Edit 6.2.2012: The sequence to be transformed should be f = 0,1,2,3,4,5... which makes the mentioning of the von Mangoldt function less necessary.
Edit 5.2.2012: I had the wrong plot of the insignal. This changes the question a lot.
In Heike's answer to the stackoverflow Riemann zeta zero spectrum question, a variant of the Fourier sine transform is described.
Clear[f] scale = 1000000; f = Range[scale] - 1; (*f = 0,1,2,3,4,5...*) xres = .002; xlist = Exp[Range[0, Log[scale], xres]]; ListLinePlot[f[[Floor[xlist]]] - xlist, ImageSize -> Large, DataRange -> {0, Log[scale]}] tmax = 100; tres = .015; ymin = -0.025; ymax = 0.015; Monitor[errList = Table[(xlist^(-1/2 + I t).(f[[Floor[xlist]]] - xlist)), {t, Range[0, tmax, tres]}];, t] g1 = ListLinePlot[Im[errList]/Length[xlist], DataRange -> {0, tmax}, PlotRange -> {ymin, ymax}, Frame -> True, Axes -> False]; g2 = Graphics[ Line[{{N[Im[ZetaZero[1]]], ymin}, {N[Im[ZetaZero[1]]], ymax}}]]; g3 = Graphics[ Line[{{N[Im[ZetaZero[2]]], ymin}, {N[Im[ZetaZero[2]]], ymax}}]]; g4 = Graphics[ Line[{{N[Im[ZetaZero[3]]], ymin}, {N[Im[ZetaZero[3]]], ymax}}]]; gw = Show[g1, g2, g3, g4, ImageSize -> Large]
Which outputs:
Notice the approximate agreement with the imaginary parts of the Riemann zeta zeros (vertical lines) and local minima in the spectrum.
The plot of the sequence $a = $(f[[Floor[xlist]]] - xlist))
for scale=20
ListLinePlot[f[[Floor[xlist]]] - xlist, ImageSize -> Large, DataRange -> {0, Log[scale]}]
is:
for scale=1000000
the plot is:
In latex the sequence to be transformed is similar, but not equal to:
$\displaystyle a = -1 - \pi + 2(\frac{\sin(e^{x})}{1} + \frac{\sin(2e^{x})}{2} + \frac{\sin(3e^{x})}{3} + ...)$
And the discrete transform itself I believe should look something like:
$\displaystyle X_{t} = \sum\limits_{x=0}^{\log(scale)} e^{-\frac{1}{2}x} \cdot \sin(t \cdot x) \cdot a$
But knowing the individual frequencies that build up the exponential saw tooth wave, does not help.
Having thought about this after rewriting the question several times, I start to think that the minima probably are at imaginary parts of Riemann zeta zeros. But even if the derivative of the spectrum would give the zeta zeros as zeros they still are calculated as zeros.