Mixing
When does $int frac{1}{(1+x^2+xy+y^2)^alpha}$ converge?
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I wish to find for which $alpha in mathbb R$ the integral $int_{mathbb R^2}frac{1}{(1+x^2+xy+y^2)^alpha}dxdy$ converges.
What I tried:
Firstly notice that the integrand is always positive since $xy leq x^2+y^2$.
If $alpha leq 0$ the integral trivially diverges. Assume $alpha >0$. Then:
$int_{mathbb R^2}frac{1}{(1+x^2+xy+y^2)^alpha}dxdy geq int_{mathbb R^2}frac{1}{(1+2x^2+2y^2)^alpha}dxdy= 2piint_{0}^{infty}frac{r}{(1+2r^2)^alpha}dr = piint_{0}^{infty}frac{1}{(1+2t)^alpha}dt = frac{pi}{2}int_{1}^{infty}frac{1}{s^alpha}ds$
This integral diverges when $alpha leq 1$ so it follows that our original integral diverges when $alpha leq 1$.
Now what? I'm stuck.
integration improper-integrals
asked Jan 1 at 22:04
Oria GruberOria Gruber
6,50732360
$endgroup$
|
show 1 more comment
$begingroup$
I wish to find for which $alpha in mathbb R$ the integral $int_{mathbb R^2}frac{1}{(1+x^2+xy+y^2)^alpha}dxdy$ converges.
What I tried:
Firstly notice that the integrand is always positive since $xy leq x^2+y^2$.
If $alpha leq 0$ the integral trivially diverges. Assume $alpha >0$. Then:
$int_{mathbb R^2}frac{1}{(1+x^2+xy+y^2)^alpha}dxdy geq int_{mathbb R^2}frac{1}{(1+2x^2+2y^2)^alpha}dxdy= 2piint_{0}^{infty}frac{r}{(1+2r^2)^alpha}dr = piint_{0}^{infty}frac{1}{(1+2t)^alpha}dt = frac{pi}{2}int_{1}^{infty}frac{1}{s^alpha}ds$
This integral diverges when $alpha leq 1$ so it follows that our original integral diverges when $alpha leq 1$.
Now what? I'm stuck.
integration improper-integrals
asked Jan 1 at 22:04
Oria GruberOria Gruber
6,50732360
$endgroup$
$begingroup$
You can actually compute the $dx$ part of the integral by completing the square and using trig sub. Also, be careful with your notation. Writing $int_{mathbb{R}^2}f(x,y);dx;dy$ is not quite right. You should either write $int_{mathbb{R}^2}f;dmu$ or else $int_{-infty}^inftyint_{-infty}^infty f(x,y);dx;dy$. The former case is harder to prove, as it will require something like Fubini Theorem.
$endgroup$
– Ben W
Jan 1 at 22:10
$begingroup$
How would you compute $int frac{1}{(1+(x+y)^2-xy)^alpha}$?
$endgroup$
– Oria Gruber
Jan 1 at 22:14
$begingroup$
Complete the square in $x$ by rewriting $1+x^2+xy+y^2$ as $(x+y/2)^2+5y^2/4$. Then use $x=(sqrt{5}/2)ytantheta$. It might not work but that is what I would try first. Note that you will have to split the $dy$ part of the integral into $y<0$ and $0<y$, and use a different sub for the $y<0$ part.
$endgroup$
– Ben W
Jan 1 at 22:17
2
$begingroup$
ummm $$ frac{1}{2} left(x^2 + y^2 right) ; leq ; x^2 + xy + y^2 ; leq ; frac{3}{2} left(x^2 + y^2 right)$$
$endgroup$
– Will Jagy
Jan 1 at 22:26
$begingroup$
Make it an answer @WillJagy, I will accept.
$endgroup$
– Oria Gruber
Jan 1 at 22:59
|
show 1 more comment
$begingroup$
I wish to find for which $alpha in mathbb R$ the integral $int_{mathbb R^2}frac{1}{(1+x^2+xy+y^2)^alpha}dxdy$ converges.
What I tried:
Firstly notice that the integrand is always positive since $xy leq x^2+y^2$.
