Computing an upper bound for the double sum $\sum_{n=1}^N \sum_{m=1}^M e^{-|an-bm|}$ I am trying to compute or find good upper upper bound for  $$\sum_{i=1}^{N_1}\sum_{j=1}^{N_2}e^{-|ai-bj|}.$$
I already tried a number of things.
For example I know that if take $N_1$ and $N_2$ do infinity the series diverges. This follows by looking at the case when $j=i$.
But I still feel that in finite case it can be computed very nicely. Just need a some trick. 
 A: Since:
$$ e^{-|c|}=\frac{1}{\pi}\int_{-\infty}^{+\infty}\frac{\cos(ct)}{1+t^2}\,dt $$
by the characteristic function of the Laplace distribution, it follows that:
$$\sum_{i=1}^{N_1}\sum_{j=1}^{N_2}e^{-|ai-bj|}=\frac{1}{\pi}\int_{-\infty}^{+\infty}\frac{1}{1+t^2}\sum_{n=1}^{N}\sum_{m=1}^{M}\cos((an-bm)t)\,dt,\tag{1}$$
where:
$$\sum_{n=1}^{N}\sum_{m=1}^{M}\cos((an-bm)t)=\Re\sum_{i=1}^{N}\sum_{j=1}^{M}e^{(an-bm)it}$$
$$\sum_{n=1}^{N}\sum_{m=1}^{M}\cos((an-bm)t)=\frac{\sin(bMt/2)}{\sin(bt/2)}\cdot\frac{\sin(aNt/2)}{\sin(at/2)}\cdot\cos((a(N+1)-b(M+1))t/2)\tag{2}.$$
Now the main issue is that the RHS of $(2)$ is a very irregular function, whose oscillations depend on the arithmetic relations between $a,b,N,M$. Anyway:
$$\sum_{i=1}^{N_1}\sum_{j=1}^{N_2}e^{-|ai-bj|}\leq\frac{N_1 N_2}{\pi}\int_{-\infty}^{+\infty}\frac{|\cos(Kt)|}{1+t^2}\,dt,$$
where $K=\left|\frac{a(N_1+1)-b(N_2+1)}{2}\right|$, and, by the Cauchy-Schwarz inequality:
$$\int_{-\infty}^{+\infty}\frac{|\cos(Kt)|}{1+t^2}\,dt\leq\sqrt{\pi\cdot\int_{-\infty}^{+\infty}\frac{\cos^2(Kt)}{1+t^2}\,dt},$$
so:

$$\sum_{i=1}^{N_1}\sum_{j=1}^{N_2}e^{-|ai-bj|}\leq N_1 N_2\sqrt{\frac{1}{2}\left(1+\frac{1}{e^{|a(N_1+1)-b(N_2+1)|}}\right)}.$$

