Trying to understand the terms given by $int_0^1 t^m sin pi t : dt$












1












$begingroup$


Reading this paper (page $3$, theorem $2$) I got confused.



First is defined the integral $I_n$ as:



begin{align}
I_n = (-1)^n frac{ pi^{2n} }{(2n)!} int_0^1 t^{2n}(1-t)^{2n} sin pi t : dt .
end{align}



Now, $I_n$ consists of terms of the shape:



begin{align}
int_0^1 t^{m} sin pi t : dt , : m leq 2n.
end{align}



Integrate by parts the integral above:



begin{align}
int_0^1 t^{m} sin pi t : dt = frac{1}{pi} - frac{m(m-1)}{pi^2}
int_0^1 t^{m-2} sin pi t : dt.
end{align}



The following conclusion I don't understand: "Hence $I_n = pi^{-1}A_n(pi^{-2})
$, where $A_n$ is an $(n-1)$st degree polynomial with integral coefficients."



I understand the above as being, $I_n = pi^{-1}( a_{n-1}pi^{-2(n-1)} + a_{n-2}pi^{-2(n-2)} + cdots + a_0).$ But this is wrong.



Can someone give a detailed explanation, please?






integration proof-explanation







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Reading this paper (page $3$, theorem $2$) I got confused.



    First is defined the integral $I_n$ as:



    begin{align}
    I_n = (-1)^n frac{ pi^{2n} }{(2n)!} int_0^1 t^{2n}(1-t)^{2n} sin pi t : dt .
    end{align}



    Now, $I_n$ consists of terms of the shape:



    begin{align}
    int_0^1 t^{m} sin pi t : dt , : m leq 2n.
    end{align}



    Integrate by parts the integral above:



    begin{align}
    int_0^1 t^{m} sin pi t : dt = frac{1}{pi} - frac{m(m-1)}{pi^2}
    int_0^1 t^{m-2} sin pi t : dt.
    end{align}



    The following conclusion I don't understand: "Hence $I_n = pi^{-1}A_n(pi^{-2})
    $    , where $A_n$ is an $(n-1)$st degree polynomial with integral coefficients."



    I understand the above as being, $I_n = pi^{-1}( a_{n-1}pi^{-2(n-1)} + a_{n-2}pi^{-2(n-2)} + cdots + a_0).$ But this is wrong.



    Can someone give a detailed explanation, please?






    integration proof-explanation







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Reading this paper (page $3$, theorem $2$) I got confused.



      First is defined the integral $I_n$ as:



      begin{align}
      I_n = (-1)^n frac{ pi^{2n} }{(2n)!} int_0^1 t^{2n}(1-t)^{2n} sin pi t : dt .
      end{align}



      Now, $I_n$ consists of terms of the shape:



      begin{align}
      int_0^1 t^{m} sin pi t : dt , : m leq 2n.
      end{align}



      Integrate by parts the integral above:



      begin{align}
      int_0^1 t^{m} sin pi t : dt = frac{1}{pi} - frac{m(m-1)}{pi^2}
      int_0^1 t^{m-2} sin pi t : dt.
      end{align}



      The following conclusion I don't understand: "Hence $I_n = pi^{-1}A_n(pi^{-2})
      $      , where $A_n$ is an $(n-1)$st degree polynomial with integral coefficients."



      I understand the above as being, $I_n = pi^{-1}( a_{n-1}pi^{-2(n-1)} + a_{n-2}pi^{-2(n-2)} + cdots + a_0).$ But this is wrong.



      Can someone give a detailed explanation, please?






      integration proof-explanation







      share|cite|improve this question











      $endgroup$




      Reading this paper (page $3$, theorem $2$) I got confused.



      First is defined the integral $I_n$ as:



      begin{align}
      I_n = (-1)^n frac{ pi^{2n} }{(2n)!} int_0^1 t^{2n}(1-t)^{2n} sin pi t : dt .
      end{align}



      Now, $I_n$ consists of terms of the shape:



      begin{align}
      int_0^1 t^{m} sin pi t : dt , : m leq 2n.
      end{align}



      Integrate by parts the integral above:



      begin{align}
      int_0^1 t^{m} sin pi t : dt = frac{1}{pi} - frac{m(m-1)}{pi^2}
      int_0^1 t^{m-2} sin pi t : dt.
      end{align}



      The following conclusion I don't understand: "Hence $I_n = pi^{-1}A_n(pi^{-2})
      $      , where $A_n$ is an $(n-1)$st degree polynomial with integral coefficients."



