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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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    $\begingroup$ Are $x$ and $y$ in $\mathbb{R}$ or a higher dimension? $\endgroup$ – anon Mar 15 '12 at 20:11
  • $\begingroup$ @anon $x$ and $y$ are in $\mathbb{R}$. $\endgroup$ – Berkheimer Mar 15 '12 at 20:37
  • $\begingroup$ Obrigado, what have you tried? $\endgroup$ – AD. Mar 15 '12 at 20:45
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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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  • $\begingroup$ 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? $\endgroup$ – Berkheimer Mar 20 '12 at 7:57
  • $\begingroup$ I've corrected typos and mistakes. If you have any questions feel free to ask. $\endgroup$ – Norbert Mar 20 '12 at 9:36
  • $\begingroup$ Thanks a lot for your help and attention. $\endgroup$ – Berkheimer Mar 20 '12 at 20:29
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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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  • $\begingroup$ $1 < r < \infty$ and $0 < \theta < {\pi \over 4}$ is not the same as the domain he's integrating over. $\endgroup$ – Zarrax Mar 15 '12 at 22:08
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    $\begingroup$ Yes but it doesn't matter since we only take care about the convergence of the integral. $\endgroup$ – Davide Giraudo Mar 15 '12 at 22:24
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    $\begingroup$ You might want to mention why the additional portion doesn't affect convergence then $\endgroup$ – Zarrax Mar 15 '12 at 23:18
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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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  • $\begingroup$ It is very similar to the proof of Zarraх. $\endgroup$ – Norbert Mar 16 '12 at 21:07

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