Prove $int_{mathbb R^{d}}f(|y|)dlambda^{d}(y)=C_{d}int_{[0,infty[}r^{d-1}f(r)dlambda^{1}(r)$
Let $f:[0,infty[ to bar{mathbb R}$ measurable, $d in mathbb N$
while $E_{d}:={x in mathbb R^d: |x| leq 1}$
Prove $int_{mathbb R^{d}}f(|y|)dlambda^{d}(y)=C_{d}int_{[0,infty[}r^{d-1}f(r)dlambda^{1}(r)$
and $C_{d}$ is constant with $C_{d}:=2int_{E_{d-1}}frac{1}{sqrt{1-|x|^2}}dlambda^{d-1}(x)$
Normally, I would write down a few of my ideas but I have no idea where to begin. I have only just acquainted myself with the $lambda^{d}$ transformation formula, but do not see its applicability in this instance.
Any help is greatly appreciated
real-analysis measure-theory lebesgue-integral transformation
This question has an open bounty worth +50
reputation from SABOY ending in 5 days.
This question has not received enough attention.
add a comment |
Let $f:[0,infty[ to bar{mathbb R}$ measurable, $d in mathbb N$
while $E_{d}:={x in mathbb R^d: |x| leq 1}$
Prove $int_{mathbb R^{d}}f(|y|)dlambda^{d}(y)=C_{d}int_{[0,infty[}r^{d-1}f(r)dlambda^{1}(r)$
and $C_{d}$ is constant with $C_{d}:=2int_{E_{d-1}}frac{1}{sqrt{1-|x|^2}}dlambda^{d-1}(x)$
Normally, I would write down a few of my ideas but I have no idea where to begin. I have only just acquainted myself with the $lambda^{d}$ transformation formula, but do not see its applicability in this instance.
Any help is greatly appreciated
real-analysis measure-theory lebesgue-integral transformation
This question has an open bounty worth +50
reputation from SABOY ending in 5 days.
This question has not received enough attention.
Write $x = zeta u,$ with $zeta > 0$ and $u$ a unitary vector (assume $x neq 0$). Apply change of variables (change of measures?) bearing in mind that Lebuesge measure decompose as $lambda otimes mu$ where $lambda$ is $r^{d-1}$ times Lebuesgue measure on $mathbf{R}_+$ and $mu$ is the surface measure on the sphere.
– Will M.
Jan 4 at 20:30
It might help to go over the general construction of the surface measure on $S^{n-1}$.
– Matematleta
Jan 4 at 23:49
Could you explain why you decided to put a bounty on this question? Why are you not satisfied with the current answers?
– saz
2 days ago
add a comment |
Let $f:[0,infty[ to bar{mathbb R}$ measurable, $d in mathbb N$
while $E_{d}:={x in mathbb R^d: |x| leq 1}$
Prove $int_{mathbb R^{d}}f(|y|)dlambda^{d}(y)=C_{d}int_{[0,infty[}r^{d-1}f(r)dlambda^{1}(r)$
and $C_{d}$ is constant with $C_{d}:=2int_{E_{d-1}}frac{1}{sqrt{1-|x|^2}}dlambda^{d-1}(x)$
Normally, I would write down a few of my ideas but I have no idea where to begin. I have only just acquainted myself with the $lambda^{d}$ transformation formula, but do not see its applicability in this instance.
Any help is greatly appreciated
real-analysis measure-theory lebesgue-integral transformation
Let $f:[0,infty[ to bar{mathbb R}$ measurable, $d in mathbb N$
while $E_{d}:={x in mathbb R^d: |x| leq 1}$
Prove $int_{mathbb R^{d}}f(|y|)dlambda^{d}(y)=C_{d}int_{[0,infty[}r^{d-1}f(r)dlambda^{1}(r)$
and $C_{d}$ is constant with $C_{d}:=2int_{E_{d-1}}frac{1}{sqrt{1-|x|^2}}dlambda^{d-1}(x)$
Normally, I would write down a few of my ideas but I have no idea where to begin. I have only just acquainted myself with the $lambda^{d}$ transformation formula, but do not see its applicability in this instance.
