Does $int_{0}^{infty}frac{r^2}{1+alpha r^4}dr$ converge?












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I want to check that $int_{mathbb R^3}frac{1}{1+x^4+y^4+z^4}dxdydz$ converges.



I moved to spherical coordinates, the integral became $int_{0}^{infty}int_{0}^{pi}int_{0}^{2pi}frac{r^2sin(theta)}{1+r^4f(theta,phi)}dphi dtheta dr leqint_{0}^{infty}int_{0}^{pi}int_{0}^{2pi}frac{r^2sin(theta)}{1+alpha r^4}dphi dtheta dr = 4pi int_{0}^{infty}frac{r^2}{1+alpha r^4}dr $



I don't know how to show this converges.



Edit: $f(theta,phi)$ is the usual spherical coordinates transform, but notice that it's bounded from below by a positive number, let that number be $alpha > 0$.






calculus integration improper-integrals







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    $begingroup$


    I want to check that $int_{mathbb R^3}frac{1}{1+x^4+y^4+z^4}dxdydz$ converges.



    I moved to spherical coordinates, the integral became $int_{0}^{infty}int_{0}^{pi}int_{0}^{2pi}frac{r^2sin(theta)}{1+r^4f(theta,phi)}dphi dtheta dr leqint_{0}^{infty}int_{0}^{pi}int_{0}^{2pi}frac{r^2sin(theta)}{1+alpha r^4}dphi dtheta dr = 4pi int_{0}^{infty}frac{r^2}{1+alpha r^4}dr $



    I don't know how to show this converges.



    Edit: $f(theta,phi)$ is the usual spherical coordinates transform, but notice that it's bounded from below by a positive number, let that number be $alpha > 0$.






    calculus integration improper-integrals







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      1












      1








      1





      $begingroup$


      I want to check that $int_{mathbb R^3}frac{1}{1+x^4+y^4+z^4}dxdydz$ converges.



      I moved to spherical coordinates, the integral became $int_{0}^{infty}int_{0}^{pi}int_{0}^{2pi}frac{r^2sin(theta)}{1+r^4f(theta,phi)}dphi dtheta dr leqint_{0}^{infty}int_{0}^{pi}int_{0}^{2pi}frac{r^2sin(theta)}{1+alpha r^4}dphi dtheta dr = 4pi int_{0}^{infty}frac{r^2}{1+alpha r^4}dr $



      I don't know how to show this converges.



      Edit: $f(theta,phi)$ is the usual spherical coordinates transform, but notice that it's bounded from below by a positive number, let that number be $alpha > 0$.






      calculus integration improper-integrals







      share|cite|improve this question









      $endgroup$




      I want to check that $int_{mathbb R^3}frac{1}{1+x^4+y^4+z^4}dxdydz$ converges.



      I moved to spherical coordinates, the integral became $int_{0}^{infty}int_{0}^{pi}int_{0}^{2pi}frac{r^2sin(theta)}{1+r^4f(theta,phi)}dphi dtheta dr leqint_{0}^{infty}int_{0}^{pi}int_{0}^{2pi}frac{r^2sin(theta)}{1+alpha r^4}dphi dtheta dr = 4pi int_{0}^{infty}frac{r^2}{1+alpha r^4}dr $



      I don't know how to show this converges.



      Edit: $f(theta,phi)$ is the usual spherical coordinates transform, but notice that it's bounded from below by a positive number, let that number be $alpha > 0$.






      calculus integration improper-integrals




      calculus integration improper-integrals






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      asked Jan 20 at 21:00









      Rick JokerRick Joker

      399111




      399111






















          3 Answers
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          $begingroup$

          Hint: Split the integral
          begin{align}
          int^infty_0 frac{r^2}{1+alpha r^4} dr = int^infty_1frac{r^2}{1+alpha r^4} dr +int^1_0 frac{r^2}{1+alpha r^4} dr
          end{align}






