For which $alphainmathbb{R}$ the integral...












0












$begingroup$


Im looking for which $alphainmathbb{R}$ the integral $int_{mathbb{R}^{2}}frac{dxdy}{left(1+x^{2}+xy+y^{2}right)^{alpha}}$
converges/diverges.



What I was looking for is an appropriate change of variables. I tried
polar coordinates $left(x,yright)=Tleft(r,thetaright)=left(rcostheta,rsinthetaright)$.
So $left|J_{T}left(r,thetaright)right|=r$ and $T^{-1}left(mathbb{R}^{2}right)=left{ left(r,thetaright)mid r>0:,:0<theta<2piright} $
up to a null set. So
$$
int_{mathbb{R}^{2}}frac{dxdy}{left(1+x^{2}+xy+y^{2}right)^{alpha}}=int_{T^{-1}left(mathbb{R}^{2}right)}frac{rdrdtheta}{left(1+r^{2}+r^{2}sinthetacosthetaright)^{alpha}}=int_{T^{-1}left(mathbb{R}^{2}right)}frac{2rdrdtheta}{left(2+r^{2}left(2+sin2thetaright)right)^{alpha}}
$$
Then with another change of variables $t=r^{2}$ we get that $dt=2rdr$
and so I got
$$
intfrac{dtdtheta}{left(2+tleft(2+sin2thetaright)right)^{alpha}}
$$
Questions:




  1. Does it help somehow?

  2. Is there a better change of variables here?

  3. How can I find the range of $alpha$ that im looking for?

  4. Is it possible maybe to block the function with and easier one to integral and then use sandwich?






multivariable-calculus multiple-integral change-of-variable







share|cite|improve this question











$endgroup$












  • $begingroup$
    Since the integrand is asymptotic to $r^{-2 alpha}$, you have convergence whenever $- 2 alpha < -1$, i.e. for $alpha >1$.
    $endgroup$
    – Crostul
    Jan 10 at 8:53










  • $begingroup$
    Is there a way to actually compute the integral?
    $endgroup$
    – Jon
    Jan 10 at 8:55










  • $begingroup$
    @Crostul There is a factor of $r^{2}$ coming in when you switch to polar coordinates.
    $endgroup$
    – Kavi Rama Murthy
    Jan 10 at 8:57












  • $begingroup$
    @Jon Explicit computation of integral may not be possible.
    $endgroup$
    – Kavi Rama Murthy
    Jan 10 at 8:59
















0












$begingroup$


Im looking for which $alphainmathbb{R}$ the integral $int_{mathbb{R}^{2}}frac{dxdy}{left(1+x^{2}+xy+y^{2}right)^{alpha}}$
converges/diverges.



What I was looking for is an appropriate change of variables. I tried
polar coordinates $left(x,yright)=Tleft(r,thetaright)=left(rcostheta,rsinthetaright)$.
So $left|J_{T}left(r,thetaright)right|=r$ and $T^{-1}left(mathbb{R}^{2}right)=left{ left(r,thetaright)mid r>0:,:0<theta<2piright} $
up to a null set. So
$$
int_{mathbb{R}^{2}}frac{dxdy}{left(1+x^{2}+xy+y^{2}right)^{alpha}}=int_{T^{-1}left(mathbb{R}^{2}right)}frac{rdrdtheta}{left(1+r^{2}+r^{2}sinthetacosthetaright)^{alpha}}=int_{T^{-1}left(mathbb{R}^{2}right)}frac{2rdrdtheta}{left(2+r^{2}left(2+sin2thetaright)right)^{alpha}}
$$
Then with another change of variables $t=r^{2}$ we get that $dt=2rdr$
and so I got
$$
intfrac{dtdtheta}{left(2+tleft(2+sin2thetaright)right)^{alpha}}
$$
Questions:




  1. Does it help somehow?

  2. Is there a better change of variables here?

  3. How can I find the range of $alpha$ that im looking for?

  4. Is it possible maybe to block the function with and easier one to integral and then use sandwich?






multivariable-calculus multiple-integral change-of-variable







share|cite|improve this question











$endgroup$












  • $begingroup$
    Since the integrand is asymptotic to $r^{-2 alpha}$, you have convergence whenever $- 2 alpha < -1$, i.e. for $alpha >1$.
    $endgroup$
    – Crostul
    Jan 10 at 8:53










  • $begingroup$
    Is there a way to actually compute the integral?
    $endgroup$
    – Jon
    Jan 10 at 8:55










  • $begingroup$
    @Crostul There is a factor of $r^{2}$ coming in when you switch to polar coordinates.
    $endgroup$
    – Kavi Rama Murthy
    Jan 10 at 8:57












  • $begingroup$
    @Jon Explicit computation of integral may not be possible.
    $endgroup$
    – Kavi Rama Murthy
    Jan 10 at 8:59














0












0








0





$begingroup$


Im looking for which $alphainmathbb{R}$ the integral $int_{mathbb{R}^{2}}frac{dxdy}{left(1+x^{2}+xy+y^{2}right)^{alpha}}$
converges/diverges.



