I'm trying to get a better understanding of when it is permissible to swtich conditionally convergent improper integrals (when Fubini inapplicable) and I looked at a case where it works:
$$\int_0^\infty \int_0^\infty e^{-xy} \sin x \, dx\, dy = \int_0^\infty \int_0^\infty e^{-xy} \sin x \, dy \, dx $$
I know the inner iterated integrals are uniformly convergent by the Weierstrass test for $x , y \in [c, \infty)$ where $c > 0$. Since $|e^{-xy} \sin x | \leqslant e^{-cy}$ for $c \leq x < \infty$ , then $\int_0^\infty e^{-xy} \sin x \, dy$ converges uniformly for $c \leq x < \infty$. Since $|e^{-xy} \sin x | \leqslant e^{-cx}$ for $c \leq y < \infty$ , then $\int_0^\infty e^{-xy} \sin x \, dx$ converges uniformly for $c \leq y < \infty$.
The Weierstrass test is not helpful to consider uniform convergence on $(0,\infty)$.
My question is how to determine if $\int_0^\infty e^{-xy} \sin x \, dy$ converges uniformly for $0 < x < \infty$ and $\int_0^\infty e^{-xy} \sin x \, dx$ converges uniformly for $0 < y < \infty$ and either prove it or disprove it.
Best Answer
Neither integral is uniformly convergent for values of the parameter in the open interval $(0,\infty)$.
For the first integral, with $y_n = (2n\pi + \pi)^{-1} \in (0,\infty)$ we have
$$\left|\int_{2n\pi}^{2n\pi+\pi} e^{-xy_n} \sin x \, dx\right|\geqslant e^{-(2n\pi+\pi) y_n}\int_{2n\pi}^{2n\pi+\pi} \sin x \, dx = 2 e^{-(2n\pi+\pi)y_n}= 2e^{-1}$$
Since the RHS does not converge to $0$ as $n \to \infty$, the Cauchy criterion for uniform convergence is violated.
For the second integral, with $x_n = 1/n \in (0,\infty)$ we have
$$\left|\int_n^\infty e^{-x_ny} \sin x_n \, dy\right| = \left|\frac{\sin x_n}{x_n} \right|e^{-nx_n} = \frac{\sin \frac{1}{n}}{\frac{1}{n}}e^{-1} \,\,\, \xrightarrow[n \to \infty]{} \,\,e^{-1},$$
and, again, violation of the Cauchy criterion precludes uniform convergence.