If $alpha leq 0$ the integral trivially diverges. Assume $alpha >0$. Then:
$int_{mathbb R^2}frac{1}{(1+x^2+xy+y^2)^alpha}dxdy geq int_{mathbb R^2}frac{1}{(1+2x^2+2y^2)^alpha}dxdy= 2piint_{0}^{infty}frac{r}{(1+2r^2)^alpha}dr = piint_{0}^{infty}frac{1}{(1+2t)^alpha}dt = frac{pi}{2}int_{1}^{infty}frac{1}{s^alpha}ds$
This integral diverges when $alpha leq 1$ so it follows that our original integral diverges when $alpha leq 1$.
Now what? I'm stuck.
integration improper-integrals
asked Jan 1 at 22:04
Oria GruberOria Gruber
6,50732360
$endgroup$
I wish to find for which $alpha in mathbb R$ the integral $int_{mathbb R^2}frac{1}{(1+x^2+xy+y^2)^alpha}dxdy$ converges.
What I tried:
Firstly notice that the integrand is always positive since $xy leq x^2+y^2$.
If $alpha leq 0$ the integral trivially diverges. Assume $alpha >0$. Then:
$int_{mathbb R^2}frac{1}{(1+x^2+xy+y^2)^alpha}dxdy geq int_{mathbb R^2}frac{1}{(1+2x^2+2y^2)^alpha}dxdy= 2piint_{0}^{infty}frac{r}{(1+2r^2)^alpha}dr = piint_{0}^{infty}frac{1}{(1+2t)^alpha}dt = frac{pi}{2}int_{1}^{infty}frac{1}{s^alpha}ds$
This integral diverges when $alpha leq 1$ so it follows that our original integral diverges when $alpha leq 1$.
Now what? I'm stuck.
integration improper-integrals
integration improper-integrals
asked Jan 1 at 22:04
Oria GruberOria Gruber
6,50732360
asked Jan 1 at 22:04
Oria GruberOria Gruber
6,50732360
asked Jan 1 at 22:04
Oria GruberOria Gruber
6,50732360
asked Jan 1 at 22:04
Oria GruberOria Gruber
6,50732360
asked Jan 1 at 22:04
Oria GruberOria Gruber
6,50732360
6,50732360
$begingroup$
You can actually compute the $dx$ part of the integral by completing the square and using trig sub. Also, be careful with your notation. Writing $int_{mathbb{R}^2}f(x,y);dx;dy$ is not quite right. You should either write $int_{mathbb{R}^2}f;dmu$ or else $int_{-infty}^inftyint_{-infty}^infty f(x,y);dx;dy$. The former case is harder to prove, as it will require something like Fubini Theorem.
$endgroup$
– Ben W
Jan 1 at 22:10
$begingroup$
How would you compute $int frac{1}{(1+(x+y)^2-xy)^alpha}$?
$endgroup$
– Oria Gruber
Jan 1 at 22:14
$begingroup$
Complete the square in $x$ by rewriting $1+x^2+xy+y^2$ as $(x+y/2)^2+5y^2/4$. Then use $x=(sqrt{5}/2)ytantheta$. It might not work but that is what I would try first. Note that you will have to split the $dy$ part of the integral into $y<0$ and $0<y$, and use a different sub for the $y<0$ part.
$endgroup$
– Ben W
Jan 1 at 22:17
2
$begingroup$
ummm $$ frac{1}{2} left(x^2 + y^2 right) ; leq ; x^2 + xy + y^2 ; leq ; frac{3}{2} left(x^2 + y^2 right)$$
$endgroup$
– Will Jagy
Jan 1 at 22:26
$begingroup$
Make it an answer @WillJagy, I will accept.
$endgroup$
– Oria Gruber
Jan 1 at 22:59
|
show 1 more comment
$begingroup$
You can actually compute the $dx$ part of the integral by completing the square and using trig sub. Also, be careful with your notation. Writing $int_{mathbb{R}^2}f(x,y);dx;dy$ is not quite right. You should either write $int_{mathbb{R}^2}f;dmu$ or else $int_{-infty}^inftyint_{-infty}^infty f(x,y);dx;dy$. The former case is harder to prove, as it will require something like Fubini Theorem.
