Is $\sum_{j=1}^\infty\sum_{n=1}^\infty\left(\frac{e^{-j/n}}{n^2}-\frac{e^{-n/j}}{j^2}\right)=\gamma ?$

Here's the solution I have. Take $1<c<2$ and consider the integral $$I=\frac1{2\pi i}\int_{c-i\infty}^{c+i\infty}\Gamma(s)\zeta(s)\zeta(2-s)\,ds.$$

Let's compute it in two ways.

First, $I$ is the sum of the residues of the integrand at all integer $s\leqslant1$ (consider integration along the boundary of $[-N-1/2,c]+i[-R,R]$ with $R>0$ and $N$ a positive integer, and take $R,N\to\infty$; the integral along the "right side of the rectangle" tends to $I$, and the remaining integrals vanish). The residue at $s=1$ (a double pole) may be computed using \begin{align*} \Gamma(s)&=1-\gamma(s-1)+o(s-1),\\ \zeta(s)&=(s-1)^{-1}+\gamma+o(1),\\ \zeta(2-s)&=-(s-1)^{-1}+\gamma+o(1), \end{align*} hence $\Gamma(s)\zeta(s)\zeta(2-s)=(s-1)^{-2}\big({-1}+\gamma(s-1)+o(s-1)\big)$ and the residue equals $\gamma$. And, for a nonnegative integer $n$, we have $$\newcommand{\res}{\operatorname*{Res}}\res_{s=-n}\Gamma(s)\zeta(s)\zeta(2-s)=\frac{(-1)^n}{n!}\zeta(-n)\zeta(n+2)=\frac{B_{n+1}}{(n+1)!}\zeta(n+2)$$ so that, using the generating function of the Bernoulli numbers, \begin{align*} I&=\gamma+\sum_{n=0}^\infty\frac{B_{n+1}}{(n+1)!}\zeta(n+2)=\gamma+\sum_{n=1}^\infty\frac{B_n}{n!}\sum_{j=1}^\infty\frac1{j^{n+1}} \\&=\gamma+\sum_{j=1}^\infty\frac1j\left(\frac{1/j}{e^{1/j}-1}-1\right)=\gamma+\sum_{j=1}^\infty\left(-\frac1j+\frac1{j^2}\sum_{n=1}^\infty e^{-n/j}\right). \end{align*}

Second (the implied uniform convergence is not hard to see; recall $1<c<2$), $$I=\sum_{j=1}^\infty I_j,\quad I_j:=\frac1{2\pi i}\int_{c-i\infty}^{c+i\infty}\Gamma(s)j^{-s}\zeta(2-s)\,ds.$$ Now $I_j$ may be computed the same way. The residue of the integrand at $s=1$ equals $-1/j$, and the residue at $s=-k$ (for a nonnegative integer $k$) is $\zeta(k+2)(-j)^k/k!$, so that $$I=\sum_{j=1}^\infty\left(-\frac1j+\sum_{k=0}^\infty\frac{(-j)^k}{k!}\sum_{n=1}^\infty\frac1{n^{k+2}}\right)=\sum_{j=1}^\infty\left(-\frac1j+\sum_{n=1}^\infty\frac{e^{-j/n}}{n^2}\right).$$

Comparing these two expressions for $I$, we get the expected result.