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












1












$begingroup$


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









$endgroup$












  • $begingroup$
    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.
    $endgroup$
    – Will M.
    Jan 4 at 20:30












  • $begingroup$
    It might help to go over the general construction of the surface measure on $S^{n-1}$.
    $endgroup$
    – Matematleta
    Jan 4 at 23:49










  • $begingroup$
    Could you explain why you decided to put a bounty on this question? Why are you not satisfied with the current answers?
    $endgroup$
    – saz
    Jan 8 at 13:18
















1












$begingroup$


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









$endgroup$












  • $begingroup$
    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.
    $endgroup$
    – Will M.
    Jan 4 at 20:30












  • $begingroup$
    It might help to go over the general construction of the surface measure on $S^{n-1}$.
    $endgroup$
    – Matematleta
    Jan 4 at 23:49










  • $begingroup$
    Could you explain why you decided to put a bounty on this question? Why are you not satisfied with the current answers?
    $endgroup$
    – saz
    Jan 8 at 13:18














1












1








1


1



$begingroup$


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









$endgroup$




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











share|cite|improve this question




share|cite|improve this question










asked Jan 4 at 19:48









SABOYSABOY

656311




656311












  • $begingroup$
    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.
    $endgroup$
    – Will M.
    Jan 4 at 20:30












  • $begingroup$
    It might help to go over the general construction of the surface measure on $S^{n-1}$.
    $endgroup$
    – Matematleta
    Jan 4 at 23:49










  • $begingroup$
    Could you explain why you decided to put a bounty on this question? Why are you not satisfied with the current answers?
    $endgroup$
    – saz
    Jan 8 at 13:18


















  • $begingroup$
    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.
    $endgroup$
    – Will M.
    Jan 4 at 20:30












  • $begingroup$
    It might help to go over the general construction of the surface measure on $S^{n-1}$.
    $endgroup$
    – Matematleta
    Jan 4 at 23:49










  • $begingroup$
    Could you explain why you decided to put a bounty on this question? Why are you not satisfied with the current answers?
    $endgroup$
    – saz
    Jan 8 at 13:18
















$begingroup$
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.
$endgroup$
– Will M.
Jan 4 at 20:30






$begingroup$
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.
$endgroup$
– Will M.
Jan 4 at 20:30














$begingroup$
It might help to go over the general construction of the surface measure on $S^{n-1}$.
$endgroup$
– Matematleta
Jan 4 at 23:49




$begingroup$
It might help to go over the general construction of the surface measure on $S^{n-1}$.
$endgroup$
– Matematleta
Jan 4 at 23:49












$begingroup$
Could you explain why you decided to put a bounty on this question? Why are you not satisfied with the current answers?
$endgroup$
– saz
Jan 8 at 13:18




$begingroup$
Could you explain why you decided to put a bounty on this question? Why are you not satisfied with the current answers?
$endgroup$
– saz
Jan 8 at 13:18










2 Answers
2






active

oldest

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3












$begingroup$

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











$endgroup$





















    2












    $begingroup$

    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











    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      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











      $endgroup$


















        3












        $begingroup$

        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











        $endgroup$
















          3












          3








          3





          $begingroup$

          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











          $endgroup$



          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

          16.7k1941




          16.7k1941























              2












              $begingroup$

              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











              $endgroup$


















                2












                $begingroup$

                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











                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  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











                  $endgroup$



                  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 Jan 8 at 12:51

























                  answered Jan 4 at 20:22









                  ncmathsadistncmathsadist

                  42.9k260103




                  42.9k260103






























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