Finding a closed form for $I_n=int_0^1 prod_{l=1}^nleft[x^2-frac{l^2}{n^2}right]dx$












6












$begingroup$


I'm now dealing with this integral and trying to give a closed form to it:
$$I_n=int_0^1 prod_{l=1}^nleft[x^2-dfrac{l^2}{n^2}right]dx$$
Where the first few values are given:
$$I_1=int_0^1 (x^2-1) dx=-dfrac23$$
$$I_2=int_0^1 left(x^2-dfrac14right)(x^2-1)dx=dfrac1{30}$$
$$I_3=int_0^1 left(x^2-frac19right)left(x^2-frac49right)(x^2-1)dx=-dfrac{136}{8505}$$
I've tried to expand the polynomial, but I could merely give a general expression for the first few and the last few coefficients, and they become complicated very fast. I have searched the Internet; did I miss something that I couldn't find them because I don't know the Theorem's name?






calculus integration







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    6












    $begingroup$


    I'm now dealing with this integral and trying to give a closed form to it:
    $$I_n=int_0^1 prod_{l=1}^nleft[x^2-dfrac{l^2}{n^2}right]dx$$
    Where the first few values are given:
    $$I_1=int_0^1 (x^2-1) dx=-dfrac23$$
    $$I_2=int_0^1 left(x^2-dfrac14right)(x^2-1)dx=dfrac1{30}$$
    $$I_3=int_0^1 left(x^2-frac19right)left(x^2-frac49right)(x^2-1)dx=-dfrac{136}{8505}$$
    I've tried to expand the polynomial, but I could merely give a general expression for the first few and the last few coefficients, and they become complicated very fast. I have searched the Internet; did I miss something that I couldn't find them because I don't know the Theorem's name?






    calculus integration







    share|cite|improve this question











    $endgroup$















      6












      6








      6


      1



      $begingroup$


      I'm now dealing with this integral and trying to give a closed form to it:
      $$I_n=int_0^1 prod_{l=1}^nleft[x^2-dfrac{l^2}{n^2}right]dx$$
      Where the first few values are given:
      $$I_1=int_0^1 (x^2-1) dx=-dfrac23$$
      $$I_2=int_0^1 left(x^2-dfrac14right)(x^2-1)dx=dfrac1{30}$$
      $$I_3=int_0^1 left(x^2-frac19right)left(x^2-frac49right)(x^2-1)dx=-dfrac{136}{8505}$$
      I've tried to expand the polynomial, but I could merely give a general expression for the first few and the last few coefficients, and they become complicated very fast. I have searched the Internet; did I miss something that I couldn't find them because I don't know the Theorem's name?






      calculus integration







      share|cite|improve this question











      $endgroup$




      I'm now dealing with this integral and trying to give a closed form to it:
      $$I_n=int_0^1 prod_{l=1}^nleft[x^2-dfrac{l^2}{n^2}right]dx$$
      Where the first few values are given:
      $$I_1=int_0^1 (x^2-1) dx=-dfrac23$$
      $$I_2=int_0^1 left(x^2-dfrac14right)(x^2-1)dx=dfrac1{30}$$
      $$I_3=int_0^1 left(x^2-frac19right)left(x^2-frac49right)(x^2-1)dx=-dfrac{136}{8505}$$
      I've tried to expand the polynomial, but I could merely give a general expression for the first few and the last few coefficients, and they become complicated very fast. I have searched the Internet; did I miss something that I couldn't find them because I don't know the Theorem's name?






      calculus integration




      calculus integration






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      share|cite|improve this question













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      edited Jan 18 at 15:59









