Can we find the exact value of a double sum with cosine without differentiation?

Question

After finding an interesting double sum

$$\sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \frac{(-1)^{m+n}}{(m+n)^2} = \frac{\pi^2}{12}-\ln 2 ,$$

I started to investigate a harder one $$\displaystyle \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{(-1)^{m+n} \cos (m+n)}{(m+n)^2} .$$

Noting that

$\displaystyle \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{(-1)^{m+n} \cos (m+n)}{(m+n)^2}=\Re f(1)\tag*{} $

where

$\displaystyle f(x)=\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{(-1)^{m+n} e^{(m+n) x i}}{(m+n)^2}\tag*{} $

Differentiating $f(x)$ once and twice yields $\displaystyle f^{\prime}(x)=\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{(-1)^{m+n} i e^{(m+n) x i}}{m+n}\tag*{} $ and

$$ \displaystyle \begin{aligned}f^{\prime \prime}(x) =\sum_{m=1}^{\infty}(-1)^m\left[\sum_{n=1}^{\infty}(-1)^{n-1} e^{(m+n) x i}\right] =\sum_{m=1}^{\infty} \frac{(-1)^m e^{(m+1) x i}}{1+e^{x i}}=-\frac{e^{2 xi }}{\left(1+e^{xi}\right)^2}\end{aligned}\tag*{} $$

Integrating back gives

$$f^{\prime}(x)-f^{\prime}(0)=\int_0^x f^{\prime \prime}(t) d t =-\int_0^x \frac{e^{2 i t}}{\left(1+e^{i t}\right)^2} d t =i \ln \left(1+e^{i x}\right)+\frac{1}{2} \tan \frac{x}{2}-i \ln 2$$

Rearranging gives us

$ \displaystyle f^{\prime}(x)=i f(0)+i \ln \left(1+e^{i x}\right)+\frac{1}{2} \tan \frac{x}{2}-i \ln 2\tag*{} $

Integrating once more gives

$$\begin{aligned} f(1)-f(0)&= \int_0^1\left[i f(0)+i \ln \left(1+e^{i t}\right)+\frac{1}{2} \tan \frac{t}{2}-i \ln 2\right] d t \end{aligned}$$

$$= i f(0)+i \int_0^1 \ln \left(1+e^{i t}\right) d t+\frac{1}{2} \int_0^1 \tan \frac{t}{2} d t-i \ln 2 $$

$$= i f(0)+i\left[-i \operatorname{Li_2}\left(-e^{-i x}\right)\right]_0^1+\ln \left(\sec \frac{1}{2}\right)-i \ln 2$$

Rearranging gives

$\displaystyle \begin{aligned}f(1)&=f(0)+i f(0)-\operatorname{Li_2}\left(-e^{-i}\right)-\frac{\pi^2}{12}+\ln \left(\sec \frac 12\right) -i \ln 2\end{aligned}\tag*{} $

$\displaystyle \boxed{\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{(-1)^{m+n} \cos (m+n)}{(m+n)^2}=\Re f(1)=-\ln 2-\Re\left(\operatorname{Li}_2\left(-e^{-i}\right)\right)+\ln \left(\sec \frac{1}{2}\right) \approx{0.0099041}} \tag*{} $

My question: Can we find the exact value of the double sum with cosine without differentiation?

Your comments and alternative solution are highly appreciated.

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Footnote:

$$f(0) =\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{(-1)^{m+n}}{(m+n)^2} =\sum_{m=1}^{\infty} \sum_{n=1}^{\infty}(-1)^{m+n-1} \int_0^1 x^{m+n-1} \ln x d x \\ =\int_0^1\left[\sum_{m=1}^{\infty} \sum_{n=1}^{\infty}(-1)^{m+n-1} x^{m+n-1}\right] \ln x d x \\ =\int_0^1 \ln x \sum_{m=1}^{\infty} \frac{(-1)^m x^m}{1+x} d x=\int_0^1 \ln x \frac{-x}{(1+x)^2} d x \\ =-\int_0^1 \frac{x^2 \ln x}{(1+x)^2} d x=\frac{\pi^2}{12}-\ln 2 $$

