$\operatorname{Li}_{2} \left(\frac{1}{e^{\pi}} \right)$ as a limit of a sum

Question

Working on the same lines as

1. This/This and
2. This

I got the following expression for the Dilogarithm $\operatorname{Li}_{2} \left(\frac{1}{e^{\pi}} \right)$:

$$\operatorname{Li}_{2} \left(\frac{1}{e^{\pi}} \right) = -\frac{\pi^2}{12} -\pi + \\ \pi \lim_{N\to \infty} \left \{ \pi N(N+1) -(2N+1)\log(2N) + \\ \sum\limits_{k=1}^{N} [-4k\tan^{-1}(2k) + \log(4k^2+1)] \right \}$$

Questions:

1. Is this type of result known?
2. Is there a way to simplify the expression further?

Disclaimer: I am not a mathematician and very often I come up with stuff that is well known to mathematicians due to my limited exposure to some topics. If this is well known or a trivial result, please advice on closing the thread, unless you think the responses may be useful to others.

$\color{red}{\mathbf{\text{Edit based on Mariusz Iwaniuk's comment}}}$

$$\operatorname{Li}_{2} \left(\frac{1}{e^{\pi}} \right) = -\frac{\pi^2}{12} + \pi \{\log(\operatorname{sinh}(\frac{\pi}{2}))+\log(2)-1\} + \\ \pi \lim_{N\to \infty} \left \{ \pi N(N+1) -2N + \\ \sum\limits_{k=1}^{N} -4k\tan^{-1}(2k) \right \}$$

Answer 1

First begin noting that by expanding at infinity, we have

$$2\pi k - 2 - 4k\arctan(2k)=\sum_{n=1}^{\infty}\frac{(-1)^n 2^{1-2n}}{(2n+1)k^{2n}}.$$ This means that \begin{align*}\lim_{N\to\infty}\left(\pi N(N+1)-2N+\sum_{k=1}^{N}-4k\arctan(2k)\right)&=\sum_{k=1}^{\infty}\left(2\pi k-2-4k\arctan(2k)\right)\\ &=\sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{(-1)^n 2^{1-2n}}{(2n+1)k^{2n}}\\ &=2\sum_{n=1}^{\infty}\frac{\zeta(2n)}{2n+1}\left(\frac{i}{2}\right)^{2n}.\end{align*} Now recall that $$1-\pi x\cot(\pi x)=2\sum_{n=1}^{\infty}\zeta(2n)x^{2n},$$ meaning that $$\int_{0}^{\frac{i}{2}}\left(1-\pi x\cot(\pi x)\right)\mathrm{d}x=2\sum_{n=1}^{\infty}\frac{\zeta(2n)}{2n+1}\left(\frac{i}{2}\right)^{2n+1}.$$ Hence, \begin{align*}\lim_{N\to\infty}\left(\pi N(N+1)-2N+\sum_{k=1}^{N}-4k\arctan(2k)\right)&=\frac{2}{i}\int_{0}^{\frac{i}{2}}\left(1-\pi x\cot(\pi x)\right)\mathrm{d}x\\ &=-2ix+\pi x^2+2ix\log\left(1-e^{2\pi i x}\right)+\frac{1}{\pi}\operatorname{Li}_2\left(e^{2\pi i x}\right)\bigg]_{0}^{\frac{i}{2}}\\ &=1+\frac{7\pi}{12}-\log\left(e^{\pi}-1\right)+\frac{1}{\pi}\operatorname{Li}_2\left(e^{-\pi}\right).\end{align*} After some rearrangement, we obtain your identity. $\square$

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To answer your two questions:

As can be seen above, it's not hard to determine this result so it wouldn't surprise me if it's already known. It cannot be simplified further as $\operatorname{Li}_2\left(e^{-\pi}\right)$ is not known to have a closed form.

Answer 2

Not an answer. Too long for comment and might distract if added in the question as edit.