$endgroup$
– Ben W
Jan 1 at 22:10
$begingroup$
How would you compute $int frac{1}{(1+(x+y)^2-xy)^alpha}$?
$endgroup$
– Oria Gruber
Jan 1 at 22:14
$begingroup$
Complete the square in $x$ by rewriting $1+x^2+xy+y^2$ as $(x+y/2)^2+5y^2/4$. Then use $x=(sqrt{5}/2)ytantheta$. It might not work but that is what I would try first. Note that you will have to split the $dy$ part of the integral into $y<0$ and $0<y$, and use a different sub for the $y<0$ part.
$endgroup$
– Ben W
Jan 1 at 22:17
2
$begingroup$
ummm $$ frac{1}{2} left(x^2 + y^2 right) ; leq ; x^2 + xy + y^2 ; leq ; frac{3}{2} left(x^2 + y^2 right)$$
$endgroup$
– Will Jagy
Jan 1 at 22:26
$begingroup$
Make it an answer @WillJagy, I will accept.
$endgroup$
– Oria Gruber
Jan 1 at 22:59
$begingroup$
You can actually compute the $dx$ part of the integral by completing the square and using trig sub. Also, be careful with your notation. Writing $int_{mathbb{R}^2}f(x,y);dx;dy$ is not quite right. You should either write $int_{mathbb{R}^2}f;dmu$ or else $int_{-infty}^inftyint_{-infty}^infty f(x,y);dx;dy$. The former case is harder to prove, as it will require something like Fubini Theorem.
$endgroup$
– Ben W
Jan 1 at 22:10
$begingroup$
You can actually compute the $dx$ part of the integral by completing the square and using trig sub. Also, be careful with your notation. Writing $int_{mathbb{R}^2}f(x,y);dx;dy$ is not quite right. You should either write $int_{mathbb{R}^2}f;dmu$ or else $int_{-infty}^inftyint_{-infty}^infty f(x,y);dx;dy$. The former case is harder to prove, as it will require something like Fubini Theorem.
$endgroup$
– Ben W
Jan 1 at 22:10
$begingroup$
How would you compute $int frac{1}{(1+(x+y)^2-xy)^alpha}$?
$endgroup$
– Oria Gruber
Jan 1 at 22:14
$begingroup$
How would you compute $int frac{1}{(1+(x+y)^2-xy)^alpha}$?
$endgroup$
– Oria Gruber
Jan 1 at 22:14
$begingroup$
Complete the square in $x$ by rewriting $1+x^2+xy+y^2$ as $(x+y/2)^2+5y^2/4$. Then use $x=(sqrt{5}/2)ytantheta$. It might not work but that is what I would try first. Note that you will have to split the $dy$ part of the integral into $y<0$ and $0<y$, and use a different sub for the $y<0$ part.
$endgroup$
– Ben W
Jan 1 at 22:17
$begingroup$
Complete the square in $x$ by rewriting $1+x^2+xy+y^2$ as $(x+y/2)^2+5y^2/4$. Then use $x=(sqrt{5}/2)ytantheta$. It might not work but that is what I would try first. Note that you will have to split the $dy$ part of the integral into $y<0$ and $0<y$, and use a different sub for the $y<0$ part.
$endgroup$
– Ben W
Jan 1 at 22:17
2
2
$begingroup$
ummm $$ frac{1}{2} left(x^2 + y^2 right) ; leq ; x^2 + xy + y^2 ; leq ; frac{3}{2} left(x^2 + y^2 right)$$
$endgroup$
– Will Jagy
Jan 1 at 22:26
$begingroup$
ummm $$ frac{1}{2} left(x^2 + y^2 right) ; leq ; x^2 + xy + y^2 ; leq ; frac{3}{2} left(x^2 + y^2 right)$$
$endgroup$
– Will Jagy
Jan 1 at 22:26
$begingroup$
Make it an answer @WillJagy, I will accept.
$endgroup$
– Oria Gruber
Jan 1 at 22:59
$begingroup$
Make it an answer @WillJagy, I will accept.
