Bernoulli polynomials link a simple differential recursion to Fourier series and definite integrals. We begin with their defining conditions, derive the identities we need, and then apply them to an integral.
Bernoulli polynomials
Define the sequence \(\{B_r(x)\}_{r=0}^{\infty}\) by
\[
\begin{aligned}
B_0(x) &= 1 \\
B_r^\prime(x) &= r B_{r-1}(x) \quad \forall r\geq 1 \\
\int_0^1 B_r(x) dx &= 0 \quad \forall r\geq 1
\end{aligned}
\]
The first few Bernoulli Polynomials are:
\[
\begin{aligned}
B_0(x) & =1 \\
B_1(x) & =x-\frac{1}{2} \\
B_2(x) & =x^2-x+\frac{1}{6} \\
B_3(x) & =x^3-\frac{3}{2}x^2+\frac{1}{2}x \\
B_4(x) & =x^4-2x^3+x^2-\frac{1}{30} \\
\end{aligned}
\]
The numbers \(B_n = B_n(0)\) are called the Bernoulli numbers. Integrating the relation \(B_r^\prime(x)=rB_r(x)\) between 0 and 1 gives:
\[
B_r(1)-B_r(0) = \int_0^1 B_r^\prime (x) dx = r\int_0^1 B_{r-1}(x) dx = 0 \quad \forall r\geq 2
\]
This motivates us to define the periodic Bernoulli polynomials by
\[
\tilde{B}_r(x) = B_r(\langle x\rangle), \quad x\in\mathbb{R}, \; r\geq 2
\]
where \(\langle x\rangle\) denotes the fractional part of \(x\). We will now compute the Fourier series of \(\tilde{B}_r(x)\), where \(r\geq 2\). The \(n\)-th Fourier coefficient is given by:
\[
\begin{aligned}
a_n &= \int_0^1 \tilde{B}_r(x) e^{-2\pi i n x} dx= \int_0^1 B_r(x) e^{-2\pi i n x} dx
\end{aligned}
\]
Let's first consider the case when \(n\neq 0\). Integration by parts, gives us:
\[
\begin{equation}\label{bernoulli-numbers-and-a-related-integral-eq-1}
\begin{aligned}
a_n &= -\frac{e^{-2\pi i n x}}{2\pi i n}B_r(x)\Big|_0^1 + \frac{1}{2\pi i n}\int_0^1 B_r^\prime (x) e^{-2\pi i n x} dx \\
&= \frac{1}{2\pi i n}\int_0^1 B_r^\prime (x) e^{-2\pi i n x} dx \\
&= \frac{r}{2\pi i n}\int_0^1 B_{r-1}(x) e^{-2\pi i n x} dx
\end{aligned}
\end{equation}
\]
The repeated use of equation \(\eqref{bernoulli-numbers-and-a-related-integral-eq-1}\) gives:
\[
\begin{aligned}
a_n &= \frac{r!}{(2\pi i n)^{r-1}} \int_0^1 B_1(x) e^{-2\pi i nx} dx \\
&= \frac{r!}{(2\pi i n)^{r-1}} \int_0^1 \left(x-\frac{1}{2} \right) e^{-2\pi i nx} dx \\
&= \frac{r!}{(2\pi i n)^{r-1}}\int_0^1 x e^{-2\pi i n x} dx \\
&= -\frac{r!}{(2\pi i n)^r}
\end{aligned}
\]
When \(n=0\), we have \(a_0 = \int_0^1 B_r(x)dx = 0\). Note that the Fourier series
\[
-r! \sum_{\substack{n=-\infty \\ n\neq 0}} \frac{e^{2\pi i n x}}{(2\pi i n)^r}
\]
converges absolutely for all \(r\geq 2\). Therefore, it converges uniformly to \(\tilde{B}_r(x)\) for all \(r\geq 2\). This leads to the following bound:
\[
\begin{equation}\label{bernoulli-numbers-and-a-related-integral-eq-2}
|\tilde{B}_r(x)| \leq \frac{2r!}{(2\pi)^r}\sum_{n=1}^\infty \frac{1}{n^r} < \frac{4r!}{(2\pi)^r} \;\; \forall r\geq 2
\end{equation}
\]
Note that the above inequality also remains valid for \(r=0\) and \(r=1\). Now, let's consider the generating function:
\[
F(x,t) = \sum_{n=0}^\infty \frac{\tilde{B}_n(x) t^n}{n!}
\]
The inequality \(\eqref{bernoulli-numbers-and-a-related-integral-eq-2}\) implies that:
\[
|F(x,t)| < 4 \sum_{n=0}^\infty \left(\frac{t}{2\pi}\right)^n
\]
Therefore, the series converges uniformly for all \(t\in [0,2\pi]\) and all \(x\). We, may, therefore differentiate term by term to obtain:
\[
\frac{\partial F(x,t)}{\partial x} = \sum_{n=1}^\infty \frac{\tilde{B}_{n-1}(x)}{(n-1)!}t^n = t F(x,t)
