Integral of the product of Bessel functions of different kinds, but with same argument












1












$begingroup$


I am looking to finding the solution to indefinite integrals
begin{align}
int z J_m(a z) Y_m(a z), text{d}z,
end{align}
and
begin{align}
int z I_m(a z) K_m(a z), text{d}z
end{align}



I know that there are solutions to these when the arguments of these Bessel functions are different, such as the solution posted in Integral of Product of Bessel functions of different kind.
Those solutions, though, are singular when the arguments are the same, as is the case I am asking for here.



Are there any ways to get a solution for these integrals? We can assume that $m$ is an integer.






integration special-functions bessel-functions







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    I am looking to finding the solution to indefinite integrals
    begin{align}
    int z J_m(a z) Y_m(a z), text{d}z,
    end{align}
    and
    begin{align}
    int z I_m(a z) K_m(a z), text{d}z
    end{align}



    I know that there are solutions to these when the arguments of these Bessel functions are different, such as the solution posted in Integral of Product of Bessel functions of different kind.
    Those solutions, though, are singular when the arguments are the same, as is the case I am asking for here.



    Are there any ways to get a solution for these integrals? We can assume that $m$ is an integer.






    integration special-functions bessel-functions







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I am looking to finding the solution to indefinite integrals
      begin{align}
      int z J_m(a z) Y_m(a z), text{d}z,
      end{align}
      and
      begin{align}
      int z I_m(a z) K_m(a z), text{d}z
      end{align}



      I know that there are solutions to these when the arguments of these Bessel functions are different, such as the solution posted in Integral of Product of Bessel functions of different kind.
      Those solutions, though, are singular when the arguments are the same, as is the case I am asking for here.



      Are there any ways to get a solution for these integrals? We can assume that $m$ is an integer.






      integration special-functions bessel-functions







      share|cite|improve this question









      $endgroup$




      I am looking to finding the solution to indefinite integrals
      begin{align}
      int z J_m(a z) Y_m(a z), text{d}z,
      end{align}
      and
      begin{align}
      int z I_m(a z) K_m(a z), text{d}z
      end{align}



      I know that there are solutions to these when the arguments of these Bessel functions are different, such as the solution posted in Integral of Product of Bessel functions of different kind.
      Those solutions, though, are singular when the arguments are the same, as is the case I am asking for here.



      Are there any ways to get a solution for these integrals? We can assume that $m$ is an integer.






      integration special-functions bessel-functions




      integration special-functions bessel-functions






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 27 at 22:54









      K LK L

      436




      436






















          1 Answer
          1






          active

          oldest

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          1












          $begingroup$

          We begin with the indefinite integral



          $$begin{align}
          int z Z_m(alpha z)B_m(beta z),dz=frac{
          beta z Z_{m}(alpha z)B_{m-1}(beta z)-alpha z Z_
          {m-1}(alpha z)B_m(beta z)}{alpha^{2}-beta^{2}}tag1
          end{align}
          $$



          for $alpha ne beta$ where $Z_m$ and $B_m$ are arbitrary Bessel functions.



          While the left-hand side $(1)$ is well-behaved for $alpha=beta$, the right-hand side is not of indeterminate form. The apparent discrepancy is reconciled by remembering that $(1)$ is an indefinite integral. Addition of an appropriate constant with respect to $z$ to the right-hand side of $(1)$ results in an indeterminate form.



          Let $V$ be defined by the limit



          $$begin{align}
          V&=lim_{alphatobeta}left(beta z Z_{m}(alpha z)B_{m-1}(beta z)-alpha z Z_
          {m-1}(alpha z)B_m(beta z)right)\\
          &=beta z left(Z_{m}(beta z)B_{m-1}(beta z)- Z_{m-1}(beta z)B_m(beta z)right)tag2
          end{align}$$



          The following Wronskians hold:



          $$begin{align}
          J_{m}(beta z)Y_{m-1}(beta z)- J_{m-1}(beta z)Y_m(beta z)&=2/(pi beta z)tag{3a}\\
          H_{m}^{(1)}(beta z)H_{m-1}^{(2)}(beta z)- H_{m-1}^{(1)}(beta z)H_m^{(2)}(beta z)&=-4i/(pi beta z)tag{3b}
          end{align}$$