$endgroup$
– Oria Gruber
Jan 1 at 22:59
|
show 1 more comment
3 Answers
3
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oldest
votes
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$$ frac{1}{2} left( x^2 + y^2 right) leq ; x^2 + xy + y^2 ; leq frac{3}{2} left( x^2 + y^2 right) $$
answered Jan 1 at 23:28
Will JagyWill Jagy
102k5100199
$endgroup$
add a comment |
$begingroup$
Let $q(x,y)$ be a positive definite binary quadratic form with associated matrix $Q$, i.e.
$$ q(x,y) = (x,y) Q (x,y)^T. $$
By the spectral theorem $Q$ can be written as $J^{-1} D J$, where $D$ is a diagonal matrix with real entries $lambda_1,lambda_2>0$ and $left|det Jright| = 1$. It follows that
$$ iint_{mathbb{R}^2}frac{1}{(1+q(x,y))^{alpha}},dx,dy=iint_{mathbb{R}^2}frac{1}{(1+(x,y)J^{-1} D J(x,y)^T)^{alpha}},dx,dy $$
equals
$$iint_{mathbb{R}^2}frac{1}{(1+(x,y)D (x,y)^T)^{alpha}},dx,dy=iint_{mathbb{R}^2}frac{1}{(1+lambda_1 x^2+lambda_2 y^2)^{alpha}},dx,dy $$
or
$$ frac{1}{sqrt{lambda_1 lambda_2}}iint_{mathbb{R}^2}frac{1}{(1+x^2+y^2)^{alpha}},dx,dy=frac{2pi}{sqrt{det Q}}int_{0}^{+infty}frac{rho}{(1+rho^2)^{alpha}},drho$$
or
$$ iint_{mathbb{R}^2}frac{1}{(1+q(x,y))^{alpha}},dx,dy=frac{pi}{(alpha-1)sqrt{det Q}} tag{Q}$$
as soon as $text{Re}(alpha)>1$. In your case $Q=left(begin{smallmatrix}1 & frac{1}{2}\ frac{1}{2} & 1end{smallmatrix}right)$, hence $det Q=frac{3}{4}$ and
$$boxed{ iint_{mathbb{R}^2}frac{dx,dy}{(1+x^2+xy+y^2)^{alpha}}=frac{2pi}{sqrt{3}(alpha-1)}qquad text{for }text{Re}(alpha)>1.} $$
In arbitrary dimension,
$$int_{mathbb{R}^n}frac{dx_1cdots dx_n}{(1+q(x_1,ldots,x_n))^{alpha}}=frac{pi^{n/2}}{(alpha-1)Gamma(n/2)sqrt{det Q}}qquad text{for }text{Re}(alpha)>1.tag{Qn}$$
answered Jan 1 at 23:02
Jack D'AurizioJack D'Aurizio
288k33280659
$endgroup$
$begingroup$
Now I am wondering: given that $sqrt{det Q}$ term, is there a direct link between the spectral theorem and the central limit theorem for a class of distributions?
$endgroup$
– Jack D'Aurizio
Jan 1 at 23:22
add a comment |
$begingroup$
Hint:
For $alpha > 0$, changing to polar coordinates,
$$frac{1}{(1+x^2+xy+y^2)^alpha} = frac{1}{(1+r^2+r^2cos theta sin theta)^alpha} = frac{1}{(1+r^2(1+sin (2theta)/2))^alpha} \ begin{cases}leqslant frac{1}{(1+r^2/2)^alpha}\ geqslant frac{1}{(1+3r^2/2)^alpha}end{cases}$$
and as $r to infty$
$$frac{r}{(1+r^2/2)^alpha} =mathcal{O}( r^{1-2alpha})$$
answered Jan 1 at 22:32
RRLRRL
49.5k42573
$endgroup$
add a comment |
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3 Answers
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3 Answers
3
active
oldest
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active
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$begingroup$
$$ frac{1}{2} left( x^2 + y^2 right) leq ; x^2 + xy + y^2 ; leq frac{3}{2} left( x^2 + y^2 right) $$
answered Jan 1 at 23:28
Will JagyWill Jagy
102k5100199
$endgroup$
add a comment |
$begingroup$
$$ frac{1}{2} left( x^2 + y^2 right) leq ; x^2 + xy + y^2 ; leq frac{3}{2} left( x^2 + y^2 right) $$
answered Jan 1 at 23:28
Will JagyWill Jagy
102k5100199
$endgroup$
add a comment |
$begingroup$
$$ frac{1}{2} left( x^2 + y^2 right) leq ; x^2 + xy + y^2 ; leq frac{3}{2} left( x^2 + y^2 right) $$
answered Jan 1 at 23:28
Will JagyWill Jagy
102k5100199
$endgroup$
$$ frac{1}{2} left( x^2 + y^2 right) leq ; x^2 + xy + y^2 ; leq frac{3}{2} left( x^2 + y^2 right) $$
answered Jan 1 at 23:28
Will JagyWill Jagy
102k5100199
answered Jan 1 at 23:28
Will JagyWill Jagy
102k5100199
answered Jan 1 at 23:28
Will JagyWill Jagy
102k5100199
answered Jan 1 at 23:28
Will JagyWill Jagy
102k5100199
102k5100199
add a comment |
add a comment |
$begingroup$
Let $q(x,y)$ be a positive definite binary quadratic form with associated matrix $Q$, i.e.
