Symmetrical sums $$\sum_{j=1}^{\infty }\left ( \sum_{n=1}^{\infty } \frac{e^{-jx/n}}{n^2}-\frac{e^{-nx/j}}{j^2}\right ),x>0$$
$$\sum_{j=1}^{\infty }\left ( \sum_{n=1}^{\infty }\frac{1}{\sqrt{nj}} \left (\frac{e^{-j/n}}{n}-\frac{e^{-n/j}}{j}  \right )\right )$$
Two sums are given. As you can see, they are symmetrical, but there is a problem. If I understand correctly, the first sum will equal $0$, but the second sum for some reason does not. Question: why is the second sum not equal to $0$ if it is symmetric?
 A: In fact, both sums are not zero: every term of every sum diverges, so we are not allowed to change the order of summation.
On the other hand, any finite summation (till any finite $N$) is zero - due to the antisymmetry (for a finite $N$ we are allowed to change the order of summation).
For example, for the first sum we have
$$\sum_{j=1}^N\left ( \sum_{n=1}^N \frac{e^{-jx/n}}{n^2}-\frac{e^{-nx/j}}{j^2}\right )=0$$
Therefore,
$$S_1(x)=\sum_{j=1}^{\infty }\left ( \sum_{n=1}^{\infty } \frac{e^{-jx/n}}{n^2}-\frac{e^{-nx/j}}{j^2}\right )=\lim_{N\to\infty}\sum_{j=N}^\infty\left ( \sum_{n=N}^\infty \frac{e^{-jx/n}}{n^2}-\frac{e^{-nx/j}}{j^2}\right )$$
$$=\lim_{N\to\infty}\frac{1}{N^2}\sum_{j=N}^\infty\left ( \sum_{n=N}^\infty \frac{e^{-(jx/N)/(n/N)}}{n^2/N^2}-\frac{e^{-(nx/N)/(j/N)}}{j^2/N^2}\right )=\int_1^\infty ds\bigg(\int_1^\infty dt\Big(\frac{e^{-sx/t}}{t^2}-\frac{e^{-tx/s}}{s^2}\Big)\bigg)$$
$$\int_1^\infty \frac{e^{-sx/t}}{t^2}dt=\frac{1-e^{-xs}}{xs}\tag{1}$$
$$\int_1^\infty\frac{e^{-tx/s}}{s^2}dt=\frac{e^{-x/s}}{xs}\tag{2}$$
Using (1) and (2), and making the change $s=\frac{1}{t}$
$$S_1(x)=\frac{1}{x}\int_1^\infty\frac{1-e^{-sx}-e^{-x/s}}{s}ds=\frac{1}{2x}\int_0^\infty\frac{1-e^{-sx}-e^{-x/s}}{s}ds$$
Integrating by part,
$$=\frac{1}{2x}\ln s\,\big(1-e^{-sx}-e^{-x/s}\big)\,\bigg|_0^\infty-\frac{1}{2}\int_0^\infty\ln s\Big(e^{-xs}-\frac{e^{-x/s}}{s^2}\Big)ds$$
$$\boxed{\,\,S_1(x)=-\int_0^\infty\ln s\,e^{-xs}ds=\frac{\gamma+\ln x}{x}\,\,}$$
A: Funny enough, but the second sum can also be presented in the closed form. More generally, let's consider
$$S_2(x)=\sum_{j=1}^{\infty }\left ( \sum_{n=1}^{\infty }\frac{1}{\sqrt{nj}} \left (\frac{e^{-jx/n}}{n}-\frac{e^{-nx/j}}{j}  \right )\right );\,\,x>0$$
Using the same approach as below for the first sum, we can present the second sum as an integral:
$$S_2(x)=\int_1^\infty\frac{ds}{\sqrt s}\int_1^\infty\frac{dt}{\sqrt t}\Big(\frac{e^{-sx/t}}{t}-\frac{e^{-tx/s}}{s}\Big)$$
Making the substitution $p=\frac{1}{t}$
$$\int_1^\infty\frac{dt}{\sqrt t}\frac{e^{-sx/t}}{t}=\int_0^1\frac{e^{-xps}}{\sqrt p}dp=\frac{2}{\sqrt s}\int_0^\sqrt se^{-xt^2}dt\tag{1}$$
$$\frac{1}{s}\int_1^\infty\frac{e^{-tx/s}}{\sqrt t}dt=\frac{2}{\sqrt s}\int_{1/\sqrt s}^\infty e^{-xt^2}dt\tag{2}$$
Using (1) and (2) and making the substitution $s\to\frac{1}{s}$
$$S_2(x)=2\int_1^\infty\frac{ds}{s}\Big(\int_0^\sqrt se^{-xt^2}dt-\int_{1/\sqrt s}^\infty e^{-xt^2}dt\Big)=\int_0^\infty\frac{ds}{s}\Big(\int_0^\sqrt se^{-xt^2}dt-\int_{1/\sqrt s}^\infty e^{-xt^2}dt\Big)$$
Integrating by part,
$$=\ln s\Big(\int_0^\sqrt se^{-xt^2}dt-\int_{1/\sqrt s}^\infty e^{-xt^2}dt\Big)\bigg|_{s=0}^{s=\infty}-\frac{1}{2}\int_0^\infty\Big(\frac{e^{-sx}}{\sqrt s}-\frac{e^{-x/s}}{s^{3/2}}\Big)\ln s\,ds$$
Making the change in the second term $s\to\frac{1}{s}$, we can present the sum in the form
$$S_2(x)=-\int_0^\infty\frac{\ln s}{\sqrt s}e^{-sx}ds$$
$$=-\frac{1}{\sqrt x}\Big(\frac{\partial}{\partial t}\,\Big|_{t=0}\int_0^\infty s^{-1/2+t}e^{-s}ds-\ln x\int_0^\infty s^{-1/2}e^{-s}ds\Big)$$
$$\boxed{\,\,S_2(x)=-\frac{1}{\sqrt x}\Big(\Gamma\bigg(\frac{1}{2}\Big)\psi\Big(\frac{1}{2}\Big)-\ln x\,\Gamma\Big(\frac{1}{2}\Big)\bigg)=\sqrt\pi\,\frac{\gamma+2\ln2+\ln x}{\sqrt x}\,\,}$$
