Integral identity involving Bernoulli polynomials












3












$begingroup$


I found the following identity on Wikipedia, and I am having a difficult time proving it.



For $m,ninBbb N$,
$$I(m,n):=int_0^1B_n(x)B_m(x)mathrm{d}x=(-1)^{n-1}frac{m!n!}{(m+n)!}b_{n+m}$$
Where $B_n(x)$ is the $n$-th Bernoulli polynomial, and $b_n=B_n(0)$ is the $n$-th Bernoulli number. Notably, when one uses $x!=Gamma(x+1)$ on the RHS and then uses the Beta function, we arrive at
$$I(m,n)=(-1)^{n-1}b_{n+m}int_0^1t^n(1-t)^mmathrm{d}t$$



Here's what I've tried.



Because $B_n(x)$ satisfies
$$B_n'(x)=nB_{n-1}(x)$$
We can integrate by parts with $mathrm{d}v=B_n(x)mathrm{d}x$:
$$I(m,n)=frac1{n+1}B_{n+1}(x)B_m(x)bigg|_0^1-frac m{n+1}int_0^1B_{n+1}(x)B_{m-1}(x)mathrm{d}x$$
$$I(m,n)=frac1{n+1}bigg(B_{n+1}(1)B_m(1)-b_{n+1}b_mbigg)-frac{m}{n+1}I(m-1,n+1)$$
Which is a start, but I don't know how to proceed. I would think that the integral should be easy given the fact that Bernoulli polynomials have so many identities, yet I can't seem to find the right one to use. Could you either point me in the right direction, or show me a complete proof? Thanks.






integration combinatorics number-theory polynomials bernoulli-polynomials







share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    I found the following identity on Wikipedia, and I am having a difficult time proving it.



    For $m,ninBbb N$,
    $$I(m,n):=int_0^1B_n(x)B_m(x)mathrm{d}x=(-1)^{n-1}frac{m!n!}{(m+n)!}b_{n+m}$$
    Where $B_n(x)$ is the $n$-th Bernoulli polynomial, and $b_n=B_n(0)$ is the $n$-th Bernoulli number. Notably, when one uses $x!=Gamma(x+1)$ on the RHS and then uses the Beta function, we arrive at
    $$I(m,n)=(-1)^{n-1}b_{n+m}int_0^1t^n(1-t)^mmathrm{d}t$$



    Here's what I've tried.



    Because $B_n(x)$ satisfies
    $$B_n'(x)=nB_{n-1}(x)$$
    We can integrate by parts with $mathrm{d}v=B_n(x)mathrm{d}x$:
    $$I(m,n)=frac1{n+1}B_{n+1}(x)B_m(x)bigg|_0^1-frac m{n+1}int_0^1B_{n+1}(x)B_{m-1}(x)mathrm{d}x$$
    $$I(m,n)=frac1{n+1}bigg(B_{n+1}(1)B_m(1)-b_{n+1}b_mbigg)-frac{m}{n+1}I(m-1,n+1)$$
    Which is a start, but I don't know how to proceed. I would think that the integral should be easy given the fact that Bernoulli polynomials have so many identities, yet I can't seem to find the right one to use. Could you either point me in the right direction, or show me a complete proof? Thanks.






    integration combinatorics number-theory polynomials bernoulli-polynomials







    share|cite|improve this question











    $endgroup$















      3












      3








      3


      2



      $begingroup$


      I found the following identity on Wikipedia, and I am having a difficult time proving it.



      For $m,ninBbb N$,
      $$I(m,n):=int_0^1B_n(x)B_m(x)mathrm{d}x=(-1)^{n-1}frac{m!n!}{(m+n)!}b_{n+m}$$
      Where $B_n(x)$ is the $n$-th Bernoulli polynomial, and $b_n=B_n(0)$ is the $n$-th Bernoulli number. Notably, when one uses $x!=Gamma(x+1)$ on the RHS and then uses the Beta function, we arrive at
      $$I(m,n)=(-1)^{n-1}b_{n+m}int_0^1t^n(1-t)^mmathrm{d}t$$



      Here's what I've tried.



      Because $B_n(x)$ satisfies
      $$B_n'(x)=nB_{n-1}(x)$$
      We can integrate by parts with $mathrm{d}v=B_n(x)mathrm{d}x$:
      $$I(m,n)=frac1{n+1}B_{n+1}(x)B_m(x)bigg|_0^1-frac m{n+1}int_0^1B_{n+1}(x)B_{m-1}(x)mathrm{d}x$$
      $$I(m,n)=frac1{n+1}bigg(B_{n+1}(1)B_m(1)-b_{n+1}b_mbigg)-frac{m}{n+1}I(m-1,n+1)$$
      Which is a start, but I don't know how to proceed. I would think that the integral should be easy given the fact that Bernoulli polynomials have so many identities, yet I can't seem to find the right one to use. Could you either point me in the right direction, or show me a complete proof? Thanks.






      integration combinatorics number-theory polynomials bernoulli-polynomials







      share|cite|improve this question











      $endgroup$




      I found the following identity on Wikipedia, and I am having a difficult time proving it.