Answer 1

Using the observation by @aschepler $$S=\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{(-1)^{m+n}\cos(m+n)}{(m+n)^2} = \sum_{k=2}^\infty \frac{(-1)^k (k-1) \cos k}{k^2}=\sum_{k=1}^\infty \frac{(-1)^k (k-1) \cos k}{k^2}$$ $$=\Re\sum_{k=1}^\infty \frac{(-1)^k e^{ik}}{k}-\frac12\sum_{k=-\infty; \,k\neq 0}^\infty\frac{(-1)^k e^{ik}}{k^2}=-\Re\ln(1+e^i)-\frac12\sum_{k=-\infty; \,k\neq 0}^\infty\frac{(-1)^k e^{ik}}{k^2}$$ $$=-\Re\ln|1+e^i|-\frac12\sum_{k=-\infty; \,k\neq 0}^\infty\frac{(-1)^k e^{ik}}{k^2}=-\frac{\ln(2+2\cos1)}2-\frac12\sum_{k=-\infty; \,k\neq 0}^\infty\frac{(-1)^k e^{ik}}{k^2}\tag{1}$$ To evaluate the last sum we integrate along a big circle in the complex plane the function $$\frac12\oint_C\frac{2\pi i}{e^{2\pi iz}-1}\frac{e^{\pi iz}e^{iz}}{z^2}dz=0=\frac{2\pi i}2\sum_{k=-\infty; \,k\neq 0}^\infty\frac{(-1)^k e^{ik}}{k^2}+\frac{2\pi i}2\underset{z=0}{\operatorname{Res}}\frac{2\pi i}{e^{2\pi iz}-1}\frac{e^{\pi iz}e^{iz}}{z^2}$$ $$\Rightarrow\,\,-\,\frac12\sum_{k=-\infty; \,k\neq 0}^\infty\frac{(-1)^k e^{ik}}{k^2}=\frac12\underset{z=0}{\operatorname{Res}}\frac{2\pi i}{e^{2\pi iz}-1}\frac{e^{\pi iz}e^{iz}}{z^2}=\frac{\pi^2}{12}-\frac14\tag{2}$$ Putting (2) into (1), $$\boxed{\,\,S=\frac{\pi^2}{12}-\frac14-\ln2-\ln\cos\frac12=0.009904093...\,\,}$$

Answer 2

Similar to the way you evaluated $f(0)$, lets use the fact $$\int_{0}^{\infty} t e^{-st} \, \mathrm dt = \frac{1}{s^{2}}, \quad \Re(s) >0. $$

We have

$$ \begin{align} \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{(-1)^{m+n} \cos (m+n)}{(m+n)^2} &= \Re \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{(-1)^{m+n} e^{i(m+n)}}{(m+n)^2} \\ &= \Re \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} (-1)^{m+n} e^{i(m+n)} \int_{0}^{\infty} x e^{-x(m+n)} \, \mathrm dx \\ &= \Re \int_{0}^{\infty} x \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} e^{i(m+n)} e^{-x(m+n)}\, \mathrm dx \\ &= \Re \int_{0}^{\infty} x \left( \sum_{m=1}^{\infty} e^{m(-x+i)} \right)^{2} \, \mathrm dx \\ &= \Re \int_{0}^{\infty} \frac{x e^{2(-x+i)}}{\left(1+e^{-x+i} \right)^{2}} \, \mathrm dx \\ &= \Re \int_{0}^{\infty} \frac{e^{2i} x}{\left(e^{x}+e^{i} \right)^{2}} \, \mathrm dx \\ &= - \Re \int_{0}^{1} \frac{t \ln(t)}{\left(t+e^{i}\right)^{2}} \, \mathrm dt \\ &= - \Re \left(-\frac{t \ln(t)}{t+e^{i}}\Bigg|_{0}^{1} + \int_{0}^{1} \frac{\ln(t)+1}{t+e^{i}} \, \mathrm dt\right) \\ &= - \Re \left(\int_{0}^{1} \frac{\ln (t)}{t+e^{i}} \, \mathrm dt + \ln(1+e^{i}) - \ln(e^{i}) \right) \\ &= - \Re \int_{0}^{1} \frac{\ln(t)}{t+e^{i}} \, \mathrm dt - \frac{1}{2} \, \ln \left(2(1+ \cos 1) \right) \\& = - \Re \int_{0}^{1} \frac{\ln(t)}{t+e^{i}} \, \mathrm dt - \ln \left(2 \cos \frac{1}{2}\right). \end{align}$$

Integrating by parts and choosing $\ln \left(\frac{t+e{i}}{e^{i}} \right)$ for $dv$, we have

$$ \begin{align} \int \frac{\ln(t)}{t+e^{i}} \, \mathrm dt &= \ln(t) \ln \left(1+ \frac{t}{e^{i}} \right)- \int \frac{\ln \left(1+ \frac{t}{e^{i}}\right)}{t} \, \mathrm dt \\ & = \ln(t) \ln \left(1+ \frac{t}{e^{i}} \right) + \operatorname{Li}_{2}(-te^{-i}) . \end{align}$$

Therefore, $$ \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{(-1)^{m+n} \cos (m+n)}{(m+n)^2} = - \Re \operatorname{Li}_{2}(-e^{-i}) - \ln \left(2 \cos \frac{1}{2} \right).$$

Since the real part of $\operatorname{Li}_{2}(-e^{-i})$ can be written as $$\Re \operatorname{Li}_{2}(-e^{-i}) = \frac{\operatorname{Li}_{2}(-e^{-i}) + \operatorname{Li}_{2}(-e^{i})}{2}, $$ it follows from the inversion formula that $$- \Re \operatorname{Li}_{2}(-e^{-i}) = \frac{1}{2} \left(\frac{\pi^{2}}{6} +\frac{\ln^{2}(e^{i})}{2} \right) = \frac{\pi^{2}}{12} - \frac{1}{4}. $$