interchange the order of integration other than Fubini and Tonelli Suppose we know the convergence of an iterated integral
$$\int_{\mathbb{R}}\int_{\mathbb{R}} f(x,y)dxdy$$. Usually to ensure the switch the order of integration, we use Fubini's theorem which requires absolute convergence or Tonelli's theorem requiring positivity. But it seems that these are only sufficient conditions, and I'm wondering if there are other assumptions on the integrals, besides the above mentioned, to ensure the convergence of the iterated integral 
$$\int_{\mathbb{R}}\int_{\mathbb{R}} f(x,y)dydx$$
and
$$\int_{\mathbb{R}}\int_{\mathbb{R}} f(x,y)dxdy=\int_{\mathbb{R}}\int_{\mathbb{R}} f(x,y)dydx?$$
I couldn't find anything new in standard references, and any answer or reference would be appreciated. 
 A: When the hypotheses of Fubini's or Tonelli's theorem are not satisfied, there is no simple "one size fits all" answer. To quote Bromwich in Theory of Infinite Series:
"It is by no means easy to determine fairly general conditions under which the equation
$$\int_a^\infty dx \int_b^\infty f(x,y) \, dy = \int_b^\infty dy \int_a^\infty f(x,y) \, dx $$
is correct."
To gain some insight, consider basic results for iterated proper and improper Riemann integrals. A good place to start is The Elements of Real Analysis by Bartle.
(1) The most elementary result (for a bounded rectangle) applies when $f:[a,b] \times [c,d] \to \mathbb{R}$ is continuous.  In that case we have
$$\int_a^ b \left(\int_c^d f(x,y) \, dy \right) \, dx = \int_c^d \left(\int_a^b f(x,y) \, dx \right) \, dy. $$
(2) Suppose one interval is infinite.  Sufficient conditions for equality of the iterated integrals are continuity of $f$ along with uniform convergence of the improper integral
$$F(y) = \int_a^\infty f(x,y) \, dx,$$
for $y \in [c,d]$, in which case we have 
$$\int_c^ d \left(\int_a^\infty f(x,y) \, dx \right) \, dy = \int_a^\infty \left(\int_c^d f(x,y) \, dy \right) \, dx. $$
Uniform convergence is by no means a necessary condition.  For example,
$$F(y) = \int_0^\infty \frac{\sin(xy)}{x} \, dx,$$
fails to converge uniformly for $y \in [0,1]$, yet
$$\int_0^ 1 \left(\int_0^\infty \frac{\sin(xy)}{x} \, dx \right) \, dy = \int_0^\infty \left(\int_0^1 \frac{\sin(xy)}{x} \, dy \right) \, dx = \frac{\pi}{2}. $$
(3) It gets more interesting when the rectangle is infinite in both dimensions.  An example where Tonelli's and Fubini's theorems do not apply --  illustrating that   uniform convergence is no longer sufficient -- is 
$$-\frac{\pi}{4} = \int_1^\infty \left(\int_1^\infty \frac{x^2-y^2}{(x^2 + y^2)^2} \, dy \right) \, dx \neq \int_1^\infty \left(\int_1^\infty  \frac{x^2-y^2}{(x^2 + y^2)^2} \, dx \right) \, dy = \frac{\pi}{4}.$$
Each of the inner integrals converge uniformly.  For example, with $c >0$
, 
$$\left|\int_c^\infty \frac{x^2-y^2}{(x^2 + y^2)^2} \, dx\right| = \frac{c}{c^2 + y^2} < \frac{1}{c},$$
and this remainder can be made arbitrarily close to zero by choosing $c$ sufficiently large for all $y \in [0, \infty)$.  Stronger conditions are needed to guarantee equality of iterated infinite integrals. A number of theorems that apply can be found in older reference books, such as Bartle.
