Integrating multiple products of order 1 Legendre functions












0












$begingroup$


I would like a closed form expression for the integrals of some products like the following:



$$
int_{-1}^1 P^1_j P^1_k P^1_l P^1_m dx
$$



and



$$
int_{-1}^1 P^0_j P^0_k P^1_l P^1_m dx,
$$



where $P^0_j(x)$ and $P^1_j(x)$ are Legendre polynomials and order 1 associated Legendre functions, respectively. From an answer to another question on this site (Legendre Polynomials Triple Product), I know of the following relationship for Legendre polynomials (i.e. order 0 functions):



$$
P^0_kP^0_l=sum_{m=|k-l|}^{k+l}pmatrix{k & l & m \ 0 & 0 & 0}^2(2m+1)P^0_m,
$$



where the Wigner $3-j$ coefficients $pmatrix{k & l & m \ 0 & 0 & 0}$ are specified in the above link.



Does anyone know of any analogous relationships for products like $P^0_kP^1_l$ and $P^1_kP^1_l$, or have any advice as to how go about deriving them? I'm trying to find some by using e.g. $P^1_j=sqrt{1-x^2}frac{dP^0_j}{dx}$, but it's not easy. Any help would be greatly appreciated!










share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    I would like a closed form expression for the integrals of some products like the following:



    $$
    int_{-1}^1 P^1_j P^1_k P^1_l P^1_m dx
    $$



    and



    $$
    int_{-1}^1 P^0_j P^0_k P^1_l P^1_m dx,
    $$



    where $P^0_j(x)$ and $P^1_j(x)$ are Legendre polynomials and order 1 associated Legendre functions, respectively. From an answer to another question on this site (Legendre Polynomials Triple Product), I know of the following relationship for Legendre polynomials (i.e. order 0 functions):



    $$
    P^0_kP^0_l=sum_{m=|k-l|}^{k+l}pmatrix{k & l & m \ 0 & 0 & 0}^2(2m+1)P^0_m,
    $$



    where the Wigner $3-j$ coefficients $pmatrix{k & l & m \ 0 & 0 & 0}$ are specified in the above link.



    Does anyone know of any analogous relationships for products like $P^0_kP^1_l$ and $P^1_kP^1_l$, or have any advice as to how go about deriving them? I'm trying to find some by using e.g. $P^1_j=sqrt{1-x^2}frac{dP^0_j}{dx}$, but it's not easy. Any help would be greatly appreciated!










    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I would like a closed form expression for the integrals of some products like the following:



      $$
      int_{-1}^1 P^1_j P^1_k P^1_l P^1_m dx
      $$



      and



      $$
      int_{-1}^1 P^0_j P^0_k P^1_l P^1_m dx,
      $$



      where $P^0_j(x)$ and $P^1_j(x)$ are Legendre polynomials and order 1 associated Legendre functions, respectively. From an answer to another question on this site (Legendre Polynomials Triple Product), I know of the following relationship for Legendre polynomials (i.e. order 0 functions):



      $$
      P^0_kP^0_l=sum_{m=|k-l|}^{k+l}pmatrix{k & l & m \ 0 & 0 & 0}^2(2m+1)P^0_m,
      $$



      where the Wigner $3-j$ coefficients $pmatrix{k & l & m \ 0 & 0 & 0}$ are specified in the above link.



      Does anyone know of any analogous relationships for products like $P^0_kP^1_l$ and $P^1_kP^1_l$, or have any advice as to how go about deriving them? I'm trying to find some by using e.g. $P^1_j=sqrt{1-x^2}frac{dP^0_j}{dx}$, but it's not easy. Any help would be greatly appreciated!










      share|cite|improve this question











      $endgroup$




      I would like a closed form expression for the integrals of some products like the following:



      $$
      int_{-1}^1 P^1_j P^1_k P^1_l P^1_m dx
      $$



      and



      $$
      int_{-1}^1 P^0_j P^0_k P^1_l P^1_m dx,
      $$



      where $P^0_j(x)$ and $P^1_j(x)$ are Legendre polynomials and order 1 associated Legendre functions, respectively. From an answer to another question on this site (Legendre Polynomials Triple Product), I know of the following relationship for Legendre polynomials (i.e. order 0 functions):



      $$
      P^0_kP^0_l=sum_{m=|k-l|}^{k+l}pmatrix{k & l & m \ 0 & 0 & 0}^2(2m+1)P^0_m,
      $$



      where the Wigner $3-j$ coefficients $pmatrix{k & l & m \ 0 & 0 & 0}$ are specified in the above link.



