By theorem. Let $f(x,y)$ be continuous on $[a,\infty) \times [c,\infty)$, and assume that the integrals $\int_a^{\infty}|f(x,y)|\,dx$ and $\int_c^\infty |f(x,y)|\,dy$ converge uniformly in every compact interval $[c,d]$ and $[a,b]$,respectively. Then
$$\int_c^\infty \int_a^\infty f(x,y)\,dx\,dy=\int_a^\infty \int_c^\infty f(x,y) \, dy \, dx.$$
I have $f(x,y)=e^{-xy}\sin x$
I want to prove that $$\int_0^{\infty}\int_0^{\infty}e^{-xy}(\sin x) \,dy \, dx = \int_0^\infty \int_0^\infty e^{-xy}(\sin x)\,dx\,dy.$$
Question : 1. How to prove $f(x,y)=e^{-xy}\sin x$ is continuous on $[0,\infty) \times [0,\infty)$.
- How to prove $\int_0^\infty e^{-xy}(\sin x)\,dx$ converge uniformly on $[0,d]$ and $\int_0^\infty e^{-xy}(\sin x)\,dy$ converge uniformly on $[0,b]$.
I know that the definition says $f_n$ converges uniformly to $f$ if given $\forall \varepsilon > 0$, $\forall n \geq N$, such that $|f_n(x) – f(x)| < \varepsilon, \forall n \geq N$.
I would be really happy if someone could help me. Thank you.
Best Answer
I'm too lazy to check some uniform convergences, so here is a sketch: $$ \int_0^{\infty}\int_0^{\infty}e^{-xy}\sin x \,dy \, dx = \int_0^\infty \frac{\sin x}{x} dx = \frac{\pi}2,\\ \int_0^\infty \int_0^\infty e^{-xy}\sin x\,dx\,dy = \int_0^\infty \mathrm{Im} \int_0^\infty e^{-xy + ix}\,dx\,dy = \int_0^\infty \mathrm{Im} \frac{1}{y -i}\,dy\\ =\int_0^\infty \frac{1}{y^2 +1 }\,dy = \frac{\pi}2. $$