$$ q(x,y) = (x,y) Q (x,y)^T. $$
By the spectral theorem $Q$ can be written as $J^{-1} D J$, where $D$ is a diagonal matrix with real entries $lambda_1,lambda_2>0$ and $left|det Jright| = 1$. It follows that
$$ iint_{mathbb{R}^2}frac{1}{(1+q(x,y))^{alpha}},dx,dy=iint_{mathbb{R}^2}frac{1}{(1+(x,y)J^{-1} D J(x,y)^T)^{alpha}},dx,dy $$
equals
$$iint_{mathbb{R}^2}frac{1}{(1+(x,y)D (x,y)^T)^{alpha}},dx,dy=iint_{mathbb{R}^2}frac{1}{(1+lambda_1 x^2+lambda_2 y^2)^{alpha}},dx,dy $$
or
$$ frac{1}{sqrt{lambda_1 lambda_2}}iint_{mathbb{R}^2}frac{1}{(1+x^2+y^2)^{alpha}},dx,dy=frac{2pi}{sqrt{det Q}}int_{0}^{+infty}frac{rho}{(1+rho^2)^{alpha}},drho$$
or
$$ iint_{mathbb{R}^2}frac{1}{(1+q(x,y))^{alpha}},dx,dy=frac{pi}{(alpha-1)sqrt{det Q}} tag{Q}$$
as soon as $text{Re}(alpha)>1$. In your case $Q=left(begin{smallmatrix}1 & frac{1}{2}\ frac{1}{2} & 1end{smallmatrix}right)$, hence $det Q=frac{3}{4}$ and
$$boxed{ iint_{mathbb{R}^2}frac{dx,dy}{(1+x^2+xy+y^2)^{alpha}}=frac{2pi}{sqrt{3}(alpha-1)}qquad text{for }text{Re}(alpha)>1.} $$
In arbitrary dimension,
$$int_{mathbb{R}^n}frac{dx_1cdots dx_n}{(1+q(x_1,ldots,x_n))^{alpha}}=frac{pi^{n/2}}{(alpha-1)Gamma(n/2)sqrt{det Q}}qquad text{for }text{Re}(alpha)>1.tag{Qn}$$
answered Jan 1 at 23:02
Jack D'AurizioJack D'Aurizio
288k33280659
$endgroup$
$begingroup$
Now I am wondering: given that $sqrt{det Q}$ term, is there a direct link between the spectral theorem and the central limit theorem for a class of distributions?
