Variant of Euler-Mascheroni constants

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For any positive integers $kgeq 1$ and $jgeq2$, let $x_j=fracj+sqrtj^2-42$. Let us define $textA_k$ to be the following constant :



$$textA_k = lim_ntoinftybigg(sum_j=2^nfracln^k(x_j)j-fracln^k+1(x_n)k+1bigg)$$



Is it possible to evaluate $textA_1$ and $textA_2$ in closed-form ?




binomial-coefficients zeta-functions eulers-constant




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  • An answer of this question will help to find the closed form of some intresting fractional part integrals.
    – Kays Tomy
    Jul 31 at 9:07














up vote
2
down vote

favorite
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For any positive integers $kgeq 1$ and $jgeq2$, let $x_j=fracj+sqrtj^2-42$. Let us define $textA_k$ to be the following constant :



$$textA_k = lim_ntoinftybigg(sum_j=2^nfracln^k(x_j)j-fracln^k+1(x_n)k+1bigg)$$



Is it possible to evaluate $textA_1$ and $textA_2$ in closed-form ?




binomial-coefficients zeta-functions eulers-constant




share|cite|improve this question





















  • An answer of this question will help to find the closed form of some intresting fractional part integrals.
    – Kays Tomy
    Jul 31 at 9:07












up vote
2
down vote

favorite
1









up vote
2
down vote

favorite
1






1





For any positive integers $kgeq 1$ and $jgeq2$, let $x_j=fracj+sqrtj^2-42$. Let us define $textA_k$ to be the following constant :



$$textA_k = lim_ntoinftybigg(sum_j=2^nfracln^k(x_j)j-fracln^k+1(x_n)k+1bigg)$$



Is it possible to evaluate $textA_1$ and $textA_2$ in closed-form ?




binomial-coefficients zeta-functions eulers-constant




share|cite|improve this question













For any positive integers $kgeq 1$ and $jgeq2$, let $x_j=fracj+sqrtj^2-42$. Let us define $textA_k$ to be the following constant :



$$textA_k = lim_ntoinftybigg(sum_j=2^nfracln^k(x_j)j-fracln^k+1(x_n)k+1bigg)$$



Is it possible to evaluate $textA_1$ and $textA_2$ in closed-form ?





binomial-coefficients zeta-functions eulers-constant





share|cite|improve this question












share|cite|improve this question




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edited Jul 31 at 1:19
























asked Jul 31 at 1:11









Kays Tomy

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  • An answer of this question will help to find the closed form of some intresting fractional part integrals.
    – Kays Tomy
    Jul 31 at 9:07
















  • An answer of this question will help to find the closed form of some intresting fractional part integrals.
    – Kays Tomy
    Jul 31 at 9:07















An answer of this question will help to find the closed form of some intresting fractional part integrals.
– Kays Tomy
Jul 31 at 9:07




An answer of this question will help to find the closed form of some intresting fractional part integrals.
– Kays Tomy
Jul 31 at 9:07










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Theorem



For any positive integers $kgeq1$ and $jgeq2$, let $x_j=fracj+sqrtj^2-42$. Let us define $(textA_k)_kgeq1$ by the following constants "which are variants of Stieltjes constants" :



$$textA_k = lim_ntoinftybigg(sum_j=2^nfracln^k(x_j)j-fracln^k+1(x_n)k+1bigg)$$



Then following a straightforward calculus we can get the following identity :



$$int_0^1int_0^1int_0^1biggx,y,z+frac1x,y,zbiggdx,dy,dz=frac138-gamma-textA_1-fractextA_22$$



And more generally we have :



$$int_0^1int_0^1dotsint_0^1biggprod_k=1^nx_k+frac1prod_k=1^nx_kbiggdx_1,dx_2dots,dx_n=frac138-gamma-sum_j=1^n-1fractextA_jj!$$



Where $gamma$ represents the Euler-Mascheroni constant and $$ denotes the fractional part. If the constants $(textA_k)_kgeq1$ are calculable in closed-form then they will be replaced in the above multiple-integrals.