Any help is greatly appreciated
real-analysis measure-theory lebesgue-integral transformation
real-analysis measure-theory lebesgue-integral transformation
asked Jan 4 at 19:48
SABOYSABOY
515311
515311
This question has an open bounty worth +50
reputation from SABOY ending in 5 days.
This question has not received enough attention.
This question has an open bounty worth +50
reputation from SABOY ending in 5 days.
This question has not received enough attention.
Write $x = zeta u,$ with $zeta > 0$ and $u$ a unitary vector (assume $x neq 0$). Apply change of variables (change of measures?) bearing in mind that Lebuesge measure decompose as $lambda otimes mu$ where $lambda$ is $r^{d-1}$ times Lebuesgue measure on $mathbf{R}_+$ and $mu$ is the surface measure on the sphere.
– Will M.
Jan 4 at 20:30
It might help to go over the general construction of the surface measure on $S^{n-1}$.
– Matematleta
Jan 4 at 23:49
Could you explain why you decided to put a bounty on this question? Why are you not satisfied with the current answers?
– saz
2 days ago
add a comment |
Write $x = zeta u,$ with $zeta > 0$ and $u$ a unitary vector (assume $x neq 0$). Apply change of variables (change of measures?) bearing in mind that Lebuesge measure decompose as $lambda otimes mu$ where $lambda$ is $r^{d-1}$ times Lebuesgue measure on $mathbf{R}_+$ and $mu$ is the surface measure on the sphere.
– Will M.
Jan 4 at 20:30
It might help to go over the general construction of the surface measure on $S^{n-1}$.
– Matematleta
Jan 4 at 23:49
Could you explain why you decided to put a bounty on this question? Why are you not satisfied with the current answers?
– saz
2 days ago
Write $x = zeta u,$ with $zeta > 0$ and $u$ a unitary vector (assume $x neq 0$). Apply change of variables (change of measures?) bearing in mind that Lebuesge measure decompose as $lambda otimes mu$ where $lambda$ is $r^{d-1}$ times Lebuesgue measure on $mathbf{R}_+$ and $mu$ is the surface measure on the sphere.
– Will M.
Jan 4 at 20:30
Write $x = zeta u,$ with $zeta > 0$ and $u$ a unitary vector (assume $x neq 0$). Apply change of variables (change of measures?) bearing in mind that Lebuesge measure decompose as $lambda otimes mu$ where $lambda$ is $r^{d-1}$ times Lebuesgue measure on $mathbf{R}_+$ and $mu$ is the surface measure on the sphere.
– Will M.
Jan 4 at 20:30
It might help to go over the general construction of the surface measure on $S^{n-1}$.
– Matematleta
Jan 4 at 23:49
It might help to go over the general construction of the surface measure on $S^{n-1}$.
– Matematleta
Jan 4 at 23:49
Could you explain why you decided to put a bounty on this question? Why are you not satisfied with the current answers?
– saz
2 days ago
Could you explain why you decided to put a bounty on this question? Why are you not satisfied with the current answers?
– saz
2 days ago
add a comment |
2 Answers
2
active
oldest
votes
It is the polar coordinate change of the Lebesgue measure. To prove the validity of the formula, we can first test it for $f$ of the form
$$
f(r) = 1_{(a,b)}(r).
$$ Then by the volume formula of $d$-dimensional ball, we can check that it's true. Since it is true for any interval $(a,b)$, the result can be extended to any measurable set $E$ by approximation, and hence to any non-negative measurable $f$. This proves the formula for general integrable $f$.
add a comment |
Change to polar coördinates. If $sigma$ is surface area measure on $S^{d - 1}$ and $s_{d-1}$ is the surface area of $S^{d-1}$, you have the following.
$$int_{mathbb{R}^d}f(|x|),dx = int_0^infty int_{S^{d-1}} f(r),dsigma, dr
= s_{d-1} int_0^infty r^{d-1}f(r), dr$$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
It is the polar coordinate change of the Lebesgue measure. To prove the validity of the formula, we can first test it for $f$ of the form
$$
f(r) = 1_{(a,b)}(r).