I tried to get a closed form of $\sum\limits_{k=1}^{N}[-4k\tan^{-1}(2k)+\log(4k^2+1)]$

$$\sum\limits_{k=1}^{N}[-4k\tan^{-1}(2k)+\log(4k^2+1)] = -4 \sum\limits_{k=1}^{N} \int\limits_{0}^{k} \tan^{-1}(2x) dx = I$$

$$I = -4 \int\limits_{0}^{\infty} \tan^{-1}(2x) S_{N}(x)$$

Here, $S_{N}(x)$ is a staircase defined as

$$ S_{N}(x) = \begin{cases} N, & 0\le x \lt 1 \\ N-1, & 1 \le x \lt 2 \\ N-2, & 2 \le x \lt 3 \\ ... \\ 1, & N-1 \le x \lt N \\ 0, & \text{elsewhere} \\ \end{cases} $$

Then I created an odd-symmetry signal $M_{N}$ by flipping $S_{N}$ horizontally around y-axis and vertically on x-axis and adding to $S_{N}$.

$$M_{N}(x) = S_{N}(x)-S_{N}(-x)$$

So, $$I=-2\int\limits_{-\infty}^{\infty} \tan^{-1}(2x) M_{N}(x)$$

Now, let us use Fourier Transforms and then use Parseval Identity. Fourier Transform formula I use is $FT[f(x)](y) = \int\limits_{-\infty}^{\infty} f(x) e^{-i2 \pi x y}dx$

With this, $$FT[\tan^{-1}(2x)](y) = -i \frac{e^{-|\pi y|}}{2y}$$ and $$FT[M_{N}(x)](y) = \frac{i}{2} \operatorname{sinc}(\pi y) \operatorname{csc}(\pi y)[\operatorname{csc}(\pi y) \sin\{(2N+1) \pi y\} -2N-1]$$

Using Parseval Identity ($\color{red}{\mathbf{\text{Edit: Plancherel/Parseval identity is}}}$ $\int\limits_{-\infty}^{\infty} f(x) \overline{g(x)} dx = \int\limits_{-\infty}^{\infty} \hat{f}(\xi) \overline{\hat{g}(\xi)} d\xi$. I had overlooked the conjugate and hence removing the negative sign I had earlier.),

$$I = \frac{1}{2}\int\limits_{-\infty}^{\infty} \frac{e^{-|\pi y|}}{y} \operatorname{sinc}(\pi y) \operatorname{csc}(\pi y)[\operatorname{csc}(\pi y) \sin\{(2N+1) \pi y\} -2N-1] dy$$

Because the integrand is even, $$I = \int\limits_{0}^{\infty} \frac{e^{-|\pi y|}}{y} \operatorname{sinc}(\pi y) \operatorname{csc}(\pi y)[\operatorname{csc}(\pi y) \sin\{(2N+1) \pi y\} -2N-1] dy$$

Substituting $\pi y = t$, $$I = \int\limits_{0}^{\infty} \frac{e^{-t}}{t} \operatorname{sinc}(t) \operatorname{csc}(t)[\operatorname{csc}(t) \sin\{(2N+1) t\} -2N-1] dt$$

Expand $\operatorname{sinc}(t)$ as $\frac{\sin(t)}{t}$ to get rid of one $\operatorname{csc}(t)$. $$I = \int\limits_{0}^{\infty} \frac{e^{-t}}{t^2} [\operatorname{csc}(t) \sin\{(2N+1) t\} -2N-1] dt$$

Note: I have verified that this integral is accurate by comparing it for small values of $N$.

I haven't proceeded much afer this. Mathematica cannot give me a closed form of this (which I know will diverge when $N \to \infty$). The hope is to get a closed form from this without summations (and most importantly not bring back the Polylog I am solving) and then do a series expansion at $N = \infty$ and plug it back into my original question in the place of my summation that I currently have. I again hope that such a series will be in terms of logarithms that might hopefully be amenable to a limit calculations. Enough of my hopes. I request all the integration masters on this site to help solving this integral if you find the above steps useful, or else, give another closed form for my original summation $\sum\limits_{k=1}^{N}[-4k\tan^{-1}(2k)+\log(4k^2+1)]$

Answer 3

Since $$I=\int \Big(\log \left(4 k^2+1\right)-4\, k\, \tan ^{-1}(2 k)\Big)\,dk$$ $$I=k \left(\log \left(4k^2+1\right)-1\right)+\left(2 k^2-\frac{1}{2}\right) \tan^{-1}(2 k)$$ using the simplest form of Euler-MacLaurin summation $$\text{rhs}=\pi \left(\frac{4937327}{4921875}-\frac{1}{2}\log (5)-\frac{1}{6} \tan ^{-1}(2)\right)+\frac{\pi }{12 n}+O\left(\frac{1}{n^2}\right)$$ This gives $$\text{rhs}=\color{red}{0.0436}55$$ corresponding to a relative error of $0.08$%.