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    Theorem



    For any positive integers $kgeq1$ and $jgeq2$, let $x_j=fracj+sqrtj^2-42$. Let us define $(textA_k)_kgeq1$ by the following constants "which are variants of Stieltjes constants" :



    $$textA_k = lim_ntoinftybigg(sum_j=2^nfracln^k(x_j)j-fracln^k+1(x_n)k+1bigg)$$



    Then following a straightforward calculus we can get the following identity :



    $$int_0^1int_0^1int_0^1biggx,y,z+frac1x,y,zbiggdx,dy,dz=frac138-gamma-textA_1-fractextA_22$$



    And more generally we have :



    $$int_0^1int_0^1dotsint_0^1biggprod_k=1^nx_k+frac1prod_k=1^nx_kbiggdx_1,dx_2dots,dx_n=frac138-gamma-sum_j=1^n-1fractextA_jj!$$



    Where $gamma$ represents the Euler-Mascheroni constant and $$ denotes the fractional part. If the constants $(textA_k)_kgeq1$ are calculable in closed-form then they will be replaced in the above multiple-integrals.






    share|cite|improve this answer

























      up vote
      0
      down vote













      Theorem



      For any positive integers $kgeq1$ and $jgeq2$, let $x_j=fracj+sqrtj^2-42$. Let us define $(textA_k)_kgeq1$ by the following constants "which are variants of Stieltjes constants" :



      $$textA_k = lim_ntoinftybigg(sum_j=2^nfracln^k(x_j)j-fracln^k+1(x_n)k+1bigg)$$



      Then following a straightforward calculus we can get the following identity :



      $$int_0^1int_0^1int_0^1biggx,y,z+frac1x,y,zbiggdx,dy,dz=frac138-gamma-textA_1-fractextA_22$$



      And more generally we have :



      $$int_0^1int_0^1dotsint_0^1biggprod_k=1^nx_k+frac1prod_k=1^nx_kbiggdx_1,dx_2dots,dx_n=frac138-gamma-sum_j=1^n-1fractextA_jj!$$



      Where $gamma$ represents the Euler-Mascheroni constant and $$ denotes the fractional part. If the constants $(textA_k)_kgeq1$ are calculable in closed-form then they will be replaced in the above multiple-integrals.






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        Theorem



        For any positive integers $kgeq1$ and $jgeq2$, let $x_j=fracj+sqrtj^2-42$. Let us define $(textA_k)_kgeq1$ by the following constants "which are variants of Stieltjes constants" :



        $$textA_k = lim_ntoinftybigg(sum_j=2^nfracln^k(x_j)j-fracln^k+1(x_n)k+1bigg)$$



        Then following a straightforward calculus we can get the following identity :



        $$int_0^1int_0^1int_0^1biggx,y,z+frac1x,y,zbiggdx,dy,dz=frac138-gamma-textA_1-fractextA_22$$



        And more generally we have :



        $$int_0^1int_0^1dotsint_0^1biggprod_k=1^nx_k+frac1prod_k=1^nx_kbiggdx_1,dx_2dots,dx_n=frac138-gamma-sum_j=1^n-1fractextA_jj!$$



        Where $gamma$ represents the Euler-Mascheroni constant and $$ denotes the fractional part. If the constants $(textA_k)_kgeq1$ are calculable in closed-form then they will be replaced in the above multiple-integrals.






        share|cite|improve this answer













        Theorem



        For any positive integers $kgeq1$ and $jgeq2$, let $x_j=fracj+sqrtj^2-42$. Let us define $(textA_k)_kgeq1$ by the following constants "which are variants of Stieltjes constants" :



        $$textA_k = lim_ntoinftybigg(sum_j=2^nfracln^k(x_j)j-fracln^k+1(x_n)k+1bigg)$$



        Then following a straightforward calculus we can get the following identity :



        $$int_0^1int_0^1int_0^1biggx,y,z+frac1x,y,zbiggdx,dy,dz=frac138-gamma-textA_1-fractextA_22$$



        And more generally we have :



        $$int_0^1int_0^1dotsint_0^1biggprod_k=1^nx_k+frac1prod_k=1^nx_kbiggdx_1,dx_2dots,dx_n=frac138-gamma-sum_j=1^n-1fractextA_jj!$$



        Where $gamma$ represents the Euler-Mascheroni constant and $$ denotes the fractional part. If the constants $(textA_k)_kgeq1$ are calculable in closed-form then they will be replaced in the above multiple-integrals.







        share|cite|improve this answer













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        answered Aug 1 at 1:11









        Kays Tomy

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