      Blue

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      asked Jan 18 at 13:10









      kelvin hong 方kelvin hong 方

      56318




      56318






















          2 Answers
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          6












          $begingroup$

          $$I_n=frac{1}{n^{2n}}int_0^1prod_{l=1}^{n}[(xn)^2-l^2]dxqquad x=frac{y}{n}$$
          $$I_n=frac{1}{n^{2n+1}}int_0^nprod_{l=1}^{n}[y^2-l^2]dy$$
          We can expand the product, without knowing the coefficients for now as following;
          $$prod_{l=1}^{n}[y^2-l^2]=sum_{j=0}^n(-1)^{n-j}y^{2j}t(n,j)$$
          $$I_n=frac{1}{n^{2n+1}}sum_{j=0}^n(-1)^{n-j}t(n,j)int_0^ny^{2j}dy$$
          $$I_n=frac{1}{n^{2n+1}}sum_{j=0}^nfrac{(-1)^{n-j}t(n,j)n^{2j+1}}{2j+1}$$
          $$I_n=frac{(-1)^n}{n^{2n}}sum_{j=0}^nfrac{(-1)^{j}t(n,j)n^{2j}}{2j+1}$$
          Now back to the coefficients, ive found it on OEIS A008955



          they are called the central factorial numbers. I think this is the closest we can get for a closed form.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Since this (I assume) can be computed in a finite number of steps, it counts as a closed form $rightarrow $ (+1).
            $endgroup$
            – clathratus
            Jan 18 at 18:45










          • $begingroup$
            Here's a (actually kinda bad) reference for central factorial. mathworld.wolfram.com/CentralFactorial.html Its the best I could find so far
            $endgroup$
            – clathratus
            Jan 18 at 18:47










          • $begingroup$
            Thank you very much, I think the Central Factorial numbers are what I want.
            $endgroup$
            – kelvin hong 方
            Jan 19 at 2:32



















          1












          $begingroup$

          You can relate the product to the rising factorials $x^overline{n} = x(x+1)dotsb (x+n-1) = sum_{k=0}^n begin{bmatrix}n\kend{bmatrix} x^k$ where the coefficients are Stirling's numbers of the first kind.
          begin{align}
          I_n
          &= int_0^1 prod_{l=1}^n left(x^2 - frac{l^2}{n^2}right);dx
          = int_0^1 prod_{l-1 =k=0}^{n-1} frac{xn - k + 1}{n}cdot prod_{k=0}^{n-1} frac{xn + k + 1}{n};dx\
          &= frac{(-1)^n }{n^{2n}}int_0^1 (-xn - 1)^{overline{n}}(xn +1)^{overline{n}};dx\
          %
          &= frac{(-1)^n }{n^{2n}}int_0^1 sum_{i=0}^nsum_{j=0}^n begin{bmatrix}n\iend{bmatrix}
          begin{bmatrix}n\jend{bmatrix} (-xn - 1)^i (xn + 1)^j;dx\
          %
          &= frac{(-1)^n }{n^{2n}} sum_{i=0}^nsum_{j=0}^n (-1)^i begin{bmatrix}n\iend{bmatrix}
          begin{bmatrix}n\jend{bmatrix} int_0^1 (xn + 1)^{i+j} ;dx\
          %
          &= frac{(-1)^n }{n^{2n}} sum_{i=0}^nsum_{j=0}^n (-1)^i begin{bmatrix}n\iend{bmatrix}
          begin{bmatrix}n\jend{bmatrix} frac{(n+1)^{i+j+1}-1}{n(i+j+1)}\
          %
          &= sum_{i=0}^nsum_{j=0}^n (-1)^{i+n}begin{bmatrix}n\iend{bmatrix}
          begin{bmatrix}n\jend{bmatrix} frac{(n+1)^{i+j+1}-1}{n^{2n+1}(i+j+1)}\
          %
          end{align}






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I think the sum is too complicated, and the Stirling numbers of the first kind are also hard to compute. But this does give me a new way to deal with this problem. Thank you.
            $endgroup$
            – kelvin hong 方
            Jan 19 at 12:14










          • $begingroup$
            It was likely obvious to you that, as per the other answer, you can expand the product as an $2n$-degree polynomial. I felt that doing a OEIS coefficient search and presenting that, without even proving the connection, fell short of an answer.
            $endgroup$
            – adfriedman
            Jan 19 at 18:58










          • $begingroup$
            Yes, to evaluate this, we still need computer, I think the simplest form is the sum, can be impossible to represented as a term.
            $endgroup$
            – kelvin hong 方
            Jan 20 at 4:05











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          2 Answers
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          2 Answers
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          6












