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Let $S=[0,1]\times[0,1]$, and $\frak m$ be the Lebesgue measure on $\Bbb R^2$. Define $f(x,y)=(y-1/2)(x-1/2)^{-3}$ if $|y-1/2|<|x-1/2|$. Otherwise $f=0$. What is $\int _S f \,{\rm d}{\frak m}$? Is it defined?

Since we can not compute the integral in orders, then how should we evaluate that integral?

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1 Answer 1

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Hint: Consider both diagonals of the square. They divide the square into four pieces. Thinking geometrically, we have that ${\rm supp}\,f$ consists of "left" and "right" pieces, say, $L$ and $R$. Once you check that $f$ is Lebesgue-integrable, you can pass to Riemann integrals and compute $$\begin{align}\int_Sf\,{\rm d}{\frak m} &= \int_Lf\,{\rm d}{\frak m} + \int_Rf\,{\rm d}{\frak m} \\ &= \lim_{b\to 1/2^-}\int_0^{b}\int_x^{1-x}\frac{y-1/2}{(x-1/2)^3}\,{\rm d}y\,{\rm d}x + \lim_{a\to 1/2^+}\int_{a}^1\int_{1-x}^x\frac{y-1/2}{(x-1/2)^3}\,{\rm d}y\,{\rm d}x. \end{align}$$

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  • $\begingroup$ WA says that each of these integrals is zero, by the way. $\endgroup$
    – Ivo Terek
    Jul 1, 2017 at 0:54
  • $\begingroup$ The $y$- integrand is odd on that interval so it makes sense it would come out to zero. However what's much less clear is that it's integrable in the first place, power counting the divergence at $1/2,1/2$. Replacing the integrands with their absolute values gives a divergence. $\endgroup$ Jul 1, 2017 at 7:57

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