      Does anyone know of any analogous relationships for products like $P^0_kP^1_l$ and $P^1_kP^1_l$, or have any advice as to how go about deriving them? I'm trying to find some by using e.g. $P^1_j=sqrt{1-x^2}frac{dP^0_j}{dx}$, but it's not easy. Any help would be greatly appreciated!







      integration legendre-polynomials legendre-functions






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 8 '18 at 23:10







      Pan Daemonium

















      asked Dec 8 '18 at 16:45









      Pan DaemoniumPan Daemonium

      526




      526






















          1 Answer
          1






          active

          oldest

          votes


















          0












          $begingroup$

          I actually found the answer so I may as well share it: The relevant paper is called "The overlap integral of three associated Legendre polynomials", by Shi-hai Dong and R.Lemus (2002). The expression is given for products of general lengths containing general order Legendre functions and can be easily written down for the above cases. For instance:



          $$
          int_{-1}^1 P^1_j P^1_k P^1_l P^1_m dx=16sqrt{j(j+1)k(k+1)l(l+1)m(m+1)}sum_alpha sum_beta sum_gamma G^{jkalpha}_{11}G^{alpha lbeta}_{21}G^{beta mgamma}_{31}timessqrt{frac{(gamma-4)!}{(gamma+4)!}}frac{((1+(-1)^gamma)Gammaleft(frac{gamma}{2}right)Gammaleft(frac{gamma+5}{2}right)}{left(frac{gamma-4}{2}right)!Gammaleft(frac{gamma+3}{2}right)}
          $$



          where I've defined



          $$
          G^{abc}_{de}=(-1)^e(2c+1)begin{pmatrix} a & b & c \ 0 & 0 & 0 end{pmatrix}begin{pmatrix} a & b & c\ d & e & -d-e end{pmatrix}
          $$



          in terms of Wigner $3-j$ symbols. Note that $G^{abc}_{de}$ is zero unless 1) $|a-b|leq cleq a+b$, 2) $cgeq d+e$, and 3) $a+b+c$ is even. Note also that the factor of $left(frac{gamma-4}{2}right)!$ is no problem for odd $gamma$, because the numerator vanishes if $gamma$ is odd.



          The other integral can be written out similarly.



          (Link to reference: https://www.sciencedirect.com/science/article/pii/S0893965902800040)






          share|cite|improve this answer











          $endgroup$













            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3031335%2fintegrating-multiple-products-of-order-1-legendre-functions%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            0












            $begingroup$

            I actually found the answer so I may as well share it: The relevant paper is called "The overlap integral of three associated Legendre polynomials", by Shi-hai Dong and R.Lemus (2002). The expression is given for products of general lengths containing general order Legendre functions and can be easily written down for the above cases. For instance:



            $$
            int_{-1}^1 P^1_j P^1_k P^1_l P^1_m dx=16sqrt{j(j+1)k(k+1)l(l+1)m(m+1)}sum_alpha sum_beta sum_gamma G^{jkalpha}_{11}G^{alpha lbeta}_{21}G^{beta mgamma}_{31}timessqrt{frac{(gamma-4)!}{(gamma+4)!}}frac{((1+(-1)^gamma)Gammaleft(frac{gamma}{2}right)Gammaleft(frac{gamma+5}{2}right)}{left(frac{gamma-4}{2}right)!Gammaleft(frac{gamma+3}{2}right)}
            $$



            where I've defined



            $$
            G^{abc}_{de}=(-1)^e(2c+1)begin{pmatrix} a & b & c \ 0 & 0 & 0 end{pmatrix}begin{pmatrix} a & b & c\ d & e & -d-e end{pmatrix}
            $$



            in terms of Wigner $3-j$ symbols. Note that $G^{abc}_{de}$ is zero unless 1) $|a-b|leq cleq a+b$, 2) $cgeq d+e$, and 3) $a+b+c$ is even. Note also that the factor of $left(frac{gamma-4}{2}right)!$ is no problem for odd $gamma$, because the numerator vanishes if $gamma$ is odd.



            The other integral can be written out similarly.



            (Link to reference: https://www.sciencedirect.com/science/article/pii/S0893965902800040)






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              I actually found the answer so I may as well share it: The relevant paper is called "The overlap integral of three associated Legendre polynomials", by Shi-hai Dong and R.Lemus (2002). The expression is given for products of general lengths containing general order Legendre functions and can be easily written down for the above cases. For instance:



              $$
              int_{-1}^1 P^1_j P^1_k P^1_l P^1_m dx=16sqrt{j(j+1)k(k+1)l(l+1)m(m+1)}sum_alpha sum_beta sum_gamma G^{jkalpha}_{11}G^{alpha lbeta}_{21}G^{beta mgamma}_{31}timessqrt{frac{(gamma-4)!}{(gamma+4)!}}frac{((1+(-1)^gamma)Gammaleft(frac{gamma}{2}right)Gammaleft(frac{gamma+5}{2}right)}{left(frac{gamma-4}{2}right)!Gammaleft(frac{gamma+3}{2}right)}
              $$