          $begingroup$

          $$I_n=frac{1}{n^{2n}}int_0^1prod_{l=1}^{n}[(xn)^2-l^2]dxqquad x=frac{y}{n}$$
          $$I_n=frac{1}{n^{2n+1}}int_0^nprod_{l=1}^{n}[y^2-l^2]dy$$
          We can expand the product, without knowing the coefficients for now as following;
          $$prod_{l=1}^{n}[y^2-l^2]=sum_{j=0}^n(-1)^{n-j}y^{2j}t(n,j)$$
          $$I_n=frac{1}{n^{2n+1}}sum_{j=0}^n(-1)^{n-j}t(n,j)int_0^ny^{2j}dy$$
          $$I_n=frac{1}{n^{2n+1}}sum_{j=0}^nfrac{(-1)^{n-j}t(n,j)n^{2j+1}}{2j+1}$$
          $$I_n=frac{(-1)^n}{n^{2n}}sum_{j=0}^nfrac{(-1)^{j}t(n,j)n^{2j}}{2j+1}$$
          Now back to the coefficients, ive found it on OEIS A008955



          they are called the central factorial numbers. I think this is the closest we can get for a closed form.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Since this (I assume) can be computed in a finite number of steps, it counts as a closed form $rightarrow $ (+1).
            $endgroup$
            – clathratus
            Jan 18 at 18:45










          • $begingroup$
            Here's a (actually kinda bad) reference for central factorial. mathworld.wolfram.com/CentralFactorial.html Its the best I could find so far
            $endgroup$
            – clathratus
            Jan 18 at 18:47










          • $begingroup$
            Thank you very much, I think the Central Factorial numbers are what I want.
            $endgroup$
            – kelvin hong 方
            Jan 19 at 2:32
















          6












          $begingroup$

          $$I_n=frac{1}{n^{2n}}int_0^1prod_{l=1}^{n}[(xn)^2-l^2]dxqquad x=frac{y}{n}$$
          $$I_n=frac{1}{n^{2n+1}}int_0^nprod_{l=1}^{n}[y^2-l^2]dy$$
          We can expand the product, without knowing the coefficients for now as following;
          $$prod_{l=1}^{n}[y^2-l^2]=sum_{j=0}^n(-1)^{n-j}y^{2j}t(n,j)$$
          $$I_n=frac{1}{n^{2n+1}}sum_{j=0}^n(-1)^{n-j}t(n,j)int_0^ny^{2j}dy$$
          $$I_n=frac{1}{n^{2n+1}}sum_{j=0}^nfrac{(-1)^{n-j}t(n,j)n^{2j+1}}{2j+1}$$
          $$I_n=frac{(-1)^n}{n^{2n}}sum_{j=0}^nfrac{(-1)^{j}t(n,j)n^{2j}}{2j+1}$$
          Now back to the coefficients, ive found it on OEIS A008955



          they are called the central factorial numbers. I think this is the closest we can get for a closed form.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Since this (I assume) can be computed in a finite number of steps, it counts as a closed form $rightarrow $ (+1).
            $endgroup$
            – clathratus
            Jan 18 at 18:45










          • $begingroup$
            Here's a (actually kinda bad) reference for central factorial. mathworld.wolfram.com/CentralFactorial.html Its the best I could find so far
            $endgroup$
            – clathratus
            Jan 18 at 18:47










          • $begingroup$
            Thank you very much, I think the Central Factorial numbers are what I want.
            $endgroup$
            – kelvin hong 方
            Jan 19 at 2:32














          6












          6








          6





          $begingroup$

          $$I_n=frac{1}{n^{2n}}int_0^1prod_{l=1}^{n}[(xn)^2-l^2]dxqquad x=frac{y}{n}$$
          $$I_n=frac{1}{n^{2n+1}}int_0^nprod_{l=1}^{n}[y^2-l^2]dy$$
          We can expand the product, without knowing the coefficients for now as following;
          $$prod_{l=1}^{n}[y^2-l^2]=sum_{j=0}^n(-1)^{n-j}y^{2j}t(n,j)$$
          $$I_n=frac{1}{n^{2n+1}}sum_{j=0}^n(-1)^{n-j}t(n,j)int_0^ny^{2j}dy$$
          $$I_n=frac{1}{n^{2n+1}}sum_{j=0}^nfrac{(-1)^{n-j}t(n,j)n^{2j+1}}{2j+1}$$
          $$I_n=frac{(-1)^n}{n^{2n}}sum_{j=0}^nfrac{(-1)^{j}t(n,j)n^{2j}}{2j+1}$$
          Now back to the coefficients, ive found it on OEIS A008955



          they are called the central factorial numbers. I think this is the closest we can get for a closed form.






