Prove $int_{mathbb R^{d}}f(|y|)dlambda^{d}(y)=C_{d}int_{[0,infty[}r^{d-1}f(r)dlambda^{1}(r)$












0














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







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This question has an open bounty worth +50
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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
















0














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







share|cite|improve this question















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














0












0








0







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







share|cite|improve this question













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






share|cite|improve this question













share|cite|improve this question











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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


















  • 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










2 Answers
2






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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 $f$.






share|cite|improve this answer































    1














    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.
    $$int_{mathbb{R}^d}f(|x|),dx = int_0^infty int_{S^{d-1}} f(r),dsigma, dr
    = s_{d-1} int_0^infty r^{d-1}f(r), dr$$






    share|cite|improve this answer























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      2 Answers
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      oldest

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      2 Answers
      2






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      active

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      active

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      2














      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 $f$.






      share|cite|improve this answer




























        2














        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 $f$.






        share|cite|improve this answer


























          2












          2








          2






          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 $f$.






          share|cite|improve this answer














          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 $f$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 4 at 20:44

























          answered Jan 4 at 20:01









          SongSong

          6,800320




          6,800320























              1














              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.
              $$int_{mathbb{R}^d}f(|x|),dx = int_0^infty int_{S^{d-1}} f(r),dsigma, dr
              = s_{d-1} int_0^infty r^{d-1}f(r), dr$$






              share|cite|improve this answer




























                1














                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.
                $$int_{mathbb{R}^d}f(|x|),dx = int_0^infty int_{S^{d-1}} f(r),dsigma, dr
                = s_{d-1} int_0^infty r^{d-1}f(r), dr$$






                share|cite|improve this answer


























                  1












                  1








                  1






                  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.
                  $$int_{mathbb{R}^d}f(|x|),dx = int_0^infty int_{S^{d-1}} f(r),dsigma, dr
                  = s_{d-1} int_0^infty r^{d-1}f(r), dr$$






                  share|cite|improve this answer














                  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.
                  $$int_{mathbb{R}^d}f(|x|),dx = int_0^infty int_{S^{d-1}} f(r),dsigma, dr
                  = s_{d-1} int_0^infty r^{d-1}f(r), dr$$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 2 days ago

























                  answered Jan 4 at 20:22









                  ncmathsadistncmathsadist

                  42.4k259102




                  42.4k259102






























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