changing the order of integration in the infinite integrals I can not understand the conditions of Theorem II in wikipedia about changing the order of integration Order of integration
It says:

Let $f(x,y)$ be continuous for $a \leq x < \infty$ and $c \leq y < \infty$, and let the integrals
$$
J(y):= \int_{a}^{\infty}f(x,y)dx \, \text{ and } \, J^{*}(x):= \int_{b}^{\infty}f(x,y)dy
$$
be respectively, uniformly convergent on every finite interval for $c \leq y \leq C$ and on every finite interval $a \leq x \leq A$.

What does it mean "uniformly convergent on every finite interval"?
Also, it says on the page that, we can not change the order of integration for the function
$$
f(x,y) = \frac{x^{2} - y^{2}}{(x^{2} + y^{2})^2}, 
$$
cause it does not satisfy this condition.
 A: As it pertains to an improper integral $\int_a^\infty f(x,y) \,dx$, uniform convergence on a finite interval $[c,C]$ means that for any $\epsilon > 0$ there exists $A(\epsilon) > a$ such that
$$\left|\int_a^b f(x,y) \, dx - \int_a^\infty f(x,y) \, dx \right|  = \left|\int_b^\infty f(x,y) \, dx \right| < \epsilon,$$
for all $b > A(\epsilon)$ and all $y \in [c,C]$.
In other words, the choice of $A(\epsilon)$ does not depend on $y$ when considering convergence with respect to a limit where $y$ is a parameter, i.e.,
$$\lim_{b \to \infty} \int_a^b f(x,y) \, dx$$
With the counterexample, $\displaystyle f(x,y) = \frac{x^{2} - y^{2}}{(x^{2} + y^{2})^2},$ we have
$$\tag{*}\int_1^\infty \left(\int_1^\infty\frac{x^{2} - y^{2}}{(x^{2} + y^{2})^2}\,dx  \right)dy = \frac{\pi}{4} \neq - \frac{\pi}{4} = \int_1^\infty \left(\int_1^\infty\frac{x^{2} - y^{2}}{(x^{2} + y^{2})^2}\,dy  \right)dx,$$
and in this case the order of integration cannot be switched. However this has nothing to do with non-uniform convergence on finite intervals.  Note that for $b> A(\epsilon) = 1/\epsilon$ and all $y$, we have
$$\left|\int_b^\infty\frac{x^{2} - y^{2}}{(x^{2} + y^{2})^2}\, dx\right| = \left|-\int_b^\infty\frac{\partial}{\partial x}\left(\frac{x}{x^{2} + y^{2}}\right)\, dx\right|= -\left. \frac{x}{x^2 + y^2}\right|_{x=b}^{x \to\infty}\\ = \frac{b}{b^2 + y^2}\leqslant \frac{1}{b} < \epsilon,$$
and the improper integral converges uniformly.
The reason for the inequality of iterated integrals in (*) is related to the other hypothesis of Theorem II, which is that one of the iterated integrals must be absolutely convergent.
