Editing Bernoulli polynomials (section)

==Integrals==
Two definite integrals relating the Bernoulli and Euler polynomials to the Bernoulli and Euler numbers are:<ref>{{cite journal |name-list-style=amp |author1=Takashi Agoh |author2=Karl Dilcher  |journal=Journal of Mathematical Analysis and Applications |volume=381 |year=2011 |pages=10–16 |title=Integrals of products of Bernoulli polynomials | doi=10.1016/j.jmaa.2011.03.061  |doi-access=free }}</ref>
*<math>\int_0^1 B_n(t) B_m(t)\,dt = (-1)^{n-1} \frac{m!\, n!}{(m+n)!} B_{n+m} \quad \text{for } m,n \geq 1 </math>
*<math>\int_0^1 E_n(t) E_m(t)\,dt = (-1)^{n} 4 (2^{m+n+2}-1)\frac{m!\,n!}{(m+n+2)!} B_{n+m+2}</math>

Another integral formula states<ref>{{cite journal | author=Elaissaoui, Lahoucine  | author2=Guennoun, Zine El Abidine | name-list-style=amp | title=Evaluation of log-tangent integrals by series involving ζ(2n+1)| journal=Integral Transforms and Special Functions | language=English  | year=2017| volume=28 | issue=6 | pages=460–475 | doi=10.1080/10652469.2017.1312366 | arxiv=1611.01274 | s2cid=119132354 }}</ref>
*<math>\int_0^{1}E_{n}\left( x +y\right)\log(\tan \frac{\pi}{2}x)\,dx= n! \sum_{k=1}^{\left\lfloor\frac {n+1}2\right\rfloor} \frac{(-1)^{k-1}}{ \pi^{2k}}   \left( 2-2^{-2k} \right)\zeta(2k+1) \frac{y^ {n+1-2k}}{(n +1- 2k)!}</math>
with the special case for <math>y=0</math>
*<math>\int_0^{1}E_{2n-1}\left( x \right)\log(\tan \frac{\pi}{2}x)\,dx=
\frac{(-1)^{n-1}(2n-1)!}{\pi^{2n}}\left( 2-2^{-2n} \right)\zeta(2n+1)</math>
*<math>\int_0^{1}B_{2n-1}\left( x \right)\log(\tan \frac{\pi}{2}x)\,dx=
\frac{(-1)^{n-1}}{\pi^{2n}}\frac{2^{2n-2}}{(2n-1)!}\sum_{k=1}^{n}( 2^{2k+1}-1 )\zeta(2k+1)\zeta(2n-2k)</math>
*<math>\int_0^{1}E_{2n}\left( x \right)\log(\tan \frac{\pi}{2}x)\,dx=\int_0^{1}B_{2n}\left( x \right)\log(\tan \frac{\pi}{2}x)\,dx=0</math>
*<math>\int_{0}^{1}{{{B}_{2n-1}}\left( x \right)\cot \left( \pi x \right)dx}=\frac{2\left( 2n-1 \right)!}{{{\left( -1 \right)}^{n-1}}{{\left( 2\pi \right)}^{2n-1}}}\zeta \left( 2n-1 \right)</math>