          share|cite|improve this answer











          $endgroup$



          $$I_n=frac{1}{n^{2n}}int_0^1prod_{l=1}^{n}[(xn)^2-l^2]dxqquad x=frac{y}{n}$$
          $$I_n=frac{1}{n^{2n+1}}int_0^nprod_{l=1}^{n}[y^2-l^2]dy$$
          We can expand the product, without knowing the coefficients for now as following;
          $$prod_{l=1}^{n}[y^2-l^2]=sum_{j=0}^n(-1)^{n-j}y^{2j}t(n,j)$$
          $$I_n=frac{1}{n^{2n+1}}sum_{j=0}^n(-1)^{n-j}t(n,j)int_0^ny^{2j}dy$$
          $$I_n=frac{1}{n^{2n+1}}sum_{j=0}^nfrac{(-1)^{n-j}t(n,j)n^{2j+1}}{2j+1}$$
          $$I_n=frac{(-1)^n}{n^{2n}}sum_{j=0}^nfrac{(-1)^{j}t(n,j)n^{2j}}{2j+1}$$
          Now back to the coefficients, ive found it on OEIS A008955



          they are called the central factorial numbers. I think this is the closest we can get for a closed form.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 18 at 23:33

























          answered Jan 18 at 17:29









          Andrew KovácsAndrew Kovács

          1264




          1264












          • $begingroup$
            Since this (I assume) can be computed in a finite number of steps, it counts as a closed form $rightarrow $ (+1).
            $endgroup$
            – clathratus
            Jan 18 at 18:45










          • $begingroup$
            Here's a (actually kinda bad) reference for central factorial. mathworld.wolfram.com/CentralFactorial.html Its the best I could find so far
            $endgroup$
            – clathratus
            Jan 18 at 18:47










          • $begingroup$
            Thank you very much, I think the Central Factorial numbers are what I want.
            $endgroup$
            – kelvin hong 方
            Jan 19 at 2:32


















          • $begingroup$
            Since this (I assume) can be computed in a finite number of steps, it counts as a closed form $rightarrow $ (+1).
            $endgroup$
            – clathratus
            Jan 18 at 18:45










          • $begingroup$
            Here's a (actually kinda bad) reference for central factorial. mathworld.wolfram.com/CentralFactorial.html Its the best I could find so far
            $endgroup$
            – clathratus
            Jan 18 at 18:47










          • $begingroup$
            Thank you very much, I think the Central Factorial numbers are what I want.
            $endgroup$
            – kelvin hong 方
            Jan 19 at 2:32
















          $begingroup$
          Since this (I assume) can be computed in a finite number of steps, it counts as a closed form $rightarrow $ (+1).
          $endgroup$
          – clathratus
          Jan 18 at 18:45




          $begingroup$
          Since this (I assume) can be computed in a finite number of steps, it counts as a closed form $rightarrow $ (+1).
          $endgroup$
          – clathratus
          Jan 18 at 18:45












          $begingroup$
          Here's a (actually kinda bad) reference for central factorial. mathworld.wolfram.com/CentralFactorial.html Its the best I could find so far
          $endgroup$
          – clathratus
          Jan 18 at 18:47




          $begingroup$
          Here's a (actually kinda bad) reference for central factorial. mathworld.wolfram.com/CentralFactorial.html Its the best I could find so far
          $endgroup$
          – clathratus
          Jan 18 at 18:47












          $begingroup$
          Thank you very much, I think the Central Factorial numbers are what I want.
          $endgroup$
          – kelvin hong 方
          Jan 19 at 2:32




          $begingroup$
          Thank you very much, I think the Central Factorial numbers are what I want.
          $endgroup$
          – kelvin hong 方
          Jan 19 at 2:32