\]
Solving the above differential equation, we get \(F(x,t) = G(t) e^{xt}\) where \(G\) is some arbitrary function of \(t\). Next, we integrate \(F(x,t)\) between 0 and 1:
\[
\begin{aligned}
\int_0^1 F(x,t) dx &= G(t) \int_0^1 e^{xt} dx \\
&= G(t) \frac{e^t-1}{t}
\end{aligned}
\]
On the other hand, note that:
\[
\begin{aligned}
\int_0^1 F(x,t) dx &= \int_0^1 \sum_{n=0}^\infty \frac{\tilde{B}_n(x) t^n}{n!} dx \\
&= 1 + \sum_{n=1}^\infty \frac{t^n}{n!}\int_0^1 B_n(x) dx \\
&= 1
\end{aligned}
\]
Therefore, we obtain \(G(t) = \frac{t}{e^t - 1}\) and
\[
\boxed{F(x,t) = \frac{t e^{xt}}{e^t-1}}
\]
An interesting property of the Bernoulli numbers is that \(B_{2n+1}=0\) for all \(n\geq 1\). To see this, consider:
\[
\frac{t}{e^t -1} + \frac{t}{2}= 1+\sum_{n=2}^\infty \frac{B_n t^n}{n!}
\]
Now, on the left hand side we have an even function of \(t\). Therefore, the coefficients of the odd powers of \(t\) on the right hand side are equal to 0. Using the Fourier series expansion, we can express the even-index Bernoulli numbers in terms of the Riemann zeta function:
\[
B_{2n} = \frac{2 (-1)^{n-1} (2n)!}{(2\pi)^{2n}} \zeta(2n)
\]
The Bernoulli polynomials satisfy the following recursive equation:
\[
{B}_n(x) = \sum_{k=0}^n \binom{n}{k} B_{n-k} x^k
\]
This can be proved by noting that:
\[
\begin{aligned}
\sum_{n=0}^\infty \frac{{B}_n(x) t^n}{n!} &= \frac{te^{xt}}{e^t-1} \\
&= e^{xt} \sum_{n=0}^\infty \frac{B_n t^n}{n!} \\
&= \sum_{m=0}^\infty \frac{(xt)^m}{m!} \sum_{n=0}^\infty \frac{B_n t^n}{n!} \\
&= \sum_{m=0}^\infty \sum_{n=0}^\infty \frac{B_n x^m t^{n+m}}{n! m!} \\
&= \sum_{n=0}^\infty \sum_{k=0}^n \frac{B_k t^n x^{n-k}}{k! (n-k)!} \\
&= \sum_{n=0}^\infty \frac{t^n}{n!}\sum_{k=0}^n \binom{n}{k} B_k x^{n-k}
\end{aligned}
\]
where \(x\in [0,1]\). Now, compare the coefficients of \(t^n\) to get the desired result. Plugging in \(x=1\), gives the identity:
\[
\sum_{k=0}^{n-1} \binom{n}{k} B_k = 0
\]
Now, let's turn our attention to the integral:
\[
I=\int_0^{\frac{\pi}{2}}\frac{\sin(2nx)}{\sin^{2n+2}(x)}\cdot \frac{1}{e^{2\pi \cot x}-1} dx
\]
where \(n\in\mathbb{N}\). We will use the following trigonometric identity:
\[
\frac{\sin(2nx)}{\sin^{2n}(x)} =(-1)^n \sum_{r=1}^{n}\binom{2n}{2r-1} (-1)^{r}\cot^{2r-1}(x)
\]
Substituting the above into the integral, gives:
\[
\begin{aligned}
I &= (-1)^n \int_0^{\pi\over 2}\left( \sum_{r=1}^{n}\binom{2n}{2r-1} (-1)^{r}\cot^{2r-1}(x)\right)\frac{\csc^2(x)}{e^{2\pi \cot x}-1}dx \\
&= (-1)^n \sum_{r=1}^{n}\binom{2n}{2r-1} (-1)^{r} \int_0^{\pi \over 2}\cot^{2r-1}(x)\frac{\csc^2(x)}{e^{2\pi \cot x}-1}dx \\
&= (-1)^n \sum_{r=1}^{n}\binom{2n}{2r-1} (-1)^{r} \int_0^\infty \frac{t^{2r-1}}{e^{2\pi t}-1}dt \\
&= (-1)^n \sum_{r=1}^{n}\binom{2n}{2r-1} (-1)^{r}\frac{(2r-1)! \zeta(2r)}{(2\pi)^r }\\
&= (-1)^n \sum_{r=1}^{n}\binom{2n}{2r-1} (-1)^{r} (-1)^{r-1} \frac{B_{2r}}{4r}\\
&= \frac{(-1)^{n-1}}{4}\sum_{r=1}^{n}\binom{2n}{2r-1}\frac{B_{2r}}{r} \\
&= \frac{(-1)^{n-1}}{2(2n+1)}\sum_{r=1}^n \binom{2n+1}{2r} B_{2r} \\
&= \frac{(-1)^{n-1}}{2(2n+1)} \left[\sum_{r=0}^{2n} \binom{2n+1}{r} B_r - \binom{2n+1}{0}B_0 - \binom{2n+1}{1} B_1\right] \\
&= \frac{(-1)^{n-1}}{2(2n+1)} \left[-\binom{2n+1}{0}B_0 - \binom{2n+1}{1} B_1\right] \\
&= \frac{(-1)^{n-1}}{4}\cdot \frac{2n-1}{2n+1}
\end{aligned}
\]