      I understand the above as being, $I_n = pi^{-1}( a_{n-1}pi^{-2(n-1)} + a_{n-2}pi^{-2(n-2)} + cdots + a_0).$ But this is wrong.



      Can someone give a detailed explanation, please?






      integration proof-explanation




      integration proof-explanation






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 18 '18 at 4:19







      Pinteco

















      asked Dec 18 '18 at 1:08









      PintecoPinteco

      765313




      765313






















          1 Answer
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          $begingroup$

          I think that there is typo in the paper; it should be "Hence $I_n = pi^{-1}A_n(pi^{-2})$ where $A_n$ is an $color{red}{n}$ degree polynomial with integral coefficients."



          I prefered to write it as
          $$I_n=frac 1 pi sum_{k=0}^n frac {a_k} {pi^{2k}}$$



          Using what is given and computing, we have
          $$left(
          begin{array}{cc}
          n & pi I_n \
          1 & 2-frac{24}{pi ^2} \
          2 & 2-frac{360}{pi ^2} +frac{3360}{pi ^4}\
          3 & 2-frac{1680}{pi ^2}+frac{151200}{pi ^4}-frac{1330560}{pi ^6} \
          4 & 2-frac{5040}{pi ^2}+frac{1663200}{pi
          ^4}-frac{121080960}{pi ^6}+frac{1037836800}{pi ^8}\
          5 & 2-frac{11880}{pi ^2}+frac{10090080}{pi ^4}-frac{2421619200}{pi ^6}+frac{158789030400}{pi
          ^8}-frac{1340885145600}{pi ^{10}} \
          6 &2 -frac{24024}{pi ^2}+frac{43243200}{pi ^4}-frac{24700515840}{pi^6}+frac{5028319296000}{pi ^8}-frac{309744468633600}{pi
          ^{10}}+frac{2590590101299200}{pi ^{12}}
          end{array}
          right)$$



          In fact, given by a CAS, there is a general expression in terms of hypergeometric function
          $$I_n=(-1)^nfrac{ pi ^{2 n+frac{3}{2}} }{2^{4 n+2},Gamma left(2 n+frac{3}{2}right)},
          _2F_3left(n+1,n+frac{3}{2};frac{3}{2},2 n+frac{3}{2},2 n+2;-frac{pi
          ^2}{4}right)$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            My question was more about how seeing the integral after integrating by parts helps you conclude that $I_n=frac 1 pi sum_{k=0}^n frac {a_k} {pi^{2k}}$. Thanks for answering though.
            $endgroup$
            – Pinteco
            Dec 19 '18 at 15:51













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          0












          $begingroup$

          I think that there is typo in the paper; it should be "Hence $I_n = pi^{-1}A_n(pi^{-2})$ where $A_n$ is an $color{red}{n}$ degree polynomial with integral coefficients."



          I prefered to write it as
          $$I_n=frac 1 pi sum_{k=0}^n frac {a_k} {pi^{2k}}$$



          Using what is given and computing, we have
          $$left(
          begin{array}{cc}
          n & pi I_n \
          1 & 2-frac{24}{pi ^2} \
          2 & 2-frac{360}{pi ^2} +frac{3360}{pi ^4}\
          3 & 2-frac{1680}{pi ^2}+frac{151200}{pi ^4}-frac{1330560}{pi ^6} \
          4 & 2-frac{5040}{pi ^2}+frac{1663200}{pi
          ^4}-frac{121080960}{pi ^6}+frac{1037836800}{pi ^8}\
          5 & 2-frac{11880}{pi ^2}+frac{10090080}{pi ^4}-frac{2421619200}{pi ^6}+frac{158789030400}{pi
          ^8}-frac{1340885145600}{pi ^{10}} \
          6 &2 -frac{24024}{pi ^2}+frac{43243200}{pi ^4}-frac{24700515840}{pi^6}+frac{5028319296000}{pi ^8}-frac{309744468633600}{pi
          ^{10}}+frac{2590590101299200}{pi ^{12}}
          end{array}
          right)$$