          1












          $begingroup$

          You can relate the product to the rising factorials $x^overline{n} = x(x+1)dotsb (x+n-1) = sum_{k=0}^n begin{bmatrix}n\kend{bmatrix} x^k$ where the coefficients are Stirling's numbers of the first kind.
          begin{align}
          I_n
          &= int_0^1 prod_{l=1}^n left(x^2 - frac{l^2}{n^2}right);dx
          = int_0^1 prod_{l-1 =k=0}^{n-1} frac{xn - k + 1}{n}cdot prod_{k=0}^{n-1} frac{xn + k + 1}{n};dx\
          &= frac{(-1)^n }{n^{2n}}int_0^1 (-xn - 1)^{overline{n}}(xn +1)^{overline{n}};dx\
          %
          &= frac{(-1)^n }{n^{2n}}int_0^1 sum_{i=0}^nsum_{j=0}^n begin{bmatrix}n\iend{bmatrix}
          begin{bmatrix}n\jend{bmatrix} (-xn - 1)^i (xn + 1)^j;dx\
          %
          &= frac{(-1)^n }{n^{2n}} sum_{i=0}^nsum_{j=0}^n (-1)^i begin{bmatrix}n\iend{bmatrix}
          begin{bmatrix}n\jend{bmatrix} int_0^1 (xn + 1)^{i+j} ;dx\
          %
          &= frac{(-1)^n }{n^{2n}} sum_{i=0}^nsum_{j=0}^n (-1)^i begin{bmatrix}n\iend{bmatrix}
          begin{bmatrix}n\jend{bmatrix} frac{(n+1)^{i+j+1}-1}{n(i+j+1)}\
          %
          &= sum_{i=0}^nsum_{j=0}^n (-1)^{i+n}begin{bmatrix}n\iend{bmatrix}
          begin{bmatrix}n\jend{bmatrix} frac{(n+1)^{i+j+1}-1}{n^{2n+1}(i+j+1)}\
          %
          end{align}






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I think the sum is too complicated, and the Stirling numbers of the first kind are also hard to compute. But this does give me a new way to deal with this problem. Thank you.
            $endgroup$
            – kelvin hong 方
            Jan 19 at 12:14










          • $begingroup$
            It was likely obvious to you that, as per the other answer, you can expand the product as an $2n$-degree polynomial. I felt that doing a OEIS coefficient search and presenting that, without even proving the connection, fell short of an answer.
            $endgroup$
            – adfriedman
            Jan 19 at 18:58










          • $begingroup$
            Yes, to evaluate this, we still need computer, I think the simplest form is the sum, can be impossible to represented as a term.
            $endgroup$
            – kelvin hong 方
            Jan 20 at 4:05
















          1












          $begingroup$

          You can relate the product to the rising factorials $x^overline{n} = x(x+1)dotsb (x+n-1) = sum_{k=0}^n begin{bmatrix}n\kend{bmatrix} x^k$ where the coefficients are Stirling's numbers of the first kind.
          begin{align}
          I_n
          &= int_0^1 prod_{l=1}^n left(x^2 - frac{l^2}{n^2}right);dx
          = int_0^1 prod_{l-1 =k=0}^{n-1} frac{xn - k + 1}{n}cdot prod_{k=0}^{n-1} frac{xn + k + 1}{n};dx\
          &= frac{(-1)^n }{n^{2n}}int_0^1 (-xn - 1)^{overline{n}}(xn +1)^{overline{n}};dx\
          %
          &= frac{(-1)^n }{n^{2n}}int_0^1 sum_{i=0}^nsum_{j=0}^n begin{bmatrix}n\iend{bmatrix}
          begin{bmatrix}n\jend{bmatrix} (-xn - 1)^i (xn + 1)^j;dx\
          %
          &= frac{(-1)^n }{n^{2n}} sum_{i=0}^nsum_{j=0}^n (-1)^i begin{bmatrix}n\iend{bmatrix}
          begin{bmatrix}n\jend{bmatrix} int_0^1 (xn + 1)^{i+j} ;dx\
          %
          &= frac{(-1)^n }{n^{2n}} sum_{i=0}^nsum_{j=0}^n (-1)^i begin{bmatrix}n\iend{bmatrix}
          begin{bmatrix}n\jend{bmatrix} frac{(n+1)^{i+j+1}-1}{n(i+j+1)}\
          %
          &= sum_{i=0}^nsum_{j=0}^n (-1)^{i+n}begin{bmatrix}n\iend{bmatrix}
          begin{bmatrix}n\jend{bmatrix} frac{(n+1)^{i+j+1}-1}{n^{2n+1}(i+j+1)}\
          %
          end{align}






