Help:For what $ alpha $ do the integrals exist?












2














Let
$ K_R (y) subset mathbb{R}^n $
be an n-dimensional Ball with radius $R>0$ and midpoint $y in mathbb{R}^n $



Far what $ alpha in mathbb{R} $ and $ n in mathbb{N}{0} $ do following integrals exist:



$ int_{K{_R(y)}} |x-y|^{alpha}dx $



$ int_{ mathbb{R}^n setminus K{_R(y)}} |x-y|^{alpha}dx $



$ int_{ mathbb{R}^n } |x-y|^{alpha}dx $



with what criteria can i solve this task?
I don't know how to begin.
Thanks for any help!!



Let's take
$ int_{K{_R(y)}} |x-y|^{alpha}dx $ ,$y=0, n=2 $



If I replace by polarcoordinates $ x= rsin phi $ :
$ int_0^{2 pi} int_0^R | r|^{alpha} r dr dphi $..is this the right way?






real-analysis integration multivariable-calculus







share|cite|improve this question
























  • Take $y=0$ and $n=2$ to start. You can use polar coordinates.
    – zhw.
    Nov 28 '18 at 18:18
















2














Let
$ K_R (y) subset mathbb{R}^n $
be an n-dimensional Ball with radius $R>0$ and midpoint $y in mathbb{R}^n $



Far what $ alpha in mathbb{R} $ and $ n in mathbb{N}{0} $ do following integrals exist:



$ int_{K{_R(y)}} |x-y|^{alpha}dx $



$ int_{ mathbb{R}^n setminus K{_R(y)}} |x-y|^{alpha}dx $



$ int_{ mathbb{R}^n } |x-y|^{alpha}dx $



with what criteria can i solve this task?
I don't know how to begin.
Thanks for any help!!



Let's take
$ int_{K{_R(y)}} |x-y|^{alpha}dx $ ,$y=0, n=2 $



If I replace by polarcoordinates $ x= rsin phi $ :
$ int_0^{2 pi} int_0^R | r|^{alpha} r dr dphi $..is this the right way?






real-analysis integration multivariable-calculus







share|cite|improve this question
























  • Take $y=0$ and $n=2$ to start. You can use polar coordinates.
    – zhw.
    Nov 28 '18 at 18:18














2












2








2







Let
$ K_R (y) subset mathbb{R}^n $
be an n-dimensional Ball with radius $R>0$ and midpoint $y in mathbb{R}^n $



Far what $ alpha in mathbb{R} $ and $ n in mathbb{N}{0} $ do following integrals exist:



$ int_{K{_R(y)}} |x-y|^{alpha}dx $



$ int_{ mathbb{R}^n setminus K{_R(y)}} |x-y|^{alpha}dx $



$ int_{ mathbb{R}^n } |x-y|^{alpha}dx $



with what criteria can i solve this task?
I don't know how to begin.
Thanks for any help!!



Let's take
$ int_{K{_R(y)}} |x-y|^{alpha}dx $ ,$y=0, n=2 $



If I replace by polarcoordinates $ x= rsin phi $ :
$ int_0^{2 pi} int_0^R | r|^{alpha} r dr dphi $..is this the right way?






real-analysis integration multivariable-calculus







share|cite|improve this question















Let
$ K_R (y) subset mathbb{R}^n $
be an n-dimensional Ball with radius $R>0$ and midpoint $y in mathbb{R}^n $



Far what $ alpha in mathbb{R} $ and $ n in mathbb{N}{0} $ do following integrals exist:



$ int_{K{_R(y)}} |x-y|^{alpha}dx $



$ int_{ mathbb{R}^n setminus K{_R(y)}} |x-y|^{alpha}dx $



$ int_{ mathbb{R}^n } |x-y|^{alpha}dx $



with what criteria can i solve this task?
I don't know how to begin.
Thanks for any help!!



Let's take
$ int_{K{_R(y)}} |x-y|^{alpha}dx $ ,$y=0, n=2 $



If I replace by polarcoordinates $ x= rsin phi $ :
$ int_0^{2 pi} int_0^R | r|^{alpha} r dr dphi $..is this the right way?






real-analysis integration multivariable-calculus




real-analysis integration multivariable-calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited Nov 28 '18 at 21:44







constant94

















asked Nov 28 '18 at 18:16









constant94constant94

1018




1018












  • Take $y=0$ and $n=2$ to start. You can use polar coordinates.
    – zhw.
    Nov 28 '18 at 18:18


















  • Take $y=0$ and $n=2$ to start. You can use polar coordinates.
    – zhw.
    Nov 28 '18 at 18:18
















Take $y=0$ and $n=2$ to start. You can use polar coordinates.
– zhw.
Nov 28 '18 at 18:18




Take $y=0$ and $n=2$ to start. You can use polar coordinates.
– zhw.
Nov 28 '18 at 18:18










1 Answer
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Let $nomega_n$ denote the $n-1$ dimensional surface area of the $n-1$-sphere.



