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The Bernoulli polynomials $B_k(.)$ are given by $$ \frac{t\:e^{xt}}{e^t-1}=\sum_{k=0}^\infty B_n(x)\frac{t^n}{n!}, \quad |t|<2\pi. \tag{1} $$ I would like to prove that $$ \int_0^1 B_n(x) dx=0, \quad n\geq1. \tag2 $$ Do you know an easy way to prove it?

I've seen that: $$ \int_0^1 B_1(x) dx=\int_0^1 \left(x-\frac12\right) dx=[\frac{x^2}{2}-\frac x2]_0^1=0 .$$

Thanks for your help.

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Use the identity $$\sum_{n=0}^\infty \frac{t^n}{n!}\int_0^1B_n(x)\;\mathrm dx=\int_0^1\frac{t\,\mathrm e^{xt}}{e^t-1}\;\mathrm dx=\frac{t}{\mathrm e^t-1}\int_0^1\mathrm e^{xt}\;\mathrm dx=\frac{t}{\mathrm e^t-1}\cdot\left.\frac{\mathrm e^{xt}}t\right|_{x=0}^{x=1}=1.$$ Likewise, for every $u$ and every $n\geqslant1$, $$\int_u^{u+1}B_n(x)\;\mathrm dx=u^n.$$

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  • $\begingroup$ Ok, thank you very much! $\endgroup$ – G. Peano Oct 13 '14 at 10:42
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I do not know how much this could help you $$\int_a^x B_n(t) dt=\frac{B_{n+1}(x)-B_{n+1}(a)}{n+1}$$ and $$B_{n}(x)=\sum _{k=0}^n \binom{n}{k}B_{n-k}~~ x^k $$

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