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I think the sum is too complicated, and the Stirling numbers of the first kind are also hard to compute. But this does give me a new way to deal with this problem. Thank you.
            $endgroup$
            – kelvin hong 方
            Jan 19 at 12:14










          • $begingroup$
            It was likely obvious to you that, as per the other answer, you can expand the product as an $2n$-degree polynomial. I felt that doing a OEIS coefficient search and presenting that, without even proving the connection, fell short of an answer.
            $endgroup$
            – adfriedman
            Jan 19 at 18:58










          • $begingroup$
            Yes, to evaluate this, we still need computer, I think the simplest form is the sum, can be impossible to represented as a term.
            $endgroup$
            – kelvin hong 方
            Jan 20 at 4:05














          1












          1








          1





          $begingroup$

          You can relate the product to the rising factorials $x^overline{n} = x(x+1)dotsb (x+n-1) = sum_{k=0}^n begin{bmatrix}n\kend{bmatrix} x^k$ where the coefficients are Stirling's numbers of the first kind.
          begin{align}
          I_n
          &= int_0^1 prod_{l=1}^n left(x^2 - frac{l^2}{n^2}right);dx
          = int_0^1 prod_{l-1 =k=0}^{n-1} frac{xn - k + 1}{n}cdot prod_{k=0}^{n-1} frac{xn + k + 1}{n};dx\
          &= frac{(-1)^n }{n^{2n}}int_0^1 (-xn - 1)^{overline{n}}(xn +1)^{overline{n}};dx\
          %
          &= frac{(-1)^n }{n^{2n}}int_0^1 sum_{i=0}^nsum_{j=0}^n begin{bmatrix}n\iend{bmatrix}
          begin{bmatrix}n\jend{bmatrix} (-xn - 1)^i (xn + 1)^j;dx\
          %
          &= frac{(-1)^n }{n^{2n}} sum_{i=0}^nsum_{j=0}^n (-1)^i begin{bmatrix}n\iend{bmatrix}
          begin{bmatrix}n\jend{bmatrix} int_0^1 (xn + 1)^{i+j} ;dx\
          %
          &= frac{(-1)^n }{n^{2n}} sum_{i=0}^nsum_{j=0}^n (-1)^i begin{bmatrix}n\iend{bmatrix}
          begin{bmatrix}n\jend{bmatrix} frac{(n+1)^{i+j+1}-1}{n(i+j+1)}\
          %
          &= sum_{i=0}^nsum_{j=0}^n (-1)^{i+n}begin{bmatrix}n\iend{bmatrix}
          begin{bmatrix}n\jend{bmatrix} frac{(n+1)^{i+j+1}-1}{n^{2n+1}(i+j+1)}\
          %
          end{align}