$endgroup$
– Jack D'Aurizio
Jan 1 at 23:22
add a comment |
$begingroup$
Let $q(x,y)$ be a positive definite binary quadratic form with associated matrix $Q$, i.e.
$$ q(x,y) = (x,y) Q (x,y)^T. $$
By the spectral theorem $Q$ can be written as $J^{-1} D J$, where $D$ is a diagonal matrix with real entries $lambda_1,lambda_2>0$ and $left|det Jright| = 1$. It follows that
$$ iint_{mathbb{R}^2}frac{1}{(1+q(x,y))^{alpha}},dx,dy=iint_{mathbb{R}^2}frac{1}{(1+(x,y)J^{-1} D J(x,y)^T)^{alpha}},dx,dy $$
equals
$$iint_{mathbb{R}^2}frac{1}{(1+(x,y)D (x,y)^T)^{alpha}},dx,dy=iint_{mathbb{R}^2}frac{1}{(1+lambda_1 x^2+lambda_2 y^2)^{alpha}},dx,dy $$
or
$$ frac{1}{sqrt{lambda_1 lambda_2}}iint_{mathbb{R}^2}frac{1}{(1+x^2+y^2)^{alpha}},dx,dy=frac{2pi}{sqrt{det Q}}int_{0}^{+infty}frac{rho}{(1+rho^2)^{alpha}},drho$$
or
$$ iint_{mathbb{R}^2}frac{1}{(1+q(x,y))^{alpha}},dx,dy=frac{pi}{(alpha-1)sqrt{det Q}} tag{Q}$$
as soon as $text{Re}(alpha)>1$. In your case $Q=left(begin{smallmatrix}1 & frac{1}{2}\ frac{1}{2} & 1end{smallmatrix}right)$, hence $det Q=frac{3}{4}$ and
$$boxed{ iint_{mathbb{R}^2}frac{dx,dy}{(1+x^2+xy+y^2)^{alpha}}=frac{2pi}{sqrt{3}(alpha-1)}qquad text{for }text{Re}(alpha)>1.} $$
In arbitrary dimension,
$$int_{mathbb{R}^n}frac{dx_1cdots dx_n}{(1+q(x_1,ldots,x_n))^{alpha}}=frac{pi^{n/2}}{(alpha-1)Gamma(n/2)sqrt{det Q}}qquad text{for }text{Re}(alpha)>1.tag{Qn}$$
answered Jan 1 at 23:02
Jack D'AurizioJack D'Aurizio
288k33280659
$endgroup$
$begingroup$
Now I am wondering: given that $sqrt{det Q}$ term, is there a direct link between the spectral theorem and the central limit theorem for a class of distributions?
$endgroup$
– Jack D'Aurizio
Jan 1 at 23:22
add a comment |
$begingroup$
Let $q(x,y)$ be a positive definite binary quadratic form with associated matrix $Q$, i.e.
$$ q(x,y) = (x,y) Q (x,y)^T. $$
By the spectral theorem $Q$ can be written as $J^{-1} D J$, where $D$ is a diagonal matrix with real entries $lambda_1,lambda_2>0$ and $left|det Jright| = 1$. It follows that
$$ iint_{mathbb{R}^2}frac{1}{(1+q(x,y))^{alpha}},dx,dy=iint_{mathbb{R}^2}frac{1}{(1+(x,y)J^{-1} D J(x,y)^T)^{alpha}},dx,dy $$
equals
$$iint_{mathbb{R}^2}frac{1}{(1+(x,y)D (x,y)^T)^{alpha}},dx,dy=iint_{mathbb{R}^2}frac{1}{(1+lambda_1 x^2+lambda_2 y^2)^{alpha}},dx,dy $$
or
$$ frac{1}{sqrt{lambda_1 lambda_2}}iint_{mathbb{R}^2}frac{1}{(1+x^2+y^2)^{alpha}},dx,dy=frac{2pi}{sqrt{det Q}}int_{0}^{+infty}frac{rho}{(1+rho^2)^{alpha}},drho$$
or
$$ iint_{mathbb{R}^2}frac{1}{(1+q(x,y))^{alpha}},dx,dy=frac{pi}{(alpha-1)sqrt{det Q}} tag{Q}$$
as soon as $text{Re}(alpha)>1$. In your case $Q=left(begin{smallmatrix}1 & frac{1}{2}\ frac{1}{2} & 1end{smallmatrix}right)$, hence $det Q=frac{3}{4}$ and
$$boxed{ iint_{mathbb{R}^2}frac{dx,dy}{(1+x^2+xy+y^2)^{alpha}}=frac{2pi}{sqrt{3}(alpha-1)}qquad text{for }text{Re}(alpha)>1.} $$
In arbitrary dimension,
$$int_{mathbb{R}^n}frac{dx_1cdots dx_n}{(1+q(x_1,ldots,x_n))^{alpha}}=frac{pi^{n/2}}{(alpha-1)Gamma(n/2)sqrt{det Q}}qquad text{for }text{Re}(alpha)>1.tag{Qn}$$
answered Jan 1 at 23:02
Jack D'AurizioJack D'Aurizio
288k33280659
$endgroup$
Let $q(x,y)$ be a positive definite binary quadratic form with associated matrix $Q$, i.e.
