About the integrabity of $f(x) = |x|^{-(n+1)}$ and $f(x) = |x|^{-n}$ on $mathbb{R}^n$












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The function $f(x) = |x|^{-(n+1)}$ on $mathbb{R}^n$ is obviously unbounded as we approach $0$. Thus, in order to prove that the function is not integrable on $mathbb{R}^n$, I use the following evaluation



$$int_{mathbb{R}^n} f(x) d mu (x) ge int_{|x| < epsilon} f(x) d mu (x) ge int_{|x| < epsilon} epsilon ^{-(n+1)} d mu (x) = cepsilon ^n timesepsilon ^{-(n+1)} = dfrac{c}{epsilon} $$
and let $epsilon to 0$



Please help me to verify my argument.



Furthermore, can the power $-(n+1)$ be replaced by $-n$? I think the answer is negative, but I can't prove it.










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    0












    $begingroup$


    The function $f(x) = |x|^{-(n+1)}$ on $mathbb{R}^n$ is obviously unbounded as we approach $0$. Thus, in order to prove that the function is not integrable on $mathbb{R}^n$, I use the following evaluation



    $$int_{mathbb{R}^n} f(x) d mu (x) ge int_{|x| < epsilon} f(x) d mu (x) ge int_{|x| < epsilon} epsilon ^{-(n+1)} d mu (x) = cepsilon ^n timesepsilon ^{-(n+1)} = dfrac{c}{epsilon} $$
    and let $epsilon to 0$



    Please help me to verify my argument.



    Furthermore, can the power $-(n+1)$ be replaced by $-n$? I think the answer is negative, but I can't prove it.










    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      The function $f(x) = |x|^{-(n+1)}$ on $mathbb{R}^n$ is obviously unbounded as we approach $0$. Thus, in order to prove that the function is not integrable on $mathbb{R}^n$, I use the following evaluation



      $$int_{mathbb{R}^n} f(x) d mu (x) ge int_{|x| < epsilon} f(x) d mu (x) ge int_{|x| < epsilon} epsilon ^{-(n+1)} d mu (x) = cepsilon ^n timesepsilon ^{-(n+1)} = dfrac{c}{epsilon} $$
      and let $epsilon to 0$



      Please help me to verify my argument.



      Furthermore, can the power $-(n+1)$ be replaced by $-n$? I think the answer is negative, but I can't prove it.










      share|cite|improve this question









      $endgroup$




      The function $f(x) = |x|^{-(n+1)}$ on $mathbb{R}^n$ is obviously unbounded as we approach $0$. Thus, in order to prove that the function is not integrable on $mathbb{R}^n$, I use the following evaluation



      $$int_{mathbb{R}^n} f(x) d mu (x) ge int_{|x| < epsilon} f(x) d mu (x) ge int_{|x| < epsilon} epsilon ^{-(n+1)} d mu (x) = cepsilon ^n timesepsilon ^{-(n+1)} = dfrac{c}{epsilon} $$
      and let $epsilon to 0$



      Please help me to verify my argument.



      Furthermore, can the power $-(n+1)$ be replaced by $-n$? I think the answer is negative, but I can't prove it.







      measure-theory proof-verification






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      asked Jan 14 at 9:14









      ElementXElementX

      404111




      404111






















          2 Answers
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          $begingroup$

          Your argument is correct. To prove that the result is true with $n+1$ replaced by $n$ note that $int frac 1 {|x|^{n}} geq sum_k int_{R_k leq |x| leq R_{k+1}} geq sum_k frac {cR_{k+1}^{n}-cR_k^{n}} {R_{k+1}^{n}}$ for any increasing sequence ${R_k}$. Choose these numbers so that $frac {R_{k+1}} {R_k} to infty$ to see that the general term of the series here does not tend to $0$ and hence the series is divergent.






          share|cite|improve this answer











          $endgroup$





















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            $begingroup$

            I suggest to use polar coordinate and changes everything into 1-dim. Note that $$
            int_0^{infty}dfrac{1}{r^k}mathtt{d} r=infty,quadforall kinmathbb{Z}
            $$

            When $k=1$, it becomes log-type and the integral also blows up.



