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.
Footnote:
$$f(0) =\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{(-1)^{m+n}}{(m+n)^2} =\sum_{m=1}^{\infty} \sum_{n=1}^{\infty}(-1)^{m+n-1} \int_0^1 x^{m+n-1} \ln x d x \\ =\int_0^1\left[\sum_{m=1}^{\infty} \sum_{n=1}^{\infty}(-1)^{m+n-1} x^{m+n-1}\right] \ln x d x \\ =\int_0^1 \ln x \sum_{m=1}^{\infty} \frac{(-1)^m x^m}{1+x} d x=\int_0^1 \ln x \frac{-x}{(1+x)^2} d x \\ =-\int_0^1 \frac{x^2 \ln x}{(1+x)^2} d x=\frac{\pi^2}{12}-\ln 2 $$
Best Answer
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), $$\boxed{\,\,S=\frac{\pi^2}{12}-\frac14-\ln2-\ln\cos\frac12=0.009904093...\,\,}$$