Computing the Lebesgue integral over a ball












0












$begingroup$


Let $alpha in mathbb{R}$ and $lambda_n$ the Lebesgue measure on $mathbb{R^n}$.



Define $f:mathbb{R^n}backslash(0)tomathbb{R}, f(x)=leftlVert xrightrVert^alpha$



For which $alpha$ is $f$ Lebesgue integrable on $B_1(0):=lbrace x in mathbb{R^n}:leftlVert xrightrVert leq 1 rbrace$



and for which $alpha$ on $mathbb{R^n}backslash B_1(0)$?



Also, how to compute $int_{B_1(0)}f dlambda_n$ and $int_{mathbb{R^n}backslash B_1(0)}f dlambda_n$ if $f$ is integrable?



I tried:



$f$ is Lebesgue integrable on $B_1(0) Leftrightarrowint_{B_1(0)}|f| dlambda_n<infty$.



So if $alpha<infty$, then $f(x)$ is Lebesgue integrable.



For computing the integrals, I thought about using the transformation formula:



$int_{B_1(0)}leftlVert xrightrVert^alpha dlambda_n=alpha cdot int_{0}^{1}r^n d lambda_r$



Here I don't know how to continue. Is this method correct or is there another way to do this?






measure-theory lebesgue-integral lebesgue-measure







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    Let $alpha in mathbb{R}$ and $lambda_n$ the Lebesgue measure on $mathbb{R^n}$.



    Define $f:mathbb{R^n}backslash(0)tomathbb{R}, f(x)=leftlVert xrightrVert^alpha$



    For which $alpha$ is $f$ Lebesgue integrable on $B_1(0):=lbrace x in mathbb{R^n}:leftlVert xrightrVert leq 1 rbrace$



    and for which $alpha$ on $mathbb{R^n}backslash B_1(0)$?



    Also, how to compute $int_{B_1(0)}f dlambda_n$ and $int_{mathbb{R^n}backslash B_1(0)}f dlambda_n$ if $f$ is integrable?



    I tried:



    $f$ is Lebesgue integrable on $B_1(0) Leftrightarrowint_{B_1(0)}|f| dlambda_n<infty$.



    So if $alpha<infty$, then $f(x)$ is Lebesgue integrable.



    For computing the integrals, I thought about using the transformation formula:



    $int_{B_1(0)}leftlVert xrightrVert^alpha dlambda_n=alpha cdot int_{0}^{1}r^n d lambda_r$



    Here I don't know how to continue. Is this method correct or is there another way to do this?






    measure-theory lebesgue-integral lebesgue-measure







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      Let $alpha in mathbb{R}$ and $lambda_n$ the Lebesgue measure on $mathbb{R^n}$.



      Define $f:mathbb{R^n}backslash(0)tomathbb{R}, f(x)=leftlVert xrightrVert^alpha$



      For which $alpha$ is $f$ Lebesgue integrable on $B_1(0):=lbrace x in mathbb{R^n}:leftlVert xrightrVert leq 1 rbrace$



      and for which $alpha$ on $mathbb{R^n}backslash B_1(0)$?



      Also, how to compute $int_{B_1(0)}f dlambda_n$ and $int_{mathbb{R^n}backslash B_1(0)}f dlambda_n$ if $f$ is integrable?



      I tried:



      $f$ is Lebesgue integrable on $B_1(0) Leftrightarrowint_{B_1(0)}|f| dlambda_n<infty$.



      So if $alpha<infty$, then $f(x)$ is Lebesgue integrable.



      For computing the integrals, I thought about using the transformation formula:



      $int_{B_1(0)}leftlVert xrightrVert^alpha dlambda_n=alpha cdot int_{0}^{1}r^n d lambda_r$



      Here I don't know how to continue. Is this method correct or is there another way to do this?






      measure-theory lebesgue-integral lebesgue-measure







      share|cite|improve this question











      $endgroup$




      Let $alpha in mathbb{R}$ and $lambda_n$ the Lebesgue measure on $mathbb{R^n}$.