      For $m,ninBbb N$,
      $$I(m,n):=int_0^1B_n(x)B_m(x)mathrm{d}x=(-1)^{n-1}frac{m!n!}{(m+n)!}b_{n+m}$$
      Where $B_n(x)$ is the $n$-th Bernoulli polynomial, and $b_n=B_n(0)$ is the $n$-th Bernoulli number. Notably, when one uses $x!=Gamma(x+1)$ on the RHS and then uses the Beta function, we arrive at
      $$I(m,n)=(-1)^{n-1}b_{n+m}int_0^1t^n(1-t)^mmathrm{d}t$$



      Here's what I've tried.



      Because $B_n(x)$ satisfies
      $$B_n'(x)=nB_{n-1}(x)$$
      We can integrate by parts with $mathrm{d}v=B_n(x)mathrm{d}x$:
      $$I(m,n)=frac1{n+1}B_{n+1}(x)B_m(x)bigg|_0^1-frac m{n+1}int_0^1B_{n+1}(x)B_{m-1}(x)mathrm{d}x$$
      $$I(m,n)=frac1{n+1}bigg(B_{n+1}(1)B_m(1)-b_{n+1}b_mbigg)-frac{m}{n+1}I(m-1,n+1)$$
      Which is a start, but I don't know how to proceed. I would think that the integral should be easy given the fact that Bernoulli polynomials have so many identities, yet I can't seem to find the right one to use. Could you either point me in the right direction, or show me a complete proof? Thanks.






      integration combinatorics number-theory polynomials bernoulli-polynomials




      integration combinatorics number-theory polynomials bernoulli-polynomials






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      edited Jan 29 at 21:28







      clathratus

















      asked Nov 28 '18 at 22:54









      clathratusclathratus

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          1 Answer
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          2












          $begingroup$

          Here are some properties,




          • for integer $k>1$, $B_k(0)=B_k(1)$,


          • $B_1(1)=-B_1(0)=dfrac 12, B_0(x)equiv1$,


          • for $kinBbb N$, $I(0,k)=displaystyleint_0^1 B_k(x)~mathrm dx=0$.



          Therefore,




          • for integers $r, s>1$,
            $$big[B_r(x)B_s(x)big]_0^1= 0,$$


          • for integer $r>1$,
            $$big[B_r(1)B_1(1)big]_0^1=frac12(B_r(1)+B_r(0))=B_r(0),$$



          Thus, when $m,nne1$, apply IBP,
          begin{align*}
          I(m,n)&=-frac m{n+1} I(m-1,n+1)\
          &=left(-frac m{n+1}right)left(-frac {m-1}{n+2}right)cdotsleft(-frac {2}{n+m-1}right)color{blue}{I(1,n+m-1)}\
          &=frac{(-1)^{m-1}m!}{(n+1)(n+2)cdots(m+n-1)}color{blue}{frac1{n+m}Big{underbrace{big[B_{m+n}(x)B_1(x)big]_0^1}_{=B_{m+n}(0)}-underbrace{I(0,n+m)}_{=0}Big}}\
          &=frac{(-1)^{m-1}m!n!}{(m+n)!}B_{m+n}(0).\
          end{align*}



          The case $m=n=1$ is followed by direct computations.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            This is perfect and amazing and beautiful. Thank you very much.
            $endgroup$
            – clathratus
            Nov 29 '18 at 19:41












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          active

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          2












          $begingroup$

          Here are some properties,




          • for integer $k>1$, $B_k(0)=B_k(1)$,


          • $B_1(1)=-B_1(0)=dfrac 12, B_0(x)equiv1$,


          • for $kinBbb N$, $I(0,k)=displaystyleint_0^1 B_k(x)~mathrm dx=0$.



          Therefore,




          • for integers $r, s>1$,
            $$big[B_r(x)B_s(x)big]_0^1= 0,$$


          • for integer $r>1$,
            $$big[B_r(1)B_1(1)big]_0^1=frac12(B_r(1)+B_r(0))=B_r(0),$$



          Thus, when $m,nne1$, apply IBP,
          begin{align*}
          I(m,n)&=-frac m{n+1} I(m-1,n+1)\
          &=left(-frac m{n+1}right)left(-frac {m-1}{n+2}right)cdotsleft(-frac {2}{n+m-1}right)color{blue}{I(1,n+m-1)}\
          &=frac{(-1)^{m-1}m!}{(n+1)(n+2)cdots(m+n-1)}color{blue}{frac1{n+m}Big{underbrace{big[B_{m+n}(x)B_1(x)big]_0^1}_{=B_{m+n}(0)}-underbrace{I(0,n+m)}_{=0}Big}}\
          &=frac{(-1)^{m-1}m!n!}{(m+n)!}B_{m+n}(0).\
          end{align*}



          The case $m=n=1$ is followed by direct computations.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            This is perfect and amazing and beautiful. Thank you very much.
            $endgroup$
            – clathratus
            Nov 29 '18 at 19:41
















          2












          $begingroup$

          Here are some properties,




          • for integer $k>1$, $B_k(0)=B_k(1)$,


          • $B_1(1)=-B_1(0)=dfrac 12, B_0(x)equiv1$,


          • for $kinBbb N$, $I(0,k)=displaystyleint_0^1 B_k(x)~mathrm dx=0$.



