## bookmark_borderBernoulli numbers and a related integral

Consider the sequence $\{B_r(x)\}_{r=0}^{\infty}$ of polynomials defined using the recursion:

\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{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 \quad \quad (1) \end{aligned} The repeated use of equation (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: $$|\tilde{B}_r(x)| \leq \frac{2r!}{(2\pi)^r}\sum_{n=1}^\infty \frac{1}{n^r} < \frac{4r!}{(2\pi)^r} \;\; \forall r\geq 2\quad\quad (2)$$

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 (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}