyle->RGBColor[1,0,0]];
Show[segments, plotSums,plotIntegrals, plotIntSegm, ImageSize->{1000,1000}, AspectRatio->1, PlotRange->Automatic]

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    $\begingroup$ The usual approach to smoothing such partial sums is not to bend the contour but to replace $1/z$ in the kernel by a different function, e.g. by the Mellin transform at $z$ of a suitable $f:\mathbb{R}\to\mathbb{R}$. Your integral corresponds to $f$ being the characteristic function of $[0,1]$. $\endgroup$
    – GH from MO
    Jun 2, 2015 at 23:01
  • $\begingroup$ Thanks GH, not being myself familiar with said usual approach, would you have a link to infos I could explore? When you say "corresponds to f being ...", do you mean that exactly the same function of x, as resulting from the proposed PPBF, can be obtained by said kernel replacement? or the two approaches are still alternative ways to said smoothing? $\endgroup$
    – Luca
    Jun 3, 2015 at 8:34
  • $\begingroup$ What I said is very standard. If your replace $1/z$ by $\hat f(z)=\int_0^\infty f(t)t^{z-1}\,dt$ on the right hand side, then on the left hand side the terms acquires a factor $f(n/x)$ (with the sum running over the positive integers $n=1,2,...$). In particular, if $f$ is the characteristic function of $[0,1]$, then you get back your original formula. $\endgroup$
    – GH from MO
    Jun 3, 2015 at 12:46
  • $\begingroup$ OK, perhaps I am starting to get it. The smoothing approach you are referring to appears similar to the one described in Exampe 3 italic bold code on Kowalsky's Blog. I am going to study it in more detail, but my first impression is that it differs from the contour bending approach I am proposing, although it may definitely achieve similar results. $\endgroup$
    – Luca
    Jun 3, 2015 at 13:43
  • $\begingroup$ Exactly, this is what I meant. Deforming the contour usually has no effect, due to Cauchy's theorem (en.wikipedia.org/wiki/Cauchy%27s_integral_theorem). $\endgroup$
    – GH from MO
    Jun 3, 2015 at 13:56

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