          Therefore,




          • for integers $r, s>1$,
            $$big[B_r(x)B_s(x)big]_0^1= 0,$$


          • for integer $r>1$,
            $$big[B_r(1)B_1(1)big]_0^1=frac12(B_r(1)+B_r(0))=B_r(0),$$



          Thus, when $m,nne1$, apply IBP,
          begin{align*}
          I(m,n)&=-frac m{n+1} I(m-1,n+1)\
          &=left(-frac m{n+1}right)left(-frac {m-1}{n+2}right)cdotsleft(-frac {2}{n+m-1}right)color{blue}{I(1,n+m-1)}\
          &=frac{(-1)^{m-1}m!}{(n+1)(n+2)cdots(m+n-1)}color{blue}{frac1{n+m}Big{underbrace{big[B_{m+n}(x)B_1(x)big]_0^1}_{=B_{m+n}(0)}-underbrace{I(0,n+m)}_{=0}Big}}\
          &=frac{(-1)^{m-1}m!n!}{(m+n)!}B_{m+n}(0).\
          end{align*}



          The case $m=n=1$ is followed by direct computations.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            This is perfect and amazing and beautiful. Thank you very much.
            $endgroup$
            – clathratus
            Nov 29 '18 at 19:41














          2












          2








          2





          $begingroup$

          Here are some properties,




          • for integer $k>1$, $B_k(0)=B_k(1)$,


          • $B_1(1)=-B_1(0)=dfrac 12, B_0(x)equiv1$,


          • for $kinBbb N$, $I(0,k)=displaystyleint_0^1 B_k(x)~mathrm dx=0$.



          Therefore,




          • for integers $r, s>1$,
            $$big[B_r(x)B_s(x)big]_0^1= 0,$$


          • for integer $r>1$,
            $$big[B_r(1)B_1(1)big]_0^1=frac12(B_r(1)+B_r(0))=B_r(0),$$



          Thus, when $m,nne1$, apply IBP,
          begin{align*}
          I(m,n)&=-frac m{n+1} I(m-1,n+1)\
          &=left(-frac m{n+1}right)left(-frac {m-1}{n+2}right)cdotsleft(-frac {2}{n+m-1}right)color{blue}{I(1,n+m-1)}\
          &=frac{(-1)^{m-1}m!}{(n+1)(n+2)cdots(m+n-1)}color{blue}{frac1{n+m}Big{underbrace{big[B_{m+n}(x)B_1(x)big]_0^1}_{=B_{m+n}(0)}-underbrace{I(0,n+m)}_{=0}Big}}\
          &=frac{(-1)^{m-1}m!n!}{(m+n)!}B_{m+n}(0).\
          end{align*}



          The case $m=n=1$ is followed by direct computations.






          share|cite|improve this answer









          $endgroup$



          Here are some properties,




          • for integer $k>1$, $B_k(0)=B_k(1)$,


          • $B_1(1)=-B_1(0)=dfrac 12, B_0(x)equiv1$,


          • for $kinBbb N$, $I(0,k)=displaystyleint_0^1 B_k(x)~mathrm dx=0$.



          Therefore,




          • for integers $r, s>1$,
            $$big[B_r(x)B_s(x)big]_0^1= 0,$$


          • for integer $r>1$,
            $$big[B_r(1)B_1(1)big]_0^1=frac12(B_r(1)+B_r(0))=B_r(0),$$



          Thus, when $m,nne1$, apply IBP,
          begin{align*}
          I(m,n)&=-frac m{n+1} I(m-1,n+1)\
          &=left(-frac m{n+1}right)left(-frac {m-1}{n+2}right)cdotsleft(-frac {2}{n+m-1}right)color{blue}{I(1,n+m-1)}\
          &=frac{(-1)^{m-1}m!}{(n+1)(n+2)cdots(m+n-1)}color{blue}{frac1{n+m}Big{underbrace{big[B_{m+n}(x)B_1(x)big]_0^1}_{=B_{m+n}(0)}-underbrace{I(0,n+m)}_{=0}Big}}\
          &=frac{(-1)^{m-1}m!n!}{(m+n)!}B_{m+n}(0).\
          end{align*}



          The case $m=n=1$ is followed by direct computations.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 29 '18 at 19:35









          TianlaluTianlalu

          3,26521238




          3,26521238












          • $begingroup$
            This is perfect and amazing and beautiful. Thank you very much.
            $endgroup$
            – clathratus
            Nov 29 '18 at 19:41


















          • $begingroup$
            This is perfect and amazing and beautiful. Thank you very much.
            $endgroup$
            – clathratus
            Nov 29 '18 at 19:41
















          $begingroup$
          This is perfect and amazing and beautiful. Thank you very much.
          $endgroup$
          – clathratus
          Nov 29 '18 at 19:41




          $begingroup$
          This is perfect and amazing and beautiful. Thank you very much.
          $endgroup$
          – clathratus
          Nov 29 '18 at 19:41


















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