$$ q(x,y) = (x,y) Q (x,y)^T. $$
By the spectral theorem $Q$ can be written as $J^{-1} D J$, where $D$ is a diagonal matrix with real entries $lambda_1,lambda_2>0$ and $left|det Jright| = 1$. It follows that
$$ iint_{mathbb{R}^2}frac{1}{(1+q(x,y))^{alpha}},dx,dy=iint_{mathbb{R}^2}frac{1}{(1+(x,y)J^{-1} D J(x,y)^T)^{alpha}},dx,dy $$
equals
$$iint_{mathbb{R}^2}frac{1}{(1+(x,y)D (x,y)^T)^{alpha}},dx,dy=iint_{mathbb{R}^2}frac{1}{(1+lambda_1 x^2+lambda_2 y^2)^{alpha}},dx,dy $$
or
$$ frac{1}{sqrt{lambda_1 lambda_2}}iint_{mathbb{R}^2}frac{1}{(1+x^2+y^2)^{alpha}},dx,dy=frac{2pi}{sqrt{det Q}}int_{0}^{+infty}frac{rho}{(1+rho^2)^{alpha}},drho$$
or
$$ iint_{mathbb{R}^2}frac{1}{(1+q(x,y))^{alpha}},dx,dy=frac{pi}{(alpha-1)sqrt{det Q}} tag{Q}$$
as soon as $text{Re}(alpha)>1$. In your case $Q=left(begin{smallmatrix}1 & frac{1}{2}\ frac{1}{2} & 1end{smallmatrix}right)$, hence $det Q=frac{3}{4}$ and
$$boxed{ iint_{mathbb{R}^2}frac{dx,dy}{(1+x^2+xy+y^2)^{alpha}}=frac{2pi}{sqrt{3}(alpha-1)}qquad text{for }text{Re}(alpha)>1.} $$
In arbitrary dimension,
$$int_{mathbb{R}^n}frac{dx_1cdots dx_n}{(1+q(x_1,ldots,x_n))^{alpha}}=frac{pi^{n/2}}{(alpha-1)Gamma(n/2)sqrt{det Q}}qquad text{for }text{Re}(alpha)>1.tag{Qn}$$
answered Jan 1 at 23:02
Jack D'AurizioJack D'Aurizio
288k33280659
edited Jan 1 at 23:15
answered Jan 1 at 23:02
Jack D'AurizioJack D'Aurizio
288k33280659
answered Jan 1 at 23:02
Jack D'AurizioJack D'Aurizio
288k33280659
answered Jan 1 at 23:02
Jack D'AurizioJack D'Aurizio
288k33280659
288k33280659
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Now I am wondering: given that $sqrt{det Q}$ term, is there a direct link between the spectral theorem and the central limit theorem for a class of distributions?
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– Jack D'Aurizio
Jan 1 at 23:22
add a comment |
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Now I am wondering: given that $sqrt{det Q}$ term, is there a direct link between the spectral theorem and the central limit theorem for a class of distributions?
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– Jack D'Aurizio
Jan 1 at 23:22
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Now I am wondering: given that $sqrt{det Q}$ term, is there a direct link between the spectral theorem and the central limit theorem for a class of distributions?
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– Jack D'Aurizio
Jan 1 at 23:22
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Now I am wondering: given that $sqrt{det Q}$ term, is there a direct link between the spectral theorem and the central limit theorem for a class of distributions?