What I was looking for is an appropriate change of variables. I tried
polar coordinates $left(x,yright)=Tleft(r,thetaright)=left(rcostheta,rsinthetaright)$.
So $left|J_{T}left(r,thetaright)right|=r$ and $T^{-1}left(mathbb{R}^{2}right)=left{ left(r,thetaright)mid r>0:,:0<theta<2piright} $
up to a null set. So
$$
int_{mathbb{R}^{2}}frac{dxdy}{left(1+x^{2}+xy+y^{2}right)^{alpha}}=int_{T^{-1}left(mathbb{R}^{2}right)}frac{rdrdtheta}{left(1+r^{2}+r^{2}sinthetacosthetaright)^{alpha}}=int_{T^{-1}left(mathbb{R}^{2}right)}frac{2rdrdtheta}{left(2+r^{2}left(2+sin2thetaright)right)^{alpha}}
$$
Then with another change of variables $t=r^{2}$ we get that $dt=2rdr$
and so I got
$$
intfrac{dtdtheta}{left(2+tleft(2+sin2thetaright)right)^{alpha}}
$$
Questions:




  1. Does it help somehow?

  2. Is there a better change of variables here?

  3. How can I find the range of $alpha$ that im looking for?

  4. Is it possible maybe to block the function with and easier one to integral and then use sandwich?






multivariable-calculus multiple-integral change-of-variable







share|cite|improve this question











$endgroup$




Im looking for which $alphainmathbb{R}$ the integral $int_{mathbb{R}^{2}}frac{dxdy}{left(1+x^{2}+xy+y^{2}right)^{alpha}}$
converges/diverges.



What I was looking for is an appropriate change of variables. I tried
polar coordinates $left(x,yright)=Tleft(r,thetaright)=left(rcostheta,rsinthetaright)$.
So $left|J_{T}left(r,thetaright)right|=r$ and $T^{-1}left(mathbb{R}^{2}right)=left{ left(r,thetaright)mid r>0:,:0<theta<2piright} $
up to a null set. So
$$
int_{mathbb{R}^{2}}frac{dxdy}{left(1+x^{2}+xy+y^{2}right)^{alpha}}=int_{T^{-1}left(mathbb{R}^{2}right)}frac{rdrdtheta}{left(1+r^{2}+r^{2}sinthetacosthetaright)^{alpha}}=int_{T^{-1}left(mathbb{R}^{2}right)}frac{2rdrdtheta}{left(2+r^{2}left(2+sin2thetaright)right)^{alpha}}
$$
Then with another change of variables $t=r^{2}$ we get that $dt=2rdr$
and so I got
$$
intfrac{dtdtheta}{left(2+tleft(2+sin2thetaright)right)^{alpha}}
$$
Questions:




  1. Does it help somehow?

  2. Is there a better change of variables here?

  3. How can I find the range of $alpha$ that im looking for?

  4. Is it possible maybe to block the function with and easier one to integral and then use sandwich?






multivariable-calculus multiple-integral change-of-variable




multivariable-calculus multiple-integral change-of-variable






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 10 at 8:55







Jon

















asked Jan 10 at 8:46









JonJon

565413




565413












  • $begingroup$
    Since the integrand is asymptotic to $r^{-2 alpha}$, you have convergence whenever $- 2 alpha < -1$, i.e. for $alpha >1$.
    $endgroup$
    – Crostul
    Jan 10 at 8:53










  • $begingroup$
    Is there a way to actually compute the integral?
    $endgroup$
    – Jon
    Jan 10 at 8:55










  • $begingroup$
    @Crostul There is a factor of $r^{2}$ coming in when you switch to polar coordinates.
    $endgroup$
    – Kavi Rama Murthy
    Jan 10 at 8:57












  • $begingroup$
    @Jon Explicit computation of integral may not be possible.
    $endgroup$
    – Kavi Rama Murthy
    Jan 10 at 8:59


















