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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.

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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.

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  • $\begingroup$ 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. $\endgroup$
    – WoodWorker
    Dec 13, 2018 at 0:07
  • $\begingroup$ Also can you take a look at this: math.stackexchange.com/q/3020423/318852 $\endgroup$
    – WoodWorker
    Dec 13, 2018 at 0:10
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    $\begingroup$ @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 $\left|\int_{\alpha_n}^{\beta_n}f(x,y_n) \, dx \right| \not\to 0$. $\endgroup$
    – RRL
    Dec 13, 2018 at 5:51
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    $\begingroup$ 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. $\endgroup$
    – RRL
    Dec 13, 2018 at 5:54

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