# Closed form for the series $\sum_{k=1}^\infty (-1)^k \ln \left( \tanh \frac{\pi k x}{2} \right)$

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 $\cosh$:

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

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

• 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. – Paramanand Singh Feb 26 '17 at 16:14
• 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 $g(x) - f(x)$. – Paramanand Singh Feb 26 '17 at 16:42
• @ParamanandSingh, oh, I didn't read your answer carefully enough. I haven't noticed that you give the general closed form – Yuriy S Feb 26 '17 at 16:45
• 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 $f(x)$). – Paramanand Singh Feb 26 '17 at 16:51

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 $$~\displaystyle\frac{x}{1-x}=\sum\limits_{k=1}^\infty x^k~$$)

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

• Where does the first identity come from? "Let's use..." – Diger Mar 22 at 11:27
• @Diger : I've added a note . ;) – user90369 Mar 22 at 12:54
• Very nice answer, thank you! I haven't checked numerically yet, but I trust you – Yuriy S Mar 25 at 8:03
• @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). – user90369 Mar 25 at 9:08