          In fact, given by a CAS, there is a general expression in terms of hypergeometric function
          $$I_n=(-1)^nfrac{ pi ^{2 n+frac{3}{2}} }{2^{4 n+2},Gamma left(2 n+frac{3}{2}right)},
          _2F_3left(n+1,n+frac{3}{2};frac{3}{2},2 n+frac{3}{2},2 n+2;-frac{pi
          ^2}{4}right)$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            My question was more about how seeing the integral after integrating by parts helps you conclude that $I_n=frac 1 pi sum_{k=0}^n frac {a_k} {pi^{2k}}$. Thanks for answering though.
            $endgroup$
            – Pinteco
            Dec 19 '18 at 15:51


















          0












          $begingroup$

          I think that there is typo in the paper; it should be "Hence $I_n = pi^{-1}A_n(pi^{-2})$ where $A_n$ is an $color{red}{n}$ degree polynomial with integral coefficients."



          I prefered to write it as
          $$I_n=frac 1 pi sum_{k=0}^n frac {a_k} {pi^{2k}}$$



          Using what is given and computing, we have
          $$left(
          begin{array}{cc}
          n & pi I_n \
          1 & 2-frac{24}{pi ^2} \
          2 & 2-frac{360}{pi ^2} +frac{3360}{pi ^4}\
          3 & 2-frac{1680}{pi ^2}+frac{151200}{pi ^4}-frac{1330560}{pi ^6} \
          4 & 2-frac{5040}{pi ^2}+frac{1663200}{pi
          ^4}-frac{121080960}{pi ^6}+frac{1037836800}{pi ^8}\
          5 & 2-frac{11880}{pi ^2}+frac{10090080}{pi ^4}-frac{2421619200}{pi ^6}+frac{158789030400}{pi
          ^8}-frac{1340885145600}{pi ^{10}} \
          6 &2 -frac{24024}{pi ^2}+frac{43243200}{pi ^4}-frac{24700515840}{pi^6}+frac{5028319296000}{pi ^8}-frac{309744468633600}{pi
          ^{10}}+frac{2590590101299200}{pi ^{12}}
          end{array}
          right)$$



          In fact, given by a CAS, there is a general expression in terms of hypergeometric function
          $$I_n=(-1)^nfrac{ pi ^{2 n+frac{3}{2}} }{2^{4 n+2},Gamma left(2 n+frac{3}{2}right)},
          _2F_3left(n+1,n+frac{3}{2};frac{3}{2},2 n+frac{3}{2},2 n+2;-frac{pi
          ^2}{4}right)$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            My question was more about how seeing the integral after integrating by parts helps you conclude that $I_n=frac 1 pi sum_{k=0}^n frac {a_k} {pi^{2k}}$. Thanks for answering though.
            $endgroup$
            – Pinteco
            Dec 19 '18 at 15:51
















          0












          0








          0





          $begingroup$

          I think that there is typo in the paper; it should be "Hence $I_n = pi^{-1}A_n(pi^{-2})$ where $A_n$ is an $color{red}{n}$ degree polynomial with integral coefficients."



          I prefered to write it as
          $$I_n=frac 1 pi sum_{k=0}^n frac {a_k} {pi^{2k}}$$



          Using what is given and computing, we have
          $$left(
          begin{array}{cc}
          n & pi I_n \
          1 & 2-frac{24}{pi ^2} \
          2 & 2-frac{360}{pi ^2} +frac{3360}{pi ^4}\
          3 & 2-frac{1680}{pi ^2}+frac{151200}{pi ^4}-frac{1330560}{pi ^6} \
          4 & 2-frac{5040}{pi ^2}+frac{1663200}{pi
          ^4}-frac{121080960}{pi ^6}+frac{1037836800}{pi ^8}\
          5 & 2-frac{11880}{pi ^2}+frac{10090080}{pi ^4}-frac{2421619200}{pi ^6}+frac{158789030400}{pi
          ^8}-frac{1340885145600}{pi ^{10}} \
          6 &2 -frac{24024}{pi ^2}+frac{43243200}{pi ^4}-frac{24700515840}{pi^6}+frac{5028319296000}{pi ^8}-frac{309744468633600}{pi
          ^{10}}+frac{2590590101299200}{pi ^{12}}
          end{array}
          right)$$



