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I am looking for a closed form of $$ S \equiv \sum_{k = 1}^{\infty} \frac{\log\left(2k + 1\right) - \log\left(2k - 1\right)}{k} $$

Notes:

  1. I do not know if there is a closed form or if I should expect a closed form.

  2. This is a convergent sum as per $\tt Wolfram$.

  3. I expect the hypothetical closed form to involve Polylog terms since my attempt $1$ ( see below ) already showed that in the integral part. I am fine with such a closed form ( it need not be in terms of known constants ). But I am not looking for more unsatisfactory result involving Hypergeometric/MeijerG etc$\ldots$ which are currently very mysterious to me :)

Context:

I was solving $\sum\limits_{k=-\infty}^{\infty} \left(\rule{0pt}{5mm}\left\vert\operatorname{Si} \left(k\right)\right\vert - \pi/2\right)$, where $\operatorname{Si}\left(x\right)$ is the Sine Integral Function.

I did a FourierTransform of $\left(\rule{0pt}{5mm} \left\vert\operatorname{Si}\left(x\right)\right\vert - \pi/2\right)$ and then tried taking the Poisson Sum. The Poisson Sum involves the sum $S$ described above. That's where this question arises.

My attempt

Since I don't know how to attack this sum directly, I thought of converting to integrals:

  1. I first tired Euler Maclaurin Expansion. I was able to compute the integral in closed form, but the remainder term $R_{p}$ again leads to the same type of sum involving the same type of logarithm. Since I am NOT looking for an approximation, I cannot avoid the remainder/error term. So I could not proceed.
  2. I briefly considered usng Abel-Plana Expansion which is purely in the form of integrals. But I could not apply that technique since the function is not analytic for $\Re\left(z\right) \ge 0$ ( as the function has branch points at $1/2$ and $-1/2$ ).

Question

Can someone show steps to derive a closed form for $S$ ?.

OR at least give me some hints or directions to explore.

Felix Marin
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Srini
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  • Evaluation of $\sum\limits_{n=0}^\infty \left(\operatorname{Si}(n)-\frac{\pi}{2}\right)$? was asked about, but no closed form was found. – Тyma Gaidash Aug 19 '24 at 17:47
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    It can be rewritten as $$\sum_k \frac{\ln(2k+1)}{k(k+1)}.$$ Don't see how that helps much. – Thomas Andrews Aug 19 '24 at 17:47
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    @TymaGaidash, thanks! I should have searched the sine integral sum first. Instead I was searching for my converted form and didn't find anything matching. Disappointing that there is not much hope of finding a closed form. In even bigger context (which I did not provide in my question), the sum involving sine integral came from me trying to answer this question: https://math.stackexchange.com/questions/4959057/is-lobachevsky-acceptable-here-int-0-infty-frac-sinxx-arcsin-cos#comment10612458_4959057 – Srini Aug 19 '24 at 17:53
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    On the other hand, the sum seems to be $\int_1^\infty\left(2-\frac\pi x\cot\left(\frac\pi{2x}\right)\right)dx\approx 1.7506$ – Тyma Gaidash Aug 19 '24 at 18:07
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    $$S=-2\int_0^\infty\frac{\sinh t}t,\ln(1-e^{-2t})dt=\pi\int_0^{\pi/2}\left(\frac 1{\sin^2x}-\frac{\cot x}x\right)dx-2=1.75063...$$ – Svyatoslav Aug 19 '24 at 18:53
  • $\tt Mathematica$ is unable to evaluate the sum. – Felix Marin Aug 19 '24 at 20:07
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    @FelixMarin Converting it to an infinite product seems better for mathematica. It gives, using ThomasAndrews expression, $S=\ln \Pi_{1,\infty} (2k+1)^{\frac{1}{k(k+1)}} = 1.750632...$, in agreement with Svyatoslav – Amos C N Aug 20 '24 at 08:26
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    A rapidly converging representation is $\sum_{s\ge 1} \frac{1}{(s-1/2)2^{2s-1}} \zeta(2s)$ where $\zeta$ is the Riemann zeta-function (rational mutliples of $\pi$). – R. J. Mathar Feb 12 '25 at 10:01

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