            So the result follows immediately. ;)






            share|cite|improve this answer









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              2 Answers
              2






              active

              oldest

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              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              1












              $begingroup$

              Your argument is correct. To prove that the result is true with $n+1$ replaced by $n$ note that $int frac 1 {|x|^{n}} geq sum_k int_{R_k leq |x| leq R_{k+1}} geq sum_k frac {cR_{k+1}^{n}-cR_k^{n}} {R_{k+1}^{n}}$ for any increasing sequence ${R_k}$. Choose these numbers so that $frac {R_{k+1}} {R_k} to infty$ to see that the general term of the series here does not tend to $0$ and hence the series is divergent.






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                Your argument is correct. To prove that the result is true with $n+1$ replaced by $n$ note that $int frac 1 {|x|^{n}} geq sum_k int_{R_k leq |x| leq R_{k+1}} geq sum_k frac {cR_{k+1}^{n}-cR_k^{n}} {R_{k+1}^{n}}$ for any increasing sequence ${R_k}$. Choose these numbers so that $frac {R_{k+1}} {R_k} to infty$ to see that the general term of the series here does not tend to $0$ and hence the series is divergent.






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Your argument is correct. To prove that the result is true with $n+1$ replaced by $n$ note that $int frac 1 {|x|^{n}} geq sum_k int_{R_k leq |x| leq R_{k+1}} geq sum_k frac {cR_{k+1}^{n}-cR_k^{n}} {R_{k+1}^{n}}$ for any increasing sequence ${R_k}$. Choose these numbers so that $frac {R_{k+1}} {R_k} to infty$ to see that the general term of the series here does not tend to $0$ and hence the series is divergent.






                  share|cite|improve this answer











                  $endgroup$



                  Your argument is correct. To prove that the result is true with $n+1$ replaced by $n$ note that $int frac 1 {|x|^{n}} geq sum_k int_{R_k leq |x| leq R_{k+1}} geq sum_k frac {cR_{k+1}^{n}-cR_k^{n}} {R_{k+1}^{n}}$ for any increasing sequence ${R_k}$. Choose these numbers so that $frac {R_{k+1}} {R_k} to infty$ to see that the general term of the series here does not tend to $0$ and hence the series is divergent.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 14 at 9:32

























                  answered Jan 14 at 9:22









                  Kavi Rama MurthyKavi Rama Murthy

                  69.1k53169




                  69.1k53169























                      0












                      $begingroup$

                      I suggest to use polar coordinate and changes everything into 1-dim. Note that $$
                      int_0^{infty}dfrac{1}{r^k}mathtt{d} r=infty,quadforall kinmathbb{Z}
                      $$

                      When $k=1$, it becomes log-type and the integral also blows up.



                      So the result follows immediately. ;)






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        I suggest to use polar coordinate and changes everything into 1-dim. Note that $$
                        int_0^{infty}dfrac{1}{r^k}mathtt{d} r=infty,quadforall kinmathbb{Z}
                        $$

                        When $k=1$, it becomes log-type and the integral also blows up.



                        So the result follows immediately. ;)






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          I suggest to use polar coordinate and changes everything into 1-dim. Note that $$
                          int_0^{infty}dfrac{1}{r^k}mathtt{d} r=infty,quadforall kinmathbb{Z}
                          $$

                          When $k=1$, it becomes log-type and the integral also blows up.



                          So the result follows immediately. ;)






                          share|cite|improve this answer









                          $endgroup$



                          I suggest to use polar coordinate and changes everything into 1-dim. Note that $$
                          int_0^{infty}dfrac{1}{r^k}mathtt{d} r=infty,quadforall kinmathbb{Z}
                          $$

                          When $k=1$, it becomes log-type and the integral also blows up.



                          So the result follows immediately. ;)







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 14 at 9:33









                          Zixiao_LiuZixiao_Liu

                          938




                          938






























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