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I am having some trouble understanding how to apply Fubini and/or Tonelli Theorems to determine whether a Lebesgue integral over $\mathbb{R}^2_+$ exists and if it is finite.

If someone could help me by showing the explicit steps for the examples below I would be grateful. I have a long list of exercises I have found online (this is self-study) and instead of posting a bunch of examples here I thought a few simple ones would help me learn how to do these problems going forward.

The examples in hand are: for each of the functions, use Fubini or Tonelli to show the existence/finiteness of the function's Lebesgue integral over $\mathbb{R}^2_+$.

$f_1(x,y)=\frac{\sin xy}{xy}$

$f_2(x,y)=e^{-(1+x^2)y}$

Many thanks in advance!

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    From [FAQ about tags](http://meta.math.stackexchange.com/questions/107/faq-for-math-stackexchange/128#128): *Try to avoid creating new tags. Instead, check if there is some synonym that already has a popular tag.* It's not easy to keep balance between too specific tags and not having enough tags, but tags for specific theorems are usually too specific, perhaps with the exception of some very important ones. (Of course, you can disagree, and there is possibility for further discussion, if needed.)2012-08-16

1 Answers 1

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  1. Assume $f_1$ is integrable in $\mathbb{R}^2_+$ (which I assume is $\mathbb{R}\times \mathbb{R}_+$), by Fubini's theorem $f_1^x(y)=f_1(x,y)$ would be (Lebesgue!) integrable for almost every $x\in \mathbb{R}$, but $ \int_{\mathbb{R}_+} f_1^x(y) dy= \int_0^{\infty} \frac{\sin(xy)}{xy} dy $ and this last is not a Lebesgue integral for $x\neq 0$ (the positive and negative parts of the function $\sin(z)/z$, when integrated give $\infty$). So $f_1$ is not integrable in $\mathbb{R}^2_+$.

  2. Since $f_2$ is positive everywhere, Tonelli's theorem guarantees that we can integrate first over $y$ and then over $x$ to get $ \int_{\mathbb{R}^2_+} f_2(x,y)d(x,y) = \int_{\mathbb{R}} \int_0^\infty e^{-(1+x^2)y}dydx= \int_{-\infty}^{\infty} \frac{1}{1+x^2} dx <\infty $

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    No, the integral does not exist, even as $\pm\infty$ since the Lebesgue integral is not defined for that function.2012-08-13