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For which values of $\alpha$ and $\beta$ does the following integral converge?

\begin{equation} \int\int_{|x|\geq 1, |y|\geq 1} \frac{1}{|x|^\alpha+ |y|^\beta} \; dx \; dy, \quad \alpha,\beta>0. \end{equation}

Thank you!

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2  
Are $x$ and $y$ in $\mathbb{R}$ or a higher dimension? –  anon Mar 15 '12 at 20:11
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This seems like a homework question; I vaguely remember it from Folland's book (or at least the class I took that used Folland's book). If it is homework, add the homework tag so we don't give away answers. –  prpl.mnky.dshwshr Mar 15 '12 at 20:20
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@EMS Why do you think that it is a homework question? I raise this question because I think that it's really good. Check Folland's book if you wish! In addition, what if it exists there? If you do not know how to solve it, the situation will not change if I put there the homework tag. –  беркай Mar 15 '12 at 20:36
    
@anon $x$ and $y$ are in $\mathbb{R}$. –  беркай Mar 15 '12 at 20:37
    
Obrigado, what have you tried? –  AD. Mar 15 '12 at 20:45

4 Answers 4

up vote 5 down vote accepted

Denote $$ D_0=\{(x,y)\in\mathbb{R}^2:|x|\geq 1, |y|\geq 1\} $$ $$ D_1=\{(x,y)\in\mathbb{R}^2:x>1, y>1\} $$ $$ D_2=\{(x,y)\in\mathbb{R}^2:|x|^\alpha+|y|^\beta\geq 1, x>0, y>0\} $$ Then denote $$ I(D)=\iint\limits_{D}\frac{dxdy}{|x|^\alpha+|y|^\beta} $$ Since the function $$ f(x,y)=\frac{1}{|x|^\alpha+|y|^\beta} $$ is even in both variables and $D_1\subseteq D_2$ we conclude $$ 0\leq I(D_0)=4I(D_1)\leq 4I(D_2) $$ so convergence of $I(D_0)$ depends on the integral $I(D_2)$. Now we make substitution $$ x=r^{\frac{1}{\alpha}}\cos^{\frac{2}{\alpha}}\varphi\qquad y=r^{\frac{1}{\beta}}\sin^{\frac{2}{\beta}}\varphi $$ in the integral $I(D_2)$. The domain $D_2$ transforms to $$ D_2'=\left\{(r,\varphi)\in\mathbb{R}_+\times\left[0,\frac{\pi}{2}\right]:r>1\right\} $$ The Jacobian of this transform is $$ J= \begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial y}{\partial r}\\ \frac{\partial x}{\partial \varphi} & \frac{\partial y}{\partial \varphi} \end{vmatrix}= \begin{vmatrix} \frac{1}{\alpha}r^{\frac{1}{\alpha}-1} \cos^{\frac{2}{\alpha}}\varphi & \frac{1}{\beta}r^{\frac{1}{\beta}-1}\sin^{\frac{2}{\beta}}\varphi\\ -\frac{2}{\alpha}r^{\frac{1}{\alpha}}\cos^{\frac{2}{\alpha}-1}\varphi\sin\varphi & \frac{2}{\beta}r^{\frac{1}{\beta}}\sin^{\frac{2}{\beta}-1}\varphi\cos\varphi \end{vmatrix}= \frac{2}{\alpha\beta}r^{\frac{1}{\alpha}+\frac{1}{\beta}-1} \cos^{\frac{2}{\alpha}-1}\varphi\sin^{\frac{2}{\beta}-1}\varphi $$ Hence $$ I(D_2)=\iint\limits_{D_2'}\frac{\frac{2}{\alpha\beta}r^{\frac{1}{\alpha}+\frac{1}{\beta}-1} \cos^{\frac{2}{\alpha}-1}\varphi\sin^{\frac{2}{\beta}-1}\varphi }{r}dr d\varphi= $$ $$ \frac{2}{\alpha\beta}\int\limits_1^{+\infty}r^{\frac{1}{\alpha}+\frac{1}{\beta}-2}dr\int\limits_0^{\frac{\pi}{2}}\cos^{\frac{2}{\alpha}-1}\varphi\sin^{\frac{2}{\beta}-1}\varphi d\varphi= \frac{2}{\alpha\beta}B\left(\frac{1}{\alpha},\frac{1}{\beta}\right) \int\limits_1^{+\infty}r^{\frac{1}{\alpha}+\frac{1}{\beta}-2}dr $$ The right hand side of this equality is finite iff the integral over variable $r$ is finite. This holds only if $\frac{1}{\alpha}+\frac{1}{\beta}-2<-1$, i.e. $$ \frac{1}{\alpha}+\frac{1}{\beta}<1 $$

