The shift of an integral variable whch is related with infinity

Question

I want to confirm. $$ \int_{-\infty}^{\infty} dx \int_{-\infty}^{\infty} dy \, f(x,y) = \lim_{a,b\to \infty} \int_{-a}^{a} dx \int_{-b}^{b} dy \, f(x,y) = \lim_{a,b\to \infty} \int_{-a}^{a} dx \int_{-b-x}^{b-x} dy \, f(x,y+x) $$ Then, is it equal to $$ \int_{-\infty}^{\infty} dx \int_{-\infty}^{\infty} dy \, f(x,y+x) \quad ? $$ When the original intagral diverges, it probably isn't.
When the original intagral converges, is it in all cases? Are there no cases where such a shift related with infinity lead to any problem?
Any answer or comment wll be welcome.
Thanks.

Answer 1

As long as everything here converges absolutely, you are fine, as your equations prove. If things do not converge absolutely, then you can indeed run into problems; really, the reason is that you chose to define $$ \int_{-\infty}^\infty f(x) \,\mathrm dx = \lim_{a\to\infty}\int_{-a}^a f(x)\,\mathrm dx. $$ This definition is no longer translation invariant: for example, if $f = \arctan$, this gives $\int_{-\infty}^\infty \arctan(x)\,\mathrm dx = 0$, but shifting the function around suddenly gives different answers. Really, it is better to say that $\int_{-\infty}^\infty f(x)\,\mathrm dx$ exists only if $\int_{-\infty}^0 f(x) \,\mathrm dx = \lim_{a \to \infty} \int_{-a}^0 f(x)\,\mathrm dx$ and $\int_0^\infty f(x)\,\mathrm dx = \lim_{a \to \infty} \int_0^a f(x)\,\mathrm dx$ both exist, and in that case it equals the sum of those values.