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My problem:

I've been given the double integral: $$\int_{0}^{+\infty}\int_{0}^{+\infty}\sin(bx)y^{p-1}e^{-xy}dxdy, \quad 0<p<2,b\in \mathbb{R}$$ I need to prove that the value of the integral is independent of the order of integration (i.e. I can either integrate first with respect to $x$ and then with respect to $y$ or first with respect to $y$ and then $x$).

My attempt to a solution:

According to Fubini’s Theorem (https://en.wikipedia.org/wiki/Fubini%27s_theorem#Tonelli.27s_theorem) a double integral may be evaluated as an iterated integral if $f(x,y)$ is integrable on the domain of integration, that is, if the following conditions are met:

  1. $f(x,y)$ is measurable. However, the function is not defined on $y=0$, so it’s not continuous on $[0, +\infty)\times[0,+\infty)$. It still is measurable?
  2. $\int|f(x,y)|d(x,y)<+\infty$. I am not sure how to check this condition. I know that $|f(x,y)|$ is a non-negative function, so I can evaluate the double integral as an iterated integral (by Tonelli's theorem for non-negative functions): $$\int_{[0, +\infty)\times[0,+\infty)}|f(x,y)|d(x,y)=\int_0^{+\infty}\int_0^{+\infty}|f(x,y)|dxdy$$ If I integrate first with respect to $y$ and then with respect to $x$ I get the following: $$\int_0^{+\infty}\bigg(\int_0^{+\infty}|\sin(bx)y^{p-1}e^{-xy}|dy\bigg)dx=\\\int_0^{+\infty}|\sin(bx)|\bigg(\int_0^{+\infty}y^{p-1}e^{-xy}dy\bigg)dx=\\\int_0^{+\infty}|\sin(bx)|\bigg(\frac{\Gamma(p)}{x^{p}}\bigg)dx=\Gamma(p)\int_0^{+\infty}\frac{|\sin(bx)|}{x^{p}}dx $$ I know that $\Gamma(p)<+\infty$ for $0<p<1$, but what about $\int_0^{+\infty}\frac{|\sin(bx)|}{x^{p}}dx$?$\int_0^{+\infty}\frac{|\sin(bx)|}{x^{p}}dx\leq\int_0^{+\infty}\frac{1}{x^{p}}dx$ doesn’t help me considering that the second integral does not converge... By integrating first with respect to $x$ and then with respect $y$ doesn’t seem to help either.

I really apologize if this is a banal question, but I am quite new to this topic.

Thanks!

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  • $\begingroup$ Read “Inside Interesting Integrals” of P.J.Nahin. At page 130 he writes about Fubini’s Integral. I think it should contain the answer to your question. $\endgroup$ – user507623 Feb 26 '18 at 13:44
  • $\begingroup$ @Pippo, Thanks for the comment! Unfortunately the author seems to assume that he can swap the order of integration, without providing any justification. $\endgroup$ – Jaacks Feb 26 '18 at 15:40
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Quick answer: 1. Yes, you could work on $(0,\infty)^2$ or add the value infinity to your function.

  1. From $\int_0^{+\infty}\frac{|\sin(bx)|}{x^{p}}dx$ you can check using l'Hopital the behaviour of $F(x):=\frac{|\sin(bx)|}{x^{p}}$ at $0$, ($F$ should be integrable at $0$ for every $p\in(0,2)$). Then deal with integrability at $\infty$ (which is easy for $p>1$).
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  • $\begingroup$ Thank you for the answer! Could you provide some more details? In particular I have doubts about how to prove the integrability at $\infty$. I agree it’s easy to prove it for $p>1$, but can we say something for $0<p\leq1$? $\endgroup$ – Jaacks Feb 26 '18 at 15:39
  • $\begingroup$ extend the argument in here math.stackexchange.com/a/1008425/112537 ? $\endgroup$ – Ton Feb 26 '18 at 17:46

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