Mixing
For which parameters the integrals are convergent?
$begingroup$
For which constants $alpha$, $beta$, $gamma$ the following two integrals are convergent:
$$
iiint_{x^2+y^2+z^2leq 1} frac{dxdydz}{x^alpha+y^beta+z^gamma}\ iiint_{x^2+y^2+z^2geq 1} frac{dxdydz}{x^alpha+y^beta+z^gamma}
$$
Attempt: I'm trying to use the so called Lebesgue layer integration method. It says that if $(X,Sigma,mu)$ is a measure space, $f:Xto[0,infty]$
a measurable function and $phi: [0,infty)to[0,infty)$ is a monotone, continuously differentiable function such that $phi(0)=0$ then $$int_X (phicirc f) dmu=int_{0}^inftymu{f>t}dt.$$
So in our case it should be $f=frac{1}{x^alpha+y^beta+z^gamma}$, $phi(t)=t$, and $mu$ is the Lebesgue measure defined on the Borel sigma algebra on $mathbb{R^3}$. The function $f$ is not non-negative but I think that can be fixed by writing it as a sum $f=f_{+}-f_{-}$. However it looks unreal to find $mu{f>t}$ with such a complicated function. I tried to calculate it but failed. Maybe I'm just trying to solve the question in a wrong way. Any ideas?
real-analysis integration improper-integrals lebesgue-integral lebesgue-measure
edited Jan 8 at 18:14
gt6989b
33.9k22455
asked Jan 8 at 18:02
MarkMark
6,763416
$endgroup$
add a comment |
$begingroup$
For which constants $alpha$, $beta$, $gamma$ the following two integrals are convergent:
$$
iiint_{x^2+y^2+z^2leq 1} frac{dxdydz}{x^alpha+y^beta+z^gamma}\ iiint_{x^2+y^2+z^2geq 1} frac{dxdydz}{x^alpha+y^beta+z^gamma}
$$
Attempt: I'm trying to use the so called Lebesgue layer integration method. It says that if $(X,Sigma,mu)$ is a measure space, $f:Xto[0,infty]$
a measurable function and $phi: [0,infty)to[0,infty)$ is a monotone, continuously differentiable function such that $phi(0)=0$ then $$int_X (phicirc f) dmu=int_{0}^inftymu{f>t}dt.$$
So in our case it should be $f=frac{1}{x^alpha+y^beta+z^gamma}$, $phi(t)=t$, and $mu$ is the Lebesgue measure defined on the Borel sigma algebra on $mathbb{R^3}$. The function $f$ is not non-negative but I think that can be fixed by writing it as a sum $f=f_{+}-f_{-}$. However it looks unreal to find $mu{f>t}$ with such a complicated function. I tried to calculate it but failed. Maybe I'm just trying to solve the question in a wrong way. Any ideas?
real-analysis integration improper-integrals lebesgue-integral lebesgue-measure
edited Jan 8 at 18:14
gt6989b
33.9k22455
asked Jan 8 at 18:02
MarkMark
6,763416
$endgroup$
add a comment |
$begingroup$
For which constants $alpha$, $beta$, $gamma$ the following two integrals are convergent:
$$
iiint_{x^2+y^2+z^2leq 1} frac{dxdydz}{x^alpha+y^beta+z^gamma}\ iiint_{x^2+y^2+z^2geq 1} frac{dxdydz}{x^alpha+y^beta+z^gamma}
$$
Attempt: I'm trying to use the so called Lebesgue layer integration method. It says that if $(X,Sigma,mu)$ is a measure space, $f:Xto[0,infty]$
a measurable function and $phi: [0,infty)to[0,infty)$ is a monotone, continuously differentiable function such that $phi(0)=0$ then $$int_X (phicirc f) dmu=int_{0}^inftymu{f>t}dt.$$
So in our case it should be $f=frac{1}{x^alpha+y^beta+z^gamma}$, $phi(t)=t$, and $mu$ is the Lebesgue measure defined on the Borel sigma algebra on $mathbb{R^3}$. The function $f$ is not non-negative but I think that can be fixed by writing it as a sum $f=f_{+}-f_{-}$. However it looks unreal to find $mu{f>t}$ with such a complicated function. I tried to calculate it but failed. Maybe I'm just trying to solve the question in a wrong way. Any ideas?
real-analysis integration improper-integrals lebesgue-integral lebesgue-measure
edited Jan 8 at 18:14
gt6989b
33.9k22455
asked Jan 8 at 18:02
MarkMark
6,763416
$endgroup$
For which constants $alpha$, $beta$, $gamma$ the following two integrals are convergent:
$$
iiint_{x^2+y^2+z^2leq 1} frac{dxdydz}{x^alpha+y^beta+z^gamma}\ iiint_{x^2+y^2+z^2geq 1} frac{dxdydz}{x^alpha+y^beta+z^gamma}
$$
Attempt: I'm trying to use the so called Lebesgue layer integration method. It says that if $(X,Sigma,mu)$ is a measure space, $f:Xto[0,infty]$
a measurable function and $phi: [0,infty)to[0,infty)$ is a monotone, continuously differentiable function such that $phi(0)=0$ then $$int_X (phicirc f) dmu=int_{0}^inftymu{f>t}dt.$$
So in our case it should be $f=frac{1}{x^alpha+y^beta+z^gamma}$, $phi(t)=t$, and $mu$ is the Lebesgue measure defined on the Borel sigma algebra on $mathbb{R^3}$. The function $f$ is not non-negative but I think that can be fixed by writing it as a sum $f=f_{+}-f_{-}$. However it looks unreal to find $mu{f>t}$ with such a complicated function. I tried to calculate it but failed. Maybe I'm just trying to solve the question in a wrong way. Any ideas?
real-analysis integration improper-integrals lebesgue-integral lebesgue-measure
real-analysis integration improper-integrals lebesgue-integral lebesgue-measure
edited Jan 8 at 18:14
gt6989b
33.9k22455
asked Jan 8 at 18:02
MarkMark
6,763416
edited Jan 8 at 18:14
gt6989b
33.9k22455
asked Jan 8 at 18:02
MarkMark
6,763416
edited Jan 8 at 18:14
gt6989b
33.9k22455
edited Jan 8 at 18:14
gt6989b
33.9k22455
edited Jan 8 at 18:14
gt6989b
33.9k22455
33.9k22455
asked Jan 8 at 18:02
MarkMark
6,763416
asked Jan 8 at 18:02
MarkMark
6,763416
asked Jan 8 at 18:02
MarkMark
6,763416
6,763416
add a comment |
add a comment |
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