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Give an example (if any) for a non-integrable function $f:\mathbb{R\times R}$ $\to$ $\mathbb{R}$ with domain in $[0,1]^2$ such that both iterated integrals exists(i.e. in both order of integration). Here is what I have got: $$ f(x,y) = \begin{cases} e^{-xy}\sin x \sin y, & \text{if }x,y \geq 0 \\ 0, & \text{otherwise } \end{cases} $$ Does this function work for my case? I found this in :http://www.mathnet.or.kr/mathnet/kms_tex/80630.pdf. It says that iterated integrals exist but not double integrals. I am not sure if this implies Riemann integral does not exist.

Mhenni Benghorbal
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user77440
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  • The example shows that $f$ is not integrable over $[0,\infty)^2$; the function is continuous on $[0,1]^2$ hence definitely integrable there. (I won't post this as an answer because I have no time to find answers to the rest of the question.) – Lord_Farin May 16 '13 at 07:08
  • so, just to be sure again, you mean f satisfies both the requirements, right? – user77440 May 16 '13 at 07:12
  • Yes, both the iterated integrals exist on $[0,1]^2$, but so does the double (= Riemann) integral. – Lord_Farin May 16 '13 at 07:13
  • I am sorry, you confused me: you say double(=Riemann) exists :) – user77440 May 16 '13 at 07:16
  • The Riemann integral over $[0,1]^2$ exists. The Riemann integral over $[0,\infty)^2$ does not. So $f$ does not provide an example. – Lord_Farin May 16 '13 at 07:40

1 Answers1

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Consider the function

$$ f(x,y)= \frac{x^2-y^2}{(x^2+y^2)^2}. $$

Now, if you evaluate the integral

$$ \int_{0}^{1}\int_{0}^{1}f(x,y)dydx = \frac{\pi}{4},$$

and if you consider the other order, you get

$$ \int_{0}^{1}\int_{0}^{1}f(x,y)dxdy = -\frac{\pi}{4}. $$

So, the iterated integrals exist, but the double integral does not.

Mhenni Benghorbal
  • 47,431