$$ Then by the volume formula of $d$-dimensional ball, we can check that it's true. Since it is true for any interval $(a,b)$, the result can be extended to any measurable set $E$ by approximation, and hence to any non-negative measurable $f$. This proves the formula for general integrable $f$.
add a comment |
It is the polar coordinate change of the Lebesgue measure. To prove the validity of the formula, we can first test it for $f$ of the form
$$
f(r) = 1_{(a,b)}(r).
$$ Then by the volume formula of $d$-dimensional ball, we can check that it's true. Since it is true for any interval $(a,b)$, the result can be extended to any measurable set $E$ by approximation, and hence to any non-negative measurable $f$. This proves the formula for general integrable $f$.
add a comment |
It is the polar coordinate change of the Lebesgue measure. To prove the validity of the formula, we can first test it for $f$ of the form
$$
f(r) = 1_{(a,b)}(r).
$$ Then by the volume formula of $d$-dimensional ball, we can check that it's true. Since it is true for any interval $(a,b)$, the result can be extended to any measurable set $E$ by approximation, and hence to any non-negative measurable $f$. This proves the formula for general integrable $f$.
It is the polar coordinate change of the Lebesgue measure. To prove the validity of the formula, we can first test it for $f$ of the form
$$
f(r) = 1_{(a,b)}(r).
$$ Then by the volume formula of $d$-dimensional ball, we can check that it's true. Since it is true for any interval $(a,b)$, the result can be extended to any measurable set $E$ by approximation, and hence to any non-negative measurable $f$. This proves the formula for general integrable $f$.
edited Jan 4 at 20:44
answered Jan 4 at 20:01
SongSong
6,800320
6,800320
add a comment |
add a comment |
Change to polar coördinates. If $sigma$ is surface area measure on $S^{d - 1}$ and $s_{d-1}$ is the surface area of $S^{d-1}$, you have the following.
$$int_{mathbb{R}^d}f(|x|),dx = int_0^infty int_{S^{d-1}} f(r),dsigma, dr
= s_{d-1} int_0^infty r^{d-1}f(r), dr$$
add a comment |
Change to polar coördinates. If $sigma$ is surface area measure on $S^{d - 1}$ and $s_{d-1}$ is the surface area of $S^{d-1}$, you have the following.
$$int_{mathbb{R}^d}f(|x|),dx = int_0^infty int_{S^{d-1}} f(r),dsigma, dr
= s_{d-1} int_0^infty r^{d-1}f(r), dr$$
add a comment |
Change to polar coördinates. If $sigma$ is surface area measure on $S^{d - 1}$ and $s_{d-1}$ is the surface area of $S^{d-1}$, you have the following.
$$int_{mathbb{R}^d}f(|x|),dx = int_0^infty int_{S^{d-1}} f(r),dsigma, dr
= s_{d-1} int_0^infty r^{d-1}f(r), dr$$
Change to polar coördinates. If $sigma$ is surface area measure on $S^{d - 1}$ and $s_{d-1}$ is the surface area of $S^{d-1}$, you have the following.
$$int_{mathbb{R}^d}f(|x|),dx = int_0^infty int_{S^{d-1}} f(r),dsigma, dr
= s_{d-1} int_0^infty r^{d-1}f(r), dr$$
edited 2 days ago
answered Jan 4 at 20:22
ncmathsadistncmathsadist
42.4k259102
42.4k259102
add a comment |
add a comment |
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Write $x = zeta u,$ with $zeta > 0$ and $u$ a unitary vector (assume $x neq 0$). Apply change of variables (change of measures?) bearing in mind that Lebuesge measure decompose as $lambda otimes mu$ where $lambda$ is $r^{d-1}$ times Lebuesgue measure on $mathbf{R}_+$ and $mu$ is the surface measure on the sphere.
– Will M.
Jan 4 at 20:30
It might help to go over the general construction of the surface measure on $S^{n-1}$.
– Matematleta
Jan 4 at 23:49
Could you explain why you decided to put a bounty on this question? Why are you not satisfied with the current answers?
– saz
2 days ago