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          • $begingroup$
            And $int_{1}^{infty} frac{r^2}{1 + alpha r^4} drleq int_{1}^{infty}frac{r^2}{alpha r^4}dr$. Literally just thought of that trick too.
            $endgroup$
            – Rick Joker
            Jan 20 at 21:05












          • $begingroup$
            Thanks anyways Jacky.
            $endgroup$
            – Rick Joker
            Jan 20 at 21:06










          • $begingroup$
            @RickJoker Yes.
            $endgroup$
            – Jacky Chong
            Jan 20 at 21:06



















          2












          $begingroup$

          At 0, this is continuous. At $infty$ it behaves like $x mapsto 1/x^2$, which integrates finitely there. The integral converges.






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            There is a problem only at $infty$. You can user equivalence:



            $$frac{r^2}{1+alpha r^4}sim_inftyfrac 1{alpha r^2},$$
            and $;displaystyleint_1^inftyfrac 1{alpha r^2},mathrm d r;$ converges.






            share|cite|improve this answer









            $endgroup$













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              3 Answers
              3






              active

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              3 Answers
              3






              active

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              active

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              active

              oldest

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              1












              $begingroup$

              Hint: Split the integral
              begin{align}
              int^infty_0 frac{r^2}{1+alpha r^4} dr = int^infty_1frac{r^2}{1+alpha r^4} dr +int^1_0 frac{r^2}{1+alpha r^4} dr
              end{align}






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                And $int_{1}^{infty} frac{r^2}{1 + alpha r^4} drleq int_{1}^{infty}frac{r^2}{alpha r^4}dr$. Literally just thought of that trick too.
                $endgroup$
                – Rick Joker
                Jan 20 at 21:05












              • $begingroup$
                Thanks anyways Jacky.
                $endgroup$
                – Rick Joker
                Jan 20 at 21:06










              • $begingroup$
                @RickJoker Yes.
                $endgroup$
                – Jacky Chong
                Jan 20 at 21:06
















              1












              $begingroup$

              Hint: Split the integral
              begin{align}
              int^infty_0 frac{r^2}{1+alpha r^4} dr = int^infty_1frac{r^2}{1+alpha r^4} dr +int^1_0 frac{r^2}{1+alpha r^4} dr
              end{align}






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                And $int_{1}^{infty} frac{r^2}{1 + alpha r^4} drleq int_{1}^{infty}frac{r^2}{alpha r^4}dr$. Literally just thought of that trick too.
                $endgroup$
                – Rick Joker
                Jan 20 at 21:05












              • $begingroup$
                Thanks anyways Jacky.
                $endgroup$
                – Rick Joker
                Jan 20 at 21:06










              • $begingroup$
                @RickJoker Yes.
                $endgroup$
                – Jacky Chong
                Jan 20 at 21:06














              1












              1








              1





              $begingroup$

              Hint: Split the integral
              begin{align}
              int^infty_0 frac{r^2}{1+alpha r^4} dr = int^infty_1frac{r^2}{1+alpha r^4} dr +int^1_0 frac{r^2}{1+alpha r^4} dr
              end{align}






              share|cite|improve this answer









              $endgroup$



              Hint: Split the integral
              begin{align}
              int^infty_0 frac{r^2}{1+alpha r^4} dr = int^infty_1frac{r^2}{1+alpha r^4} dr +int^1_0 frac{r^2}{1+alpha r^4} dr
              end{align}







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Jan 20 at 21:04









              Jacky ChongJacky Chong

              19.1k21129




              19.1k21129












              • $begingroup$
                And $int_{1}^{infty} frac{r^2}{1 + alpha r^4} drleq int_{1}^{infty}frac{r^2}{alpha r^4}dr$. Literally just thought of that trick too.
                $endgroup$
                – Rick Joker
                Jan 20 at 21:05