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– Jack D'Aurizio
Jan 1 at 23:22
add a comment |
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Hint:
For $alpha > 0$, changing to polar coordinates,
$$frac{1}{(1+x^2+xy+y^2)^alpha} = frac{1}{(1+r^2+r^2cos theta sin theta)^alpha} = frac{1}{(1+r^2(1+sin (2theta)/2))^alpha} \ begin{cases}leqslant frac{1}{(1+r^2/2)^alpha}\ geqslant frac{1}{(1+3r^2/2)^alpha}end{cases}$$
and as $r to infty$
$$frac{r}{(1+r^2/2)^alpha} =mathcal{O}( r^{1-2alpha})$$
answered Jan 1 at 22:32
RRLRRL
49.5k42573
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add a comment |
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Hint:
For $alpha > 0$, changing to polar coordinates,
$$frac{1}{(1+x^2+xy+y^2)^alpha} = frac{1}{(1+r^2+r^2cos theta sin theta)^alpha} = frac{1}{(1+r^2(1+sin (2theta)/2))^alpha} \ begin{cases}leqslant frac{1}{(1+r^2/2)^alpha}\ geqslant frac{1}{(1+3r^2/2)^alpha}end{cases}$$
and as $r to infty$
$$frac{r}{(1+r^2/2)^alpha} =mathcal{O}( r^{1-2alpha})$$
answered Jan 1 at 22:32
RRLRRL
49.5k42573
$endgroup$
add a comment |
$begingroup$
Hint:
For $alpha > 0$, changing to polar coordinates,
$$frac{1}{(1+x^2+xy+y^2)^alpha} = frac{1}{(1+r^2+r^2cos theta sin theta)^alpha} = frac{1}{(1+r^2(1+sin (2theta)/2))^alpha} \ begin{cases}leqslant frac{1}{(1+r^2/2)^alpha}\ geqslant frac{1}{(1+3r^2/2)^alpha}end{cases}$$
and as $r to infty$
$$frac{r}{(1+r^2/2)^alpha} =mathcal{O}( r^{1-2alpha})$$
answered Jan 1 at 22:32
RRLRRL
49.5k42573
$endgroup$
Hint:
For $alpha > 0$, changing to polar coordinates,
$$frac{1}{(1+x^2+xy+y^2)^alpha} = frac{1}{(1+r^2+r^2cos theta sin theta)^alpha} = frac{1}{(1+r^2(1+sin (2theta)/2))^alpha} \ begin{cases}leqslant frac{1}{(1+r^2/2)^alpha}\ geqslant frac{1}{(1+3r^2/2)^alpha}end{cases}$$
and as $r to infty$
$$frac{r}{(1+r^2/2)^alpha} =mathcal{O}( r^{1-2alpha})$$
answered Jan 1 at 22:32
RRLRRL
49.5k42573
edited Jan 2 at 20:01
answered Jan 1 at 22:32
RRLRRL
49.5k42573
answered Jan 1 at 22:32
RRLRRL
49.5k42573
answered Jan 1 at 22:32
RRLRRL
49.5k42573
49.5k42573
add a comment |
add a comment |
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You can actually compute the $dx$ part of the integral by completing the square and using trig sub. Also, be careful with your notation. Writing $int_{mathbb{R}^2}f(x,y);dx;dy$ is not quite right. You should either write $int_{mathbb{R}^2}f;dmu$ or else $int_{-infty}^inftyint_{-infty}^infty f(x,y);dx;dy$. The former case is harder to prove, as it will require something like Fubini Theorem.
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– Ben W
Jan 1 at 22:10
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How would you compute $int frac{1}{(1+(x+y)^2-xy)^alpha}$?
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– Oria Gruber
Jan 1 at 22:14
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Complete the square in $x$ by rewriting $1+x^2+xy+y^2$ as $(x+y/2)^2+5y^2/4$. Then use $x=(sqrt{5}/2)ytantheta$. It might not work but that is what I would try first. Note that you will have to split the $dy$ part of the integral into $y<0$ and $0<y$, and use a different sub for the $y<0$ part.
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– Ben W
Jan 1 at 22:17
2
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ummm $$ frac{1}{2} left(x^2 + y^2 right) ; leq ; x^2 + xy + y^2 ; leq ; frac{3}{2} left(x^2 + y^2 right)$$
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– Will Jagy
Jan 1 at 22:26
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Make it an answer @WillJagy, I will accept.
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– Oria Gruber
Jan 1 at 22:59