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Is there a closed form for: $$f(x)=\sum_{k=1}^\infty (-1)^k \ln \left( \tanh \frac{\pi k x}{2} \right)=2\sum_{n=0}^\infty \frac{1}{2n+1}\frac{1}{e^{\pi (2n+1) x}+1}$$

This sum originated from a recent question, where we have:

$$f(1)= -\frac{1}{\pi}\int_0^1 \ln \left( \ln \frac{1}{x} \right) \frac{dx}{1+x^2}=\ln \frac{\Gamma (3/4)}{\pi^{1/4}}$$

If we differentiate w.r.t. $x$, we obtain:

$$f'(x)=\sum_{k=1}^\infty (-1)^k \frac{\pi k}{\sinh \pi k x}$$

There is again a closed form for $x=1$ (obtained numerically):

$$f'(1)=-\frac{1}{4}$$

So, is there a closed form or at least an integral definition for arbitrary $x>0$?


The series converges absolutely (numerically at least):

$$\sum_{k=1}^\infty \ln \left( \tanh \frac{\pi k x}{2} \right)< \infty$$

Thus, this series can also be expressed as a logarithm of an infinite product:

$$f(x)=\ln \prod_{k=1}^\infty \tanh (\pi k x) - \ln \prod_{k=1}^\infty \tanh \left( \pi (k-1/2) x \right)$$

$$e^{f(x)}= \prod_{k=1}^\infty \frac{\tanh (\pi k x)}{\tanh \left( \pi (k-1/2) x \right)}$$

This by the way leads to:

$$\prod_{k=1}^\infty \frac{\tanh (\pi k)}{\tanh \left( \pi (k-1/2) \right)}=\frac{\pi^{1/4}}{\Gamma(3/4)}$$

I feel like there is a way to use the infinite product form for $\sinh$ and $\cosh$:

$$\sinh (\pi x)=\pi x \prod_{n=1}^\infty \left(1+\frac{x^2}{n^2} \right)$$

$$\cosh (\pi x)=\prod_{n=1}^\infty \left(1+\frac{x^2}{(n-1/2)^2} \right)$$

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    $\begingroup$ In my answer to the linked question I have shown that the series on right equals $$\frac{1}{2}\log\vartheta_{3}(q)$$ where $q = e^{-\pi x}$. I don't think there is any closed form different from theta functions and their cousins elliptic integrals. $\endgroup$ – Paramanand Singh Feb 26 '17 at 16:14
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    $\begingroup$ One can get a closed form if one considers the function $$g(x) = \sum_{n = 1}^{\infty}\log\tanh \frac{n\pi x}{2} = \log\vartheta_{4}(q)$$ (see math.stackexchange.com/a/1793756/72031) and then $$g(x) - f(x) = \frac{1}{4}\log\frac{\vartheta_{4}^{4}(q)}{\vartheta_{3}^{4}(q)} = \frac{1}{4}\log(1 - k^{2})$$ so that if $x = \sqrt{r}, r \in\mathbb{Q}^{+}$ then $k$ is algebraic and we have a closed form for $g(x) - f(x)$. $\endgroup$ – Paramanand Singh Feb 26 '17 at 16:42
  • $\begingroup$ @ParamanandSingh, oh, I didn't read your answer carefully enough. I haven't noticed that you give the general closed form $\endgroup$ – Yuriy S Feb 26 '17 at 16:45
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    $\begingroup$ The general form is proved there, but it is evaluated in terms of Gamma values only for specific value of $q = e^{-\pi}$ (which is specific to that integral like your $f(1)$ instead of general $f(x)$). $\endgroup$ – Paramanand Singh Feb 26 '17 at 16:51
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Let’s use $~\displaystyle\prod\limits_{k=1}^\infty (1+z^k)(1-z^{2k-1}) =1~$ . $\enspace$ (It's explained in a note below.)

For $~z:=q^2~$ and $~q:=e^{-\pi x}~$ with $~x>0~$ we get

$\displaystyle e^{f(x)} = \prod\limits_{k=1}^\infty\frac{\tanh(k\pi x)}{\tanh((k-\frac{1}{2})\pi x)} = \prod\limits_{k=1}^\infty\frac{ \frac{q^{-k}-q^k}{q^{-k}+q^k} }{ \frac{q^{\frac{1}{2}-k}-q^{k-\frac{1}{2} }}{q^{\frac{1}{2}-k}+q^{k-\frac{1}{2}}} } = \prod\limits_{k=1}^\infty\frac{(1-q^{2k})(1+q^{2k-1})}{(1+q^{2k})(1-q^{2k-1})} =$

$\displaystyle = \prod\limits_{k=1}^\infty (1-q^{2k})(1+q^{2k-1})^2 = \sum\limits_{k=-\infty}^{+\infty} q^{k^2} = \vartheta(0;ix)$

The “closed form” for $\,f\,$ is:

$$f(x) = \ln\vartheta(0;ix)$$

Please see e.g. Theta function .


Note:

$\displaystyle\prod\limits_{k=1}^\infty (1+z^k)(1-z^{2k-1}) =1$

$\Leftrightarrow\hspace{2cm}$ (logarithm)

$\displaystyle \sum\limits_{k=1}^\infty \sum\limits_{v=1}^\infty \frac{(-1)^{v-1}z^{kv}}{v} = \sum\limits_{k=1}^\infty \ln(1+z^k) = -\sum\limits_{k=1}^\infty \ln(1-z^{2k-1}) = \sum\limits_{k=1}^\infty \sum\limits_{v=1}^\infty \frac{z^{(2k-1)v}}{v}$

$\Leftrightarrow\hspace{2cm}$ (exchanging the sum symbols which is valid for $~|z|<1~$

$\hspace{2.7cm}$ and using $~\displaystyle\frac{x}{1-x}=\sum\limits_{k=1}^\infty x^k~$)

$\displaystyle\sum\limits_{v=1}^\infty \frac{(-1)^{v-1}}{v}\frac{z^v}{1-z^v} = \sum\limits_{v=1}^\infty \frac{1}{v}\frac{z^v}{1-z^v} - 2\sum\limits_{v=1}^\infty \frac{1}{2v}\frac{z^{2v}}{1-z^{2v}} = \sum\limits_{v=1}^\infty \frac{1}{v}\frac{z^v}{1-z^{2v}}$

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  • $\begingroup$ Where does the first identity come from? "Let's use..." $\endgroup$ – Diger Mar 22 at 11:27
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    $\begingroup$ @Diger : I've added a note . ;) $\endgroup$ – user90369 Mar 22 at 12:54
  • $\begingroup$ Very nice answer, thank you! I haven't checked numerically yet, but I trust you $\endgroup$ – Yuriy S Mar 25 at 8:03
  • $\begingroup$ @YuriyS : You are welcome! And thank you for paying attention to my answer although my answer is 2 years too late. ;) Explicit values can be seen e.g. at the web page for Theta function (the link above). $\endgroup$ – user90369 Mar 25 at 9:08

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