          Given $(3a)$ and $(3b)$, it is evident that $V$, as given on the right-hand side of $(2)$, is independent of $z$. Hence, we subtract $V$ from the numerator of $(1)$ and obtain



          $$begin{align}
          int z Z_m(alpha z)B_m(beta z),dz=frac{
          beta z Z_{m}(alpha z)B_{m-1}(beta z)-alpha z Z_
          {m-1}(alpha z)B_m(beta z)-V}{alpha^{2}-beta^{2}}tag4
          end{align}
          $$



          which is clearly of indeterminate form.



          It is left as an exercise for the reader to evaluate the limit as $alphato beta$ of $(4)$ using L'Hospital's Rule to obtain the result



          $$begin{align}
          int z Z_m(alpha z)B_m(beta z),dz&=frac12 z^2left(Z_m(beta z)B_m(beta z)-Z_{m-1}(beta z)B_{m-1}(beta z)right)\\&-frac{mz}{2beta}left(Z_m(beta z)B_{m-1}(beta z)+Z_{m-1}(beta z)B_{m}(beta z)right)
          end{align}
          $$






          share|cite|improve this answer











          $endgroup$













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            $begingroup$

            We begin with the indefinite integral



            $$begin{align}
            int z Z_m(alpha z)B_m(beta z),dz=frac{
            beta z Z_{m}(alpha z)B_{m-1}(beta z)-alpha z Z_
            {m-1}(alpha z)B_m(beta z)}{alpha^{2}-beta^{2}}tag1
            end{align}
            $$



            for $alpha ne beta$ where $Z_m$ and $B_m$ are arbitrary Bessel functions.



            While the left-hand side $(1)$ is well-behaved for $alpha=beta$, the right-hand side is not of indeterminate form. The apparent discrepancy is reconciled by remembering that $(1)$ is an indefinite integral. Addition of an appropriate constant with respect to $z$ to the right-hand side of $(1)$ results in an indeterminate form.



            Let $V$ be defined by the limit



            $$begin{align}
            V&=lim_{alphatobeta}left(beta z Z_{m}(alpha z)B_{m-1}(beta z)-alpha z Z_
            {m-1}(alpha z)B_m(beta z)right)\\
            &=beta z left(Z_{m}(beta z)B_{m-1}(beta z)- Z_{m-1}(beta z)B_m(beta z)right)tag2
            end{align}$$



            The following Wronskians hold:



            $$begin{align}
            J_{m}(beta z)Y_{m-1}(beta z)- J_{m-1}(beta z)Y_m(beta z)&=2/(pi beta z)tag{3a}\\
            H_{m}^{(1)}(beta z)H_{m-1}^{(2)}(beta z)- H_{m-1}^{(1)}(beta z)H_m^{(2)}(beta z)&=-4i/(pi beta z)tag{3b}
            end{align}$$



            Given $(3a)$ and $(3b)$, it is evident that $V$, as given on the right-hand side of $(2)$, is independent of $z$. Hence, we subtract $V$ from the numerator of $(1)$ and obtain



            $$begin{align}
            int z Z_m(alpha z)B_m(beta z),dz=frac{
            beta z Z_{m}(alpha z)B_{m-1}(beta z)-alpha z Z_
            {m-1}(alpha z)B_m(beta z)-V}{alpha^{2}-beta^{2}}tag4
            end{align}
            $$



            which is clearly of indeterminate form.



            It is left as an exercise for the reader to evaluate the limit as $alphato beta$ of $(4)$ using L'Hospital's Rule to obtain the result



            $$begin{align}
            int z Z_m(alpha z)B_m(beta z),dz&=frac12 z^2left(Z_m(beta z)B_m(beta z)-Z_{m-1}(beta z)B_{m-1}(beta z)right)\\&-frac{mz}{2beta}left(Z_m(beta z)B_{m-1}(beta z)+Z_{m-1}(beta z)B_{m}(beta z)right)
            end{align}
            $$






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              We begin with the indefinite integral



              $$begin{align}
              int z Z_m(alpha z)B_m(beta z),dz=frac{
              beta z Z_{m}(alpha z)B_{m-1}(beta z)-alpha z Z_
              {m-1}(alpha z)B_m(beta z)}{alpha^{2}-beta^{2}}tag1
              end{align}
              $$



              for $alpha ne beta$ where $Z_m$ and $B_m$ are arbitrary Bessel functions.



