Consider the sequence $\\{B_r(x)\\}_{r=0}^{\\infty}$ of polynomials defined using the recursion:\n
\nThe first few Bernoulli Polynomials are:\n$$\n\\begin{aligned}\nB_0(x) & =1 \\\\\nB_1(x) & =x-\\frac{1}{2} \\\\\nB_2(x) & =x^2-x+\\frac{1}{6} \\\\\nB_3(x) & =x^3-\\frac{3}{2}x^2+\\frac{1}{2}x \\\\\nB_4(x) & =x^4-2x^3+x^2-\\frac{1}{30} \\\\\n\\end{aligned}\n$$\n<\/p>\n
\nThe 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:\n$$B_r(1)-B_r(0) = \\int_0^1 B_r^\\prime (x) dx = r\\int_0^1 B_{r-1}(x) dx = 0 \\quad \\forall r\\geq 2$$\nThis motivates us to define the periodic Bernoulli polynomials by\n$$\\tilde{B}_r(x) = B_r(\\langle x\\rangle), \\quad x\\in\\mathbb{R}, \\; r\\geq 2$$\nwhere $\\langle x\\rangle$ 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:\n$$\n\\begin{aligned}\na_n &= \\int_0^1 \\tilde{B}_r(x) e^{-2\\pi i n x} dx= \\int_0^1 B_r(x) e^{-2\\pi i n x} dx\n\\end{aligned}\n$$\nLet’s first consider the case when $n\\neq 0$. Integration by parts, gives us:\n$$\n\\begin{aligned}\na_n &= -\\frac{e^{-2\\pi i n x}}{2\\pi i n}B_r(x)\\Big|_0^1 + \\frac{1}{2\\pi i n}\\int_0^1 B_r^\\prime (x) e^{-2\\pi i n x} dx \\\\ \n&= \\frac{1}{2\\pi i n}\\int_0^1 B_r^\\prime (x) e^{-2\\pi i n x} dx \\\\\n&= \\frac{r}{2\\pi i n}\\int_0^1 B_{r-1}(x) e^{-2\\pi i n x} dx \\quad \\quad (1)\n\\end{aligned}\n$$\nThe repeated use of equation (1) gives:\n$$\n\\begin{aligned}\na_n &= \\frac{r!}{(2\\pi i n)^{r-1}} \\int_0^1 B_1(x) e^{-2\\pi i nx} dx \\\\\n&= \\frac{r!}{(2\\pi i n)^{r-1}} \\int_0^1 \\left(x-\\frac{1}{2} \\right) e^{-2\\pi i nx} dx \\\\\n&= \\frac{r!}{(2\\pi i n)^{r-1}}\\int_0^1 x e^{-2\\pi i n x} dx \\\\\n&= -\\frac{r!}{(2\\pi i n)^r}\n\\end{aligned}\n$$\nWhen $n=0$, 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:\n$$|\\tilde{B}_r(x)| \\leq \\frac{2r!}{(2\\pi)^r}\\sum_{n=1}^\\infty \\frac{1}{n^r} < \\frac{4r!}{(2\\pi)^r} \\;\\; \\forall r\\geq 2\\quad\\quad (2)$$\n\n<\/p>\n
\nNote that the above inequality also remains valid for $r=0$ and $r=1$. Now, let’s consider the generating function:\n$$F(x,t) = \\sum_{n=0}^\\infty \\frac{\\tilde{B}_n(x) t^n}{n!} $$\nThe inequality (2) implies that:\n$$\n|F(x,t)| < 4 \\sum_{n=0}^\\infty \\left(\\frac{t}{2\\pi}\\right)^n\n$$\nTherefore, the series converges uniformly for all $t\\in [0,2\\pi]$ and all $x$. We, may, therefore differentiate term by term to obtain:\n$$\n\\frac{\\partial F(x,t)}{\\partial x} = \\sum_{n=1}^\\infty \\frac{\\tilde{B}_{n-1}(x)}{(n-1)!