  • $begingroup$
    Since the integrand is asymptotic to $r^{-2 alpha}$, you have convergence whenever $- 2 alpha < -1$, i.e. for $alpha >1$.
    $endgroup$
    – Crostul
    Jan 10 at 8:53










  • $begingroup$
    Is there a way to actually compute the integral?
    $endgroup$
    – Jon
    Jan 10 at 8:55










  • $begingroup$
    @Crostul There is a factor of $r^{2}$ coming in when you switch to polar coordinates.
    $endgroup$
    – Kavi Rama Murthy
    Jan 10 at 8:57












  • $begingroup$
    @Jon Explicit computation of integral may not be possible.
    $endgroup$
    – Kavi Rama Murthy
    Jan 10 at 8:59
















$begingroup$
Since the integrand is asymptotic to $r^{-2 alpha}$, you have convergence whenever $- 2 alpha < -1$, i.e. for $alpha >1$.
$endgroup$
– Crostul
Jan 10 at 8:53




$begingroup$
Since the integrand is asymptotic to $r^{-2 alpha}$, you have convergence whenever $- 2 alpha < -1$, i.e. for $alpha >1$.
$endgroup$
– Crostul
Jan 10 at 8:53












$begingroup$
Is there a way to actually compute the integral?
$endgroup$
– Jon
Jan 10 at 8:55




$begingroup$
Is there a way to actually compute the integral?
$endgroup$
– Jon
Jan 10 at 8:55












$begingroup$
@Crostul There is a factor of $r^{2}$ coming in when you switch to polar coordinates.
$endgroup$
– Kavi Rama Murthy
Jan 10 at 8:57






$begingroup$
@Crostul There is a factor of $r^{2}$ coming in when you switch to polar coordinates.
$endgroup$
– Kavi Rama Murthy
Jan 10 at 8:57














$begingroup$
@Jon Explicit computation of integral may not be possible.
$endgroup$
– Kavi Rama Murthy
Jan 10 at 8:59




$begingroup$
@Jon Explicit computation of integral may not be possible.
$endgroup$
– Kavi Rama Murthy
Jan 10 at 8:59










1 Answer
1






active

oldest

votes


















0












$begingroup$

First note that if the integral over one of the four quadrants converge then the integral over the whole plane converges. [Use the transformations $x to -x$ and $y to -y$ for this]. So integrate over $x,y>0$ Note that $x^{2}+xy+y^{2} geq x^{2}+y^{2}$ and $x^{2}+xy+y^{2} leq 2(x^{2}+y^{2})$ since $2xy leq x^{2}+y^{2}$. Now using polar coordinates you see that the integral converges iff $frac {r} {r^{2alpha}}$ is integrable near $infty$ which is true iff $alpha >1$. Note: I have used the inequalities above just to get rid of $sin (theta)$ and $cos (theta)$. When you have a function of $r$ alone it is easy to see when the integral converges.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    The Jacobian of the polar transformation is $r$, not $r^2$. Then we integrate $dtheta$, a fixed total angle - it should be $frac{r}{r^{2alpha}}$ we're testing for integrability at $infty$.
    $endgroup$
    – jmerry
    Jan 10 at 9:07










  • $begingroup$
    @jmerry Thanks for the correction. I have edited my answer.
    $endgroup$
    – Kavi Rama Murthy
    Jan 10 at 9:11











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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

First note that if the integral over one of the four quadrants converge then the integral over the whole plane converges. [Use the transformations $x to -x$ and $y to -y$ for this]. So integrate over $x,y>0$ Note that $x^{2}+xy+y^{2} geq x^{2}+y^{2}$ and $x^{2}+xy+y^{2} leq 2(x^{2}+y^{2})$ since $2xy leq x^{2}+y^{2}$. Now using polar coordinates you see that the integral converges iff $frac {r} {r^{2alpha}}$ is integrable near $infty$ which is true iff $alpha >1$. Note: I have used the inequalities above just to get rid of $sin (theta)$ and $cos (theta)$. When you have a function of $r$ alone it is easy to see when the integral converges.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    The Jacobian of the polar transformation is $r$, not $r^2$. Then we integrate $dtheta$, a fixed total angle - it should be $frac{r}{r^{2alpha}}$ we're testing for integrability at $infty$.
    $endgroup$
    – jmerry
    Jan 10 at 9:07










  • $begingroup$
    @jmerry Thanks for the correction. I have edited my answer.
    $endgroup$
    – Kavi Rama Murthy
    Jan 10 at 9:11
