              While the left-hand side $(1)$ is well-behaved for $alpha=beta$, the right-hand side is not of indeterminate form. The apparent discrepancy is reconciled by remembering that $(1)$ is an indefinite integral. Addition of an appropriate constant with respect to $z$ to the right-hand side of $(1)$ results in an indeterminate form.



              Let $V$ be defined by the limit



              $$begin{align}
              V&=lim_{alphatobeta}left(beta z Z_{m}(alpha z)B_{m-1}(beta z)-alpha z Z_
              {m-1}(alpha z)B_m(beta z)right)\\
              &=beta z left(Z_{m}(beta z)B_{m-1}(beta z)- Z_{m-1}(beta z)B_m(beta z)right)tag2
              end{align}$$



              The following Wronskians hold:



              $$begin{align}
              J_{m}(beta z)Y_{m-1}(beta z)- J_{m-1}(beta z)Y_m(beta z)&=2/(pi beta z)tag{3a}\\
              H_{m}^{(1)}(beta z)H_{m-1}^{(2)}(beta z)- H_{m-1}^{(1)}(beta z)H_m^{(2)}(beta z)&=-4i/(pi beta z)tag{3b}
              end{align}$$



              Given $(3a)$ and $(3b)$, it is evident that $V$, as given on the right-hand side of $(2)$, is independent of $z$. Hence, we subtract $V$ from the numerator of $(1)$ and obtain



              $$begin{align}
              int z Z_m(alpha z)B_m(beta z),dz=frac{
              beta z Z_{m}(alpha z)B_{m-1}(beta z)-alpha z Z_
              {m-1}(alpha z)B_m(beta z)-V}{alpha^{2}-beta^{2}}tag4
              end{align}
              $$



              which is clearly of indeterminate form.



              It is left as an exercise for the reader to evaluate the limit as $alphato beta$ of $(4)$ using L'Hospital's Rule to obtain the result



              $$begin{align}
              int z Z_m(alpha z)B_m(beta z),dz&=frac12 z^2left(Z_m(beta z)B_m(beta z)-Z_{m-1}(beta z)B_{m-1}(beta z)right)\\&-frac{mz}{2beta}left(Z_m(beta z)B_{m-1}(beta z)+Z_{m-1}(beta z)B_{m}(beta z)right)
              end{align}
              $$






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                We begin with the indefinite integral



                $$begin{align}
                int z Z_m(alpha z)B_m(beta z),dz=frac{
                beta z Z_{m}(alpha z)B_{m-1}(beta z)-alpha z Z_
                {m-1}(alpha z)B_m(beta z)}{alpha^{2}-beta^{2}}tag1
                end{align}
                $$



                for $alpha ne beta$ where $Z_m$ and $B_m$ are arbitrary Bessel functions.



                While the left-hand side $(1)$ is well-behaved for $alpha=beta$, the right-hand side is not of indeterminate form. The apparent discrepancy is reconciled by remembering that $(1)$ is an indefinite integral. Addition of an appropriate constant with respect to $z$ to the right-hand side of $(1)$ results in an indeterminate form.



                Let $V$ be defined by the limit



                $$begin{align}
                V&=lim_{alphatobeta}left(beta z Z_{m}(alpha z)B_{m-1}(beta z)-alpha z Z_
                {m-1}(alpha z)B_m(beta z)right)\\
                &=beta z left(Z_{m}(beta z)B_{m-1}(beta z)- Z_{m-1}(beta z)B_m(beta z)right)tag2
                end{align}$$



                The following Wronskians hold:



                $$begin{align}
                J_{m}(beta z)Y_{m-1}(beta z)- J_{m-1}(beta z)Y_m(beta z)&=2/(pi beta z)tag{3a}\\
                H_{m}^{(1)}(beta z)H_{m-1}^{(2)}(beta z)- H_{m-1}^{(1)}(beta z)H_m^{(2)}(beta z)&=-4i/(pi beta z)tag{3b}
                end{align}$$