              where I've defined



              $$
              G^{abc}_{de}=(-1)^e(2c+1)begin{pmatrix} a & b & c \ 0 & 0 & 0 end{pmatrix}begin{pmatrix} a & b & c\ d & e & -d-e end{pmatrix}
              $$



              in terms of Wigner $3-j$ symbols. Note that $G^{abc}_{de}$ is zero unless 1) $|a-b|leq cleq a+b$, 2) $cgeq d+e$, and 3) $a+b+c$ is even. Note also that the factor of $left(frac{gamma-4}{2}right)!$ is no problem for odd $gamma$, because the numerator vanishes if $gamma$ is odd.



              The other integral can be written out similarly.



              (Link to reference: https://www.sciencedirect.com/science/article/pii/S0893965902800040)






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                I actually found the answer so I may as well share it: The relevant paper is called "The overlap integral of three associated Legendre polynomials", by Shi-hai Dong and R.Lemus (2002). The expression is given for products of general lengths containing general order Legendre functions and can be easily written down for the above cases. For instance:



                $$
                int_{-1}^1 P^1_j P^1_k P^1_l P^1_m dx=16sqrt{j(j+1)k(k+1)l(l+1)m(m+1)}sum_alpha sum_beta sum_gamma G^{jkalpha}_{11}G^{alpha lbeta}_{21}G^{beta mgamma}_{31}timessqrt{frac{(gamma-4)!}{(gamma+4)!}}frac{((1+(-1)^gamma)Gammaleft(frac{gamma}{2}right)Gammaleft(frac{gamma+5}{2}right)}{left(frac{gamma-4}{2}right)!Gammaleft(frac{gamma+3}{2}right)}
                $$



                where I've defined



                $$
                G^{abc}_{de}=(-1)^e(2c+1)begin{pmatrix} a & b & c \ 0 & 0 & 0 end{pmatrix}begin{pmatrix} a & b & c\ d & e & -d-e end{pmatrix}
                $$



                in terms of Wigner $3-j$ symbols. Note that $G^{abc}_{de}$ is zero unless 1) $|a-b|leq cleq a+b$, 2) $cgeq d+e$, and 3) $a+b+c$ is even. Note also that the factor of $left(frac{gamma-4}{2}right)!$ is no problem for odd $gamma$, because the numerator vanishes if $gamma$ is odd.



                The other integral can be written out similarly.



                (Link to reference: https://www.sciencedirect.com/science/article/pii/S0893965902800040)






                share|cite|improve this answer











                $endgroup$



                I actually found the answer so I may as well share it: The relevant paper is called "The overlap integral of three associated Legendre polynomials", by Shi-hai Dong and R.Lemus (2002). The expression is given for products of general lengths containing general order Legendre functions and can be easily written down for the above cases. For instance:



                $$
                int_{-1}^1 P^1_j P^1_k P^1_l P^1_m dx=16sqrt{j(j+1)k(k+1)l(l+1)m(m+1)}sum_alpha sum_beta sum_gamma G^{jkalpha}_{11}G^{alpha lbeta}_{21}G^{beta mgamma}_{31}timessqrt{frac{(gamma-4)!}{(gamma+4)!}}frac{((1+(-1)^gamma)Gammaleft(frac{gamma}{2}right)Gammaleft(frac{gamma+5}{2}right)}{left(frac{gamma-4}{2}right)!Gammaleft(frac{gamma+3}{2}right)}
                $$



                where I've defined



                $$
                G^{abc}_{de}=(-1)^e(2c+1)begin{pmatrix} a & b & c \ 0 & 0 & 0 end{pmatrix}begin{pmatrix} a & b & c\ d & e & -d-e end{pmatrix}
                $$



                in terms of Wigner $3-j$ symbols. Note that $G^{abc}_{de}$ is zero unless 1) $|a-b|leq cleq a+b$, 2) $cgeq d+e$, and 3) $a+b+c$ is even. Note also that the factor of $left(frac{gamma-4}{2}right)!$ is no problem for odd $gamma$, because the numerator vanishes if $gamma$ is odd.



                The other integral can be written out similarly.



                (Link to reference: https://www.sciencedirect.com/science/article/pii/S0893965902800040)







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 8 '18 at 23:33

























                answered Dec 8 '18 at 23:27









                Pan DaemoniumPan Daemonium

                526




                526






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3031335%2fintegrating-multiple-products-of-order-1-legendre-functions%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    Popular posts from this blog

                    Plaza Victoria

                    Puebla de Zaragoza

                    Musa