0












$begingroup$

First note that if the integral over one of the four quadrants converge then the integral over the whole plane converges. [Use the transformations $x to -x$ and $y to -y$ for this]. So integrate over $x,y>0$ Note that $x^{2}+xy+y^{2} geq x^{2}+y^{2}$ and $x^{2}+xy+y^{2} leq 2(x^{2}+y^{2})$ since $2xy leq x^{2}+y^{2}$. Now using polar coordinates you see that the integral converges iff $frac {r} {r^{2alpha}}$ is integrable near $infty$ which is true iff $alpha >1$. Note: I have used the inequalities above just to get rid of $sin (theta)$ and $cos (theta)$. When you have a function of $r$ alone it is easy to see when the integral converges.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    The Jacobian of the polar transformation is $r$, not $r^2$. Then we integrate $dtheta$, a fixed total angle - it should be $frac{r}{r^{2alpha}}$ we're testing for integrability at $infty$.
    $endgroup$
    – jmerry
    Jan 10 at 9:07










  • $begingroup$
    @jmerry Thanks for the correction. I have edited my answer.
    $endgroup$
    – Kavi Rama Murthy
    Jan 10 at 9:11














0












0








0





$begingroup$

First note that if the integral over one of the four quadrants converge then the integral over the whole plane converges. [Use the transformations $x to -x$ and $y to -y$ for this]. So integrate over $x,y>0$ Note that $x^{2}+xy+y^{2} geq x^{2}+y^{2}$ and $x^{2}+xy+y^{2} leq 2(x^{2}+y^{2})$ since $2xy leq x^{2}+y^{2}$. Now using polar coordinates you see that the integral converges iff $frac {r} {r^{2alpha}}$ is integrable near $infty$ which is true iff $alpha >1$. Note: I have used the inequalities above just to get rid of $sin (theta)$ and $cos (theta)$. When you have a function of $r$ alone it is easy to see when the integral converges.






share|cite|improve this answer











$endgroup$



First note that if the integral over one of the four quadrants converge then the integral over the whole plane converges. [Use the transformations $x to -x$ and $y to -y$ for this]. So integrate over $x,y>0$ Note that $x^{2}+xy+y^{2} geq x^{2}+y^{2}$ and $x^{2}+xy+y^{2} leq 2(x^{2}+y^{2})$ since $2xy leq x^{2}+y^{2}$. Now using polar coordinates you see that the integral converges iff $frac {r} {r^{2alpha}}$ is integrable near $infty$ which is true iff $alpha >1$. Note: I have used the inequalities above just to get rid of $sin (theta)$ and $cos (theta)$. When you have a function of $r$ alone it is easy to see when the integral converges.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 10 at 9:10

























answered Jan 10 at 8:55









Kavi Rama MurthyKavi Rama Murthy

54.8k32055




54.8k32055












  • $begingroup$
    The Jacobian of the polar transformation is $r$, not $r^2$. Then we integrate $dtheta$, a fixed total angle - it should be $frac{r}{r^{2alpha}}$ we're testing for integrability at $infty$.
    $endgroup$
    – jmerry
    Jan 10 at 9:07










  • $begingroup$
    @jmerry Thanks for the correction. I have edited my answer.
    $endgroup$
    – Kavi Rama Murthy
    Jan 10 at 9:11


















  • $begingroup$
    The Jacobian of the polar transformation is $r$, not $r^2$. Then we integrate $dtheta$, a fixed total angle - it should be $frac{r}{r^{2alpha}}$ we're testing for integrability at $infty$.
    $endgroup$
    – jmerry
    Jan 10 at 9:07










  • $begingroup$
    @jmerry Thanks for the correction. I have edited my answer.
    $endgroup$
    – Kavi Rama Murthy
    Jan 10 at 9:11
















$begingroup$
The Jacobian of the polar transformation is $r$, not $r^2$. Then we integrate $dtheta$, a fixed total angle - it should be $frac{r}{r^{2alpha}}$ we're testing for integrability at $infty$.
$endgroup$
– jmerry
Jan 10 at 9:07




$begingroup$
The Jacobian of the polar transformation is $r$, not $r^2$. Then we integrate $dtheta$, a fixed total angle - it should be $frac{r}{r^{2alpha}}$ we're testing for integrability at $infty$.
$endgroup$
– jmerry
Jan 10 at 9:07












$begingroup$
@jmerry Thanks for the correction. I have edited my answer.
$endgroup$
– Kavi Rama Murthy
Jan 10 at 9:11




$begingroup$
@jmerry Thanks for the correction. I have edited my answer.
$endgroup$
– Kavi Rama Murthy
Jan 10 at 9:11


















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