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Let $A$ be the set in $\mathbb{R}^2$ defined by $$A = \left\{(x,y)\left| x \gt 1\text{ and }0\lt y\lt\frac{1}{x}\right.\right\}.$$ Calculate $$\int\!\!\int_A\left(\frac{1}{x}\right)y^{1/2}dA$$ if it exists.

*Important: There's only 1 integral sub A, this is not a double integral.

My proof:

So our integral will have bounds of $x$ from $1$ to $\infty$ and $y$ will have bounds from $0$ to $\frac{1}{x}$.

So, we have an integral of $1$ to $\infty$ of $1/x\; dx$ and an integral of $1$ to $1/x$ of $y^{1/2}\; dy$

and we have an integral of $1$ to $\infty$ of $$\left(\frac{2}{3}\left(\frac{1}{x}\right)^{3/2}-\frac{2}{3}\right)\frac{1}{x}\,dx$$ and our final answer diverges so it doesn't exist.

But how to say this rigorously/correctly?

Thanks

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I've $\LaTeX$-ed your post. Please go over it and make sure I did not change the meaning; a lot of your fractions were ambiguous and I'm not sure if I got them right. –  Arturo Magidin Feb 22 '12 at 6:21
5  
If you find that integral ugly, you should start adjusting your beauty-meter for what's coming... Life has much worse integrals in store for you! –  Mariano Suárez-Alvarez Feb 22 '12 at 6:22
    
The integral exists (but $dA$ should read $dxdy$). –  Did Feb 22 '12 at 6:24
    
Many Thanks Arturo! There should only be 1 integral for the region for A. –  James R. Feb 22 '12 at 6:35

2 Answers 2

up vote 1 down vote accepted

When you convert it to the iterated integral, $y$ should range from $0$ to $\frac1x$, not from $1$ to $\frac1x$. Thus, your first few steps should be as follows:

$$\begin{align*} \int_A\left(\frac1x\right)y^{1/2}dA&=\int_1^\infty\int_0^{1/x}\left(\frac1x\right)y^{1/2}dydx\\ &=\frac23\int_1^\infty\left(\frac1x\right)[y^{3/2}]_0^{1/x}dx\\ &=\frac23\int_1^\infty\left(\frac1x\right)\left(\frac1x\right)^{3/2}dx \end{align*}$$

(where I’ve pulled out the constant factor to get it out of the way). In other words, you shouldn’t have $\frac23\left(\frac1x\right)^{3/2}-\frac23$, but just $\frac23\left(\frac1x\right)^{3/2}$. Now simplify to

$$\frac23\int_1^\infty\left(\frac1x\right)\left(\frac1x\right)^{3/2}dx=\frac23\int_1^\infty x^{-5/2}dx$$

and continue from there. Be a little careful: this is an improper integral, so you should be rewriting it as

$$\frac23\lim_{a\to\infty}\int_1^a x^{-5/2}dx\;.$$

You should now find that it does indeed exist.

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It turns out that there is no existence problem. In order to be ultra cautious, let's integrate over the bounded region $A_M$ that has $1\le x\le M$. When we express our integral in the usual way as an iterated integral, we get $$\iint_{A_M}\frac{y^{1/2}}{x}dA=\int_{x=1}^M \left(\int_{y=0}^{1/x} \frac{y^{1/2}}{x} dy\right) dx.$$ The inner integral is $\dfrac{2}{3x^{5/2}}.$

Now integrate with respect to $x$, from $1$ to $M$. We get $$\frac{4}{9}\left(1-\frac{1}{M^{3/2}}\right).$$ As $M\to\infty$, this approaches $\dfrac{4}{9}$.

There would have been trouble if the region $A$ extended from $x=0$ on.

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