                Given $(3a)$ and $(3b)$, it is evident that $V$, as given on the right-hand side of $(2)$, is independent of $z$. Hence, we subtract $V$ from the numerator of $(1)$ and obtain



                $$begin{align}
                int z Z_m(alpha z)B_m(beta z),dz=frac{
                beta z Z_{m}(alpha z)B_{m-1}(beta z)-alpha z Z_
                {m-1}(alpha z)B_m(beta z)-V}{alpha^{2}-beta^{2}}tag4
                end{align}
                $$



                which is clearly of indeterminate form.



                It is left as an exercise for the reader to evaluate the limit as $alphato beta$ of $(4)$ using L'Hospital's Rule to obtain the result



                $$begin{align}
                int z Z_m(alpha z)B_m(beta z),dz&=frac12 z^2left(Z_m(beta z)B_m(beta z)-Z_{m-1}(beta z)B_{m-1}(beta z)right)\\&-frac{mz}{2beta}left(Z_m(beta z)B_{m-1}(beta z)+Z_{m-1}(beta z)B_{m}(beta z)right)
                end{align}
                $$






                share|cite|improve this answer











                $endgroup$



                We begin with the indefinite integral



                $$begin{align}
                int z Z_m(alpha z)B_m(beta z),dz=frac{
                beta z Z_{m}(alpha z)B_{m-1}(beta z)-alpha z Z_
                {m-1}(alpha z)B_m(beta z)}{alpha^{2}-beta^{2}}tag1
                end{align}
                $$



                for $alpha ne beta$ where $Z_m$ and $B_m$ are arbitrary Bessel functions.



                While the left-hand side $(1)$ is well-behaved for $alpha=beta$, the right-hand side is not of indeterminate form. The apparent discrepancy is reconciled by remembering that $(1)$ is an indefinite integral. Addition of an appropriate constant with respect to $z$ to the right-hand side of $(1)$ results in an indeterminate form.



                Let $V$ be defined by the limit



                $$begin{align}
                V&=lim_{alphatobeta}left(beta z Z_{m}(alpha z)B_{m-1}(beta z)-alpha z Z_
                {m-1}(alpha z)B_m(beta z)right)\\
                &=beta z left(Z_{m}(beta z)B_{m-1}(beta z)- Z_{m-1}(beta z)B_m(beta z)right)tag2
                end{align}$$



                The following Wronskians hold:



                $$begin{align}
                J_{m}(beta z)Y_{m-1}(beta z)- J_{m-1}(beta z)Y_m(beta z)&=2/(pi beta z)tag{3a}\\
                H_{m}^{(1)}(beta z)H_{m-1}^{(2)}(beta z)- H_{m-1}^{(1)}(beta z)H_m^{(2)}(beta z)&=-4i/(pi beta z)tag{3b}
                end{align}$$



                Given $(3a)$ and $(3b)$, it is evident that $V$, as given on the right-hand side of $(2)$, is independent of $z$. Hence, we subtract $V$ from the numerator of $(1)$ and obtain



                $$begin{align}
                int z Z_m(alpha z)B_m(beta z),dz=frac{
                beta z Z_{m}(alpha z)B_{m-1}(beta z)-alpha z Z_
                {m-1}(alpha z)B_m(beta z)-V}{alpha^{2}-beta^{2}}tag4
                end{align}
                $$



                which is clearly of indeterminate form.



                It is left as an exercise for the reader to evaluate the limit as $alphato beta$ of $(4)$ using L'Hospital's Rule to obtain the result



                $$begin{align}
                int z Z_m(alpha z)B_m(beta z),dz&=frac12 z^2left(Z_m(beta z)B_m(beta z)-Z_{m-1}(beta z)B_{m-1}(beta z)right)\\&-frac{mz}{2beta}left(Z_m(beta z)B_{m-1}(beta z)+Z_{m-1}(beta z)B_{m}(beta z)right)
                end{align}
                $$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 28 at 2:27

























                answered Jan 28 at 0:36









                Mark ViolaMark Viola

                134k1278176




                134k1278176






























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