          share|cite|improve this answer









          $endgroup$



          You can relate the product to the rising factorials $x^overline{n} = x(x+1)dotsb (x+n-1) = sum_{k=0}^n begin{bmatrix}n\kend{bmatrix} x^k$ where the coefficients are Stirling's numbers of the first kind.
          begin{align}
          I_n
          &= int_0^1 prod_{l=1}^n left(x^2 - frac{l^2}{n^2}right);dx
          = int_0^1 prod_{l-1 =k=0}^{n-1} frac{xn - k + 1}{n}cdot prod_{k=0}^{n-1} frac{xn + k + 1}{n};dx\
          &= frac{(-1)^n }{n^{2n}}int_0^1 (-xn - 1)^{overline{n}}(xn +1)^{overline{n}};dx\
          %
          &= frac{(-1)^n }{n^{2n}}int_0^1 sum_{i=0}^nsum_{j=0}^n begin{bmatrix}n\iend{bmatrix}
          begin{bmatrix}n\jend{bmatrix} (-xn - 1)^i (xn + 1)^j;dx\
          %
          &= frac{(-1)^n }{n^{2n}} sum_{i=0}^nsum_{j=0}^n (-1)^i begin{bmatrix}n\iend{bmatrix}
          begin{bmatrix}n\jend{bmatrix} int_0^1 (xn + 1)^{i+j} ;dx\
          %
          &= frac{(-1)^n }{n^{2n}} sum_{i=0}^nsum_{j=0}^n (-1)^i begin{bmatrix}n\iend{bmatrix}
          begin{bmatrix}n\jend{bmatrix} frac{(n+1)^{i+j+1}-1}{n(i+j+1)}\
          %
          &= sum_{i=0}^nsum_{j=0}^n (-1)^{i+n}begin{bmatrix}n\iend{bmatrix}
          begin{bmatrix}n\jend{bmatrix} frac{(n+1)^{i+j+1}-1}{n^{2n+1}(i+j+1)}\
          %
          end{align}







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 19 at 8:43









          adfriedmanadfriedman

          3,171169




          3,171169












          • $begingroup$
            I think the sum is too complicated, and the Stirling numbers of the first kind are also hard to compute. But this does give me a new way to deal with this problem. Thank you.
            $endgroup$
            – kelvin hong 方
            Jan 19 at 12:14










          • $begingroup$
            It was likely obvious to you that, as per the other answer, you can expand the product as an $2n$-degree polynomial. I felt that doing a OEIS coefficient search and presenting that, without even proving the connection, fell short of an answer.
            $endgroup$
            – adfriedman
            Jan 19 at 18:58










          • $begingroup$
            Yes, to evaluate this, we still need computer, I think the simplest form is the sum, can be impossible to represented as a term.
            $endgroup$
            – kelvin hong 方
            Jan 20 at 4:05


















          • $begingroup$
            I think the sum is too complicated, and the Stirling numbers of the first kind are also hard to compute. But this does give me a new way to deal with this problem. Thank you.
            $endgroup$
            – kelvin hong 方
            Jan 19 at 12:14










          • $begingroup$
            It was likely obvious to you that, as per the other answer, you can expand the product as an $2n$-degree polynomial. I felt that doing a OEIS coefficient search and presenting that, without even proving the connection, fell short of an answer.
            $endgroup$
            – adfriedman
            Jan 19 at 18:58










          • $begingroup$
            Yes, to evaluate this, we still need computer, I think the simplest form is the sum, can be impossible to represented as a term.
            $endgroup$
            – kelvin hong 方
            Jan 20 at 4:05
















          $begingroup$
          I think the sum is too complicated, and the Stirling numbers of the first kind are also hard to compute. But this does give me a new way to deal with this problem. Thank you.
          $endgroup$
          – kelvin hong 方
          Jan 19 at 12:14




          $begingroup$
          I think the sum is too complicated, and the Stirling numbers of the first kind are also hard to compute. But this does give me a new way to deal with this problem. Thank you.
          $endgroup$
          – kelvin hong 方
          Jan 19 at 12:14












          $begingroup$
          It was likely obvious to you that, as per the other answer, you can expand the product as an $2n$-degree polynomial. I felt that doing a OEIS coefficient search and presenting that, without even proving the connection, fell short of an answer.
          $endgroup$
          – adfriedman
          Jan 19 at 18:58




          $begingroup$
          It was likely obvious to you that, as per the other answer, you can expand the product as an $2n$-degree polynomial. I felt that doing a OEIS coefficient search and presenting that, without even proving the connection, fell short of an answer.
          $endgroup$
          – adfriedman
          Jan 19 at 18:58












          $begingroup$
          Yes, to evaluate this, we still need computer, I think the simplest form is the sum, can be impossible to represented as a term.
          $endgroup$
          – kelvin hong 方
          Jan 20 at 4:05




          $begingroup$
          Yes, to evaluate this, we still need computer, I think the simplest form is the sum, can be impossible to represented as a term.
          $endgroup$
          – kelvin hong 方
          Jan 20 at 4:05


















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