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I now realize that there is a typo in the Jacobian. Besides, you wrote that $D_2 \subseteq D_1$. Shouldn't it be vice versa? If not, could you please explain this inclusion? –  беркай Mar 20 '12 at 7:57
    
I've corrected typos and mistakes. If you have any questions feel free to ask. –  userNaN Mar 20 '12 at 9:36
    
Thanks a lot for your help and attention. –  беркай Mar 20 '12 at 20:29

Break the above integral into two parts, the first where $y^{\beta} < x^{\alpha}$ (equivalently $y < x^{\alpha \over \beta}$) and the second where $y^{\beta} > x^{\alpha}$. On the first part, the integrand is within a factor of $2$ of $x^{\alpha}$, so the overall integral is within a factor of $2$ of $$\int_1^{\infty}\int_{1 < y < x^{\alpha \over \beta}} {1 \over x^{\alpha}}\,dy\,dx$$ $$\int_1^{\infty} x^{{\alpha \over \beta} - \alpha} - x^{-\alpha}\,dx$$ This converges iff the two exponents appearing in the above are less than $-1$, so convergence occurs if and only if ${\alpha \over \beta} - \alpha < -1$. Equivalently, $\alpha + \beta < \alpha\beta$. Note this is symmetric in $\alpha$ and $\beta$, so the other part of the integral converges for the same values of $\alpha$ and $\beta$. Hence your original integral converges if and only if $\alpha + \beta < \alpha\beta$.

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We make the substitution $t_1=:=x^{\alpha/2}$ and $t_2:=y^{\beta/2}$, then we can write, possibly with an equality $+\infty=+\infty$ that $$\iint_{x,y\geq 1}\frac{dxdy}{x^{\alpha}+y^{\alpha}}=\frac 4{\alpha\beta}\int_{t_1,t_2\geq 1}\frac{t_1^{2/\alpha-1}t_2^{2/\beta-1}}{t_1^2+t_2^2}dt_1dt_2.$$ Then we use polar coordinates. The convergence of the initial integral is equivalent to the convergence of $$J(\alpha,\beta)=\int_0^{\frac{\pi}4}\cos\theta^{2/\alpha-1}\sin\theta^{2/\beta-1}\int_1^{+\infty}r^{2/\alpha+2/\beta-2+1-2}drd\theta$$ since we only added a compact portion and the integrand is continuous on this compact portion. Since $1\geq \cos\theta\geq \frac 1{\sqrt 2}$ and $2/\beta-1>-1$ the integral in $\theta$ is convergent, whereas the integral in $r$ is convergent if and only if $2\left(\alpha^{-1}+\beta^{-1}\right)-3<-1$ hence $\alpha^{-1}+\beta^{-1}<1$.

We conclude that the integral is convergent if and only if $\alpha^{-1}+\beta^{-1}<1$.

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$1 < r < \infty$ and $0 < \theta < {\pi \over 4}$ is not the same as the domain he's integrating over. –  Zarrax Mar 15 '12 at 22:08
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Yes but it doesn't matter since we only take care about the convergence of the integral. –  Davide Giraudo Mar 15 '12 at 22:24
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You might want to mention why the additional portion doesn't affect convergence then –  Zarrax Mar 15 '12 at 23:18

I'll try to make a different proof. First, we have:

$$I := \iint_{|x| \geq 1, \ |y| \geq 1} \frac{1}{|x|^\alpha+|y|^\beta} \ dx \ dy = 4 \int_{x \geq 1} \int_{y \geq 1} \frac{1}{x^\alpha+y^\beta} \ dy \ dx.$$

Factorizing by $x^\alpha$, and then using the change of variables $u:=x^{-\alpha / \beta} y$, yields:

$$I = 4 \int_{x \geq 1} \frac{1}{x^\alpha} \int_{y \geq 1} \frac{1}{1 + \frac{y^\beta}{x^\alpha}} \ dy \ dx = 4 \int_{x \geq 1} x^{\frac{\alpha}{\beta}-\alpha} \int_{u \geq x^{-\alpha / \beta}} \frac{1}{1 + u^\beta} \ du \ dx.$$

If $\beta \leq 1$, then this integral diverges. Let us assume that $\beta >1$. Then:

$$I \leq 4 \int_{x \geq 1} x^{\frac{\alpha}{\beta}-\alpha} \int_0^{+\infty} \frac{1}{1 + u^\beta} \ du \ dx = 4 \int_0^{+\infty} \frac{1}{1 + u^\beta} \ du \cdot \int_{x \geq 1} x^{\frac{\alpha}{\beta}-\alpha} \ dx.$$

Hence, $I$ is finite if and only if $\beta >1$ and $\alpha (1-1/\beta) > 1$, which is equivalent to the criterion:

$$\frac{1}{\alpha}+\frac{1}{\beta}<1.$$

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It is very similar to the proof of Zarraх. –  userNaN Mar 16 '12 at 21:07

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