          In fact, given by a CAS, there is a general expression in terms of hypergeometric function
          $$I_n=(-1)^nfrac{ pi ^{2 n+frac{3}{2}} }{2^{4 n+2},Gamma left(2 n+frac{3}{2}right)},
          _2F_3left(n+1,n+frac{3}{2};frac{3}{2},2 n+frac{3}{2},2 n+2;-frac{pi
          ^2}{4}right)$$






          share|cite|improve this answer









          $endgroup$



          I think that there is typo in the paper; it should be "Hence $I_n = pi^{-1}A_n(pi^{-2})$ where $A_n$ is an $color{red}{n}$ degree polynomial with integral coefficients."



          I prefered to write it as
          $$I_n=frac 1 pi sum_{k=0}^n frac {a_k} {pi^{2k}}$$



          Using what is given and computing, we have
          $$left(
          begin{array}{cc}
          n & pi I_n \
          1 & 2-frac{24}{pi ^2} \
          2 & 2-frac{360}{pi ^2} +frac{3360}{pi ^4}\
          3 & 2-frac{1680}{pi ^2}+frac{151200}{pi ^4}-frac{1330560}{pi ^6} \
          4 & 2-frac{5040}{pi ^2}+frac{1663200}{pi
          ^4}-frac{121080960}{pi ^6}+frac{1037836800}{pi ^8}\
          5 & 2-frac{11880}{pi ^2}+frac{10090080}{pi ^4}-frac{2421619200}{pi ^6}+frac{158789030400}{pi
          ^8}-frac{1340885145600}{pi ^{10}} \
          6 &2 -frac{24024}{pi ^2}+frac{43243200}{pi ^4}-frac{24700515840}{pi^6}+frac{5028319296000}{pi ^8}-frac{309744468633600}{pi
          ^{10}}+frac{2590590101299200}{pi ^{12}}
          end{array}
          right)$$



          In fact, given by a CAS, there is a general expression in terms of hypergeometric function
          $$I_n=(-1)^nfrac{ pi ^{2 n+frac{3}{2}} }{2^{4 n+2},Gamma left(2 n+frac{3}{2}right)},
          _2F_3left(n+1,n+frac{3}{2};frac{3}{2},2 n+frac{3}{2},2 n+2;-frac{pi
          ^2}{4}right)$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 18 '18 at 5:40









          Claude LeiboviciClaude Leibovici

          124k1158135




          124k1158135












          • $begingroup$
            My question was more about how seeing the integral after integrating by parts helps you conclude that $I_n=frac 1 pi sum_{k=0}^n frac {a_k} {pi^{2k}}$. Thanks for answering though.
            $endgroup$
            – Pinteco
            Dec 19 '18 at 15:51




















          • $begingroup$
            My question was more about how seeing the integral after integrating by parts helps you conclude that $I_n=frac 1 pi sum_{k=0}^n frac {a_k} {pi^{2k}}$. Thanks for answering though.
            $endgroup$
            – Pinteco
            Dec 19 '18 at 15:51


















          $begingroup$
          My question was more about how seeing the integral after integrating by parts helps you conclude that $I_n=frac 1 pi sum_{k=0}^n frac {a_k} {pi^{2k}}$. Thanks for answering though.
          $endgroup$
          – Pinteco
          Dec 19 '18 at 15:51






          $begingroup$
          My question was more about how seeing the integral after integrating by parts helps you conclude that $I_n=frac 1 pi sum_{k=0}^n frac {a_k} {pi^{2k}}$. Thanks for answering though.
          $endgroup$
          – Pinteco
          Dec 19 '18 at 15:51




















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