# Uniform convergence of iterated improper integrals on $(0,\infty)$

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.

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.

• Thank you! I'm a bit unsure how to prove non-uniform convergence. But now I'm confused why the integral switch is justified if these are not uniformly convergent improper integrals. Dec 13, 2018 at 0:07
• Also can you take a look at this: math.stackexchange.com/q/3020423/318852 Dec 13, 2018 at 0:10
• @WoodWorker: Proving non-uniform convergence basically means we find $\epsilon_0 >0$ so that for any $C > 0$ no matter how large there exists $c_2 > c_1 > C$ and $y_C$ such that $\left|\int_{c_1}^{c_2}f(x,y_C) \, dx \right| > \epsilon_0$ -- which follows if as I showed there are sequences $\beta_n > \alpha_n$ that diverge to $+\infty$ and $y_n$ where $\left|\int_{\alpha_n}^{\beta_n}f(x,y_n) \, dx \right| \not\to 0$.
– RRL
Dec 13, 2018 at 5:51
• Uniform convergence of $\int_0^\infty f(x,y) \, dx$ and $\int_0^\infty f(x,y) \, dy$ are neither necessary (as we see here) nor sufficient for switching these improper integrals. It can be justified however if $F(x) = \int_0^\infty f(x,y) \, dy$ is such that $\int_0^\infty F(x) \, dx$ is uniformly convergent.
– RRL
Dec 13, 2018 at 5:54