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

For $m,n\in\Bbb N$, $$I(m,n):=\int_0^1B_n(x)B_m(x)\mathrm{d}x=(-1)^{n-1}\frac{m!n!}{(m+n)!}b_{n+m}$$ 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
$$I(m,n)=(-1)^{n-1}b_{n+m}\int_0^1t^n(1-t)^m\mathrm{d}t$$

Here's what I've tried.

Because $B_n(x)$ satisfies $$B_n'(x)=nB_{n-1}(x)$$ We can integrate by parts with $\mathrm{d}v=B_n(x)\mathrm{d}x$: $$I(m,n)=\frac1{n+1}B_{n+1}(x)B_m(x)\bigg|_0^1-\frac m{n+1}\int_0^1B_{n+1}(x)B_{m-1}(x)\mathrm{d}x$$ $$I(m,n)=\frac1{n+1}\bigg(B_{n+1}(1)B_m(1)-b_{n+1}b_m\bigg)-\frac{m}{n+1}I(m-1,n+1)$$ 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.

clathratus
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1 Answers1

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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 $k\in\Bbb N$, $I(0,k)=\displaystyle\int_0^1 B_k(x)~\mathrm dx=0$.

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 $m,n\ne1$, 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)\cdots\left(-\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.

Tianlalu
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