              • $begingroup$
                Thanks anyways Jacky.
                $endgroup$
                – Rick Joker
                Jan 20 at 21:06










              • $begingroup$
                @RickJoker Yes.
                $endgroup$
                – Jacky Chong
                Jan 20 at 21:06


















              • $begingroup$
                And $int_{1}^{infty} frac{r^2}{1 + alpha r^4} drleq int_{1}^{infty}frac{r^2}{alpha r^4}dr$. Literally just thought of that trick too.
                $endgroup$
                – Rick Joker
                Jan 20 at 21:05












              • $begingroup$
                Thanks anyways Jacky.
                $endgroup$
                – Rick Joker
                Jan 20 at 21:06










              • $begingroup$
                @RickJoker Yes.
                $endgroup$
                – Jacky Chong
                Jan 20 at 21:06
















              $begingroup$
              And $int_{1}^{infty} frac{r^2}{1 + alpha r^4} drleq int_{1}^{infty}frac{r^2}{alpha r^4}dr$. Literally just thought of that trick too.
              $endgroup$
              – Rick Joker
              Jan 20 at 21:05






              $begingroup$
              And $int_{1}^{infty} frac{r^2}{1 + alpha r^4} drleq int_{1}^{infty}frac{r^2}{alpha r^4}dr$. Literally just thought of that trick too.
              $endgroup$
              – Rick Joker
              Jan 20 at 21:05














              $begingroup$
              Thanks anyways Jacky.
              $endgroup$
              – Rick Joker
              Jan 20 at 21:06




              $begingroup$
              Thanks anyways Jacky.
              $endgroup$
              – Rick Joker
              Jan 20 at 21:06












              $begingroup$
              @RickJoker Yes.
              $endgroup$
              – Jacky Chong
              Jan 20 at 21:06




              $begingroup$
              @RickJoker Yes.
              $endgroup$
              – Jacky Chong
              Jan 20 at 21:06











              2












              $begingroup$

              At 0, this is continuous. At $infty$ it behaves like $x mapsto 1/x^2$, which integrates finitely there. The integral converges.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                At 0, this is continuous. At $infty$ it behaves like $x mapsto 1/x^2$, which integrates finitely there. The integral converges.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  At 0, this is continuous. At $infty$ it behaves like $x mapsto 1/x^2$, which integrates finitely there. The integral converges.






                  share|cite|improve this answer









                  $endgroup$



                  At 0, this is continuous. At $infty$ it behaves like $x mapsto 1/x^2$, which integrates finitely there. The integral converges.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 20 at 21:10









                  ncmathsadistncmathsadist

                  42.9k260103




                  42.9k260103























                      1












                      $begingroup$

                      There is a problem only at $infty$. You can user equivalence:



                      $$frac{r^2}{1+alpha r^4}sim_inftyfrac 1{alpha r^2},$$
                      and $;displaystyleint_1^inftyfrac 1{alpha r^2},mathrm d r;$ converges.






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        There is a problem only at $infty$. You can user equivalence:



                        $$frac{r^2}{1+alpha r^4}sim_inftyfrac 1{alpha r^2},$$
                        and $;displaystyleint_1^inftyfrac 1{alpha r^2},mathrm d r;$ converges.






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          There is a problem only at $infty$. You can user equivalence:



                          $$frac{r^2}{1+alpha r^4}sim_inftyfrac 1{alpha r^2},$$
                          and $;displaystyleint_1^inftyfrac 1{alpha r^2},mathrm d r;$ converges.






                          share|cite|improve this answer









                          $endgroup$



                          There is a problem only at $infty$. You can user equivalence:



                          $$frac{r^2}{1+alpha r^4}sim_inftyfrac 1{alpha r^2},$$
                          and $;displaystyleint_1^inftyfrac 1{alpha r^2},mathrm d r;$ converges.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Jan 20 at 21:10









                          BernardBernard

                          122k740116




                          122k740116






























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