      Define $f:mathbb{R^n}backslash(0)tomathbb{R}, f(x)=leftlVert xrightrVert^alpha$



      For which $alpha$ is $f$ Lebesgue integrable on $B_1(0):=lbrace x in mathbb{R^n}:leftlVert xrightrVert leq 1 rbrace$



      and for which $alpha$ on $mathbb{R^n}backslash B_1(0)$?



      Also, how to compute $int_{B_1(0)}f dlambda_n$ and $int_{mathbb{R^n}backslash B_1(0)}f dlambda_n$ if $f$ is integrable?



      I tried:



      $f$ is Lebesgue integrable on $B_1(0) Leftrightarrowint_{B_1(0)}|f| dlambda_n<infty$.



      So if $alpha<infty$, then $f(x)$ is Lebesgue integrable.



      For computing the integrals, I thought about using the transformation formula:



      $int_{B_1(0)}leftlVert xrightrVert^alpha dlambda_n=alpha cdot int_{0}^{1}r^n d lambda_r$



      Here I don't know how to continue. Is this method correct or is there another way to do this?






      measure-theory lebesgue-integral lebesgue-measure




      measure-theory lebesgue-integral lebesgue-measure






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 1 '18 at 16:52







      Olsgur

















      asked Dec 1 '18 at 13:48









      OlsgurOlsgur

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      564






















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          $begingroup$

          Consider spherical shells $S_{a,b}:={xin{mathbb R}^n,|,aleq|x|leq b}$. One then has
          $$int_{S_{a,b}}|x|^alpha {rm d}(x)=omega_{n-1}int_a^b r^alpha>r^{n+1}>dr ,$$
          whereby $omega_{n-1}$ denotes the $(n-1)$-dimensional surface area of $S^{n-1}subset{mathbb R}^n$. Now see what happens when $ato0+$ or $btoinfty$ for various values of $n$ and $alpha$.






          share|cite|improve this answer









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            $begingroup$

            Consider spherical shells $S_{a,b}:={xin{mathbb R}^n,|,aleq|x|leq b}$. One then has
            $$int_{S_{a,b}}|x|^alpha {rm d}(x)=omega_{n-1}int_a^b r^alpha>r^{n+1}>dr ,$$
            whereby $omega_{n-1}$ denotes the $(n-1)$-dimensional surface area of $S^{n-1}subset{mathbb R}^n$. Now see what happens when $ato0+$ or $btoinfty$ for various values of $n$ and $alpha$.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              Consider spherical shells $S_{a,b}:={xin{mathbb R}^n,|,aleq|x|leq b}$. One then has
              $$int_{S_{a,b}}|x|^alpha {rm d}(x)=omega_{n-1}int_a^b r^alpha>r^{n+1}>dr ,$$
              whereby $omega_{n-1}$ denotes the $(n-1)$-dimensional surface area of $S^{n-1}subset{mathbb R}^n$. Now see what happens when $ato0+$ or $btoinfty$ for various values of $n$ and $alpha$.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                Consider spherical shells $S_{a,b}:={xin{mathbb R}^n,|,aleq|x|leq b}$. One then has
                $$int_{S_{a,b}}|x|^alpha {rm d}(x)=omega_{n-1}int_a^b r^alpha>r^{n+1}>dr ,$$
                whereby $omega_{n-1}$ denotes the $(n-1)$-dimensional surface area of $S^{n-1}subset{mathbb R}^n$. Now see what happens when $ato0+$ or $btoinfty$ for various values of $n$ and $alpha$.






                share|cite|improve this answer









                $endgroup$



                Consider spherical shells $S_{a,b}:={xin{mathbb R}^n,|,aleq|x|leq b}$. One then has
                $$int_{S_{a,b}}|x|^alpha {rm d}(x)=omega_{n-1}int_a^b r^alpha>r^{n+1}>dr ,$$
                whereby $omega_{n-1}$ denotes the $(n-1)$-dimensional surface area of $S^{n-1}subset{mathbb R}^n$. Now see what happens when $ato0+$ or $btoinfty$ for various values of $n$ and $alpha$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 1 '18 at 17:03









                Christian BlatterChristian Blatter

                173k7113326




                173k7113326






























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