Your integrals, respectively, are
$$int_{K_R(y)} |x-y|^alpha , dx= int_0^R int_{|x-y| = r} |x-y|^alpha , dx , dr = n omega_n int_0^R |x-y|^{n+alpha} , dr$$
$$int_{mathbb R^n setminus K_R(y)} |x-y|^alpha , dx= int_R^infty int_{|x-y| = r} |x-y|^alpha , dx , dr = n omega_n int_R^infty |x-y|^{n+alpha} , dr$$
$$int_{mathbb R^n} |x-y|^alpha , dx= int_0^infty int_{|x-y| = r} |x-y|^alpha , dx , dr = n omega_n int_0^infty|x-y|^{n+alpha} , dr$$
where $dx$ is the usual $n$-dimensional volume element in the integrals on the left and the $n-1$ dimensional surface area element in the integrals in the middle.



The remaining improper integrals are straight forward to work out.






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    1 Answer
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    1 Answer
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    active

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    active

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    1














    Let $nomega_n$ denote the $n-1$ dimensional surface area of the $n-1$-sphere.



    Your integrals, respectively, are
    $$int_{K_R(y)} |x-y|^alpha , dx= int_0^R int_{|x-y| = r} |x-y|^alpha , dx , dr = n omega_n int_0^R |x-y|^{n+alpha} , dr$$
    $$int_{mathbb R^n setminus K_R(y)} |x-y|^alpha , dx= int_R^infty int_{|x-y| = r} |x-y|^alpha , dx , dr = n omega_n int_R^infty |x-y|^{n+alpha} , dr$$
    $$int_{mathbb R^n} |x-y|^alpha , dx= int_0^infty int_{|x-y| = r} |x-y|^alpha , dx , dr = n omega_n int_0^infty|x-y|^{n+alpha} , dr$$
    where $dx$ is the usual $n$-dimensional volume element in the integrals on the left and the $n-1$ dimensional surface area element in the integrals in the middle.



    The remaining improper integrals are straight forward to work out.






    share|cite|improve this answer


























      1














      Let $nomega_n$ denote the $n-1$ dimensional surface area of the $n-1$-sphere.



      Your integrals, respectively, are
      $$int_{K_R(y)} |x-y|^alpha , dx= int_0^R int_{|x-y| = r} |x-y|^alpha , dx , dr = n omega_n int_0^R |x-y|^{n+alpha} , dr$$
      $$int_{mathbb R^n setminus K_R(y)} |x-y|^alpha , dx= int_R^infty int_{|x-y| = r} |x-y|^alpha , dx , dr = n omega_n int_R^infty |x-y|^{n+alpha} , dr$$
      $$int_{mathbb R^n} |x-y|^alpha , dx= int_0^infty int_{|x-y| = r} |x-y|^alpha , dx , dr = n omega_n int_0^infty|x-y|^{n+alpha} , dr$$
      where $dx$ is the usual $n$-dimensional volume element in the integrals on the left and the $n-1$ dimensional surface area element in the integrals in the middle.



      The remaining improper integrals are straight forward to work out.






      share|cite|improve this answer
























        1












        1








        1






        Let $nomega_n$ denote the $n-1$ dimensional surface area of the $n-1$-sphere.



        Your integrals, respectively, are
        $$int_{K_R(y)} |x-y|^alpha , dx= int_0^R int_{|x-y| = r} |x-y|^alpha , dx , dr = n omega_n int_0^R |x-y|^{n+alpha} , dr$$
        $$int_{mathbb R^n setminus K_R(y)} |x-y|^alpha , dx= int_R^infty int_{|x-y| = r} |x-y|^alpha , dx , dr = n omega_n int_R^infty |x-y|^{n+alpha} , dr$$
        $$int_{mathbb R^n} |x-y|^alpha , dx= int_0^infty int_{|x-y| = r} |x-y|^alpha , dx , dr = n omega_n int_0^infty|x-y|^{n+alpha} , dr$$
        where $dx$ is the usual $n$-dimensional volume element in the integrals on the left and the $n-1$ dimensional surface area element in the integrals in the middle.



        The remaining improper integrals are straight forward to work out.






        share|cite|improve this answer












        Let $nomega_n$ denote the $n-1$ dimensional surface area of the $n-1$-sphere.



        Your integrals, respectively, are
        $$int_{K_R(y)} |x-y|^alpha , dx= int_0^R int_{|x-y| = r} |x-y|^alpha , dx , dr = n omega_n int_0^R |x-y|^{n+alpha} , dr$$
        $$int_{mathbb R^n setminus K_R(y)} |x-y|^alpha , dx= int_R^infty int_{|x-y| = r} |x-y|^alpha , dx , dr = n omega_n int_R^infty |x-y|^{n+alpha} , dr$$
        $$int_{mathbb R^n} |x-y|^alpha , dx= int_0^infty int_{|x-y| = r} |x-y|^alpha , dx , dr = n omega_n int_0^infty|x-y|^{n+alpha} , dr$$
        where $dx$ is the usual $n$-dimensional volume element in the integrals on the left and the $n-1$ dimensional surface area element in the integrals in the middle.



        The remaining improper integrals are straight forward to work out.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 28 '18 at 21:52









        Umberto P.Umberto P.

        38.5k13064




        38.5k13064






























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