}t^n = t F(x,t)\n$$\nSolving 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:\n$$\n\\begin{aligned}\n\\int_0^1 F(x,t) dx &= G(t) \\int_0^1 e^{xt} dx \\\\\n&= G(t) \\frac{e^t-1}{t}\n\\end{aligned}\n$$\nOn the other hand, note that:\n$$\\begin{aligned}\n\\int_0^1 F(x,t) dx &= \\int_0^1 \\sum_{n=0}^\\infty \\frac{\\tilde{B}_n(x) t^n}{n!} dx \\\\\n&= 1 + \\sum_{n=1}^\\infty \\frac{t^n}{n!}\\int_0^1 B_n(x) dx \\\\\n&= 1\n\\end{aligned}\n$$\nTherefore, we obtain $G(t) = \\frac{t}{e^t – 1}$ and \n$$\\boxed{F(x,t) = \\frac{t e^{xt}}{e^t-1}}$$\n<\/p>\n\n\n\n
\nAn interesting property of the Bernoulli numbers is that $B_{2n+1}=0$ for all $n\\geq 1$. To see this, consider:\n$$\\frac{t}{e^t -1} + \\frac{t}{2}= 1+\\sum_{n=2}^\\infty \\frac{B_n t^n}{n!}$$\nNow, 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:\n$$B_{2n} = \\frac{2 (-1)^{n-1} (2n)!}{(2\\pi)^{2n}} \\zeta(2n)$$ \n<\/p>\n\n\n\n
\nThe Bernoulli polynomials satisfy the following recursive equation:\n$$\n{B}_n(x) = \\sum_{k=0}^n \\binom{n}{k} B_{n-k} x^k\n$$\nThis can be proved by noting that:\n$$\n\\begin{aligned}\n\\sum_{n=0}^\\infty \\frac{{B}_n(x) t^n}{n!} &= \\frac{te^{xt}}{e^t-1} \\\\\n&= e^{xt} \\sum_{n=0}^\\infty \\frac{B_n t^n}{n!} \\\\\n&= \\sum_{m=0}^\\infty \\frac{(xt)^m}{m!} \\sum_{n=0}^\\infty \\frac{B_n t^n}{n!} \\\\\n&= \\sum_{m=0}^\\infty \\sum_{n=0}^\\infty \\frac{B_n x^m t^{n+m}}{n! m!} \\\\\n&= \\sum_{n=0}^\\infty \\sum_{k=0}^n \\frac{B_k t^n x^{n-k}}{k! (n-k)!} \\\\\n&= \\sum_{n=0}^\\infty \\frac{t^n}{n!}\\sum_{k=0}^n \\binom{n}{k} B_k x^{n-k}\n\\end{aligned} \n$$\nwhere $x\\in [0,1]$. Now, compare the coefficients of $t^n$ to get the desired result. Plugging in $x=1$, gives the identity:\n$$\n\\sum_{k=0}^{n-1} \\binom{n}{k} B_k = 0\n$$\n<\/p>\n\n\n\n
\nNow, let’s turn our attention to the integral:\n$$I=\\int_0^{\\frac{\\pi}{2}}\\frac{\\sin(2nx)}{\\sin^{2n+2}(x)}\\cdot \\frac{1}{e^{2\\pi \\cot x}-1} dx$$\nwhere $n\\in\\mathbb{N}$. We will use the following trigonometric identity:\n$$\n\\frac{\\sin(2nx)}{\\sin^{2n}(x)} =(-1)^n \\sum_{r=1}^{n}\\binom{2n}{2r-1} (-1)^{r}\\cot^{2r-1}(x)\n$$\nSubstituting the above into the integral, gives:\n$$\n\\begin{aligned}\nI &= (-1)^n \\int_0^{\\pi\\over 2}\\left( \\sum_{r=1}^{n}\\binom{2n}{2r-1} (-1)^{r}\\cot^{2r-1}(x)\\right)\\frac{\\csc^2(x)}{e^{2\\pi \\cot x}-1}dx \\\\ \n&= (-1)^n \\sum_{r=1}^{n}\\binom{2n}{2r-1} (-1)^{r} \\int_0^{\\pi \\over 2}\\cot^{2r-1}(x)\\frac{\\csc^2(x)}{e^{2\\pi \\cot x}-1}dx \\\\ \n&= (-1)^n \\sum_{r=1}^{n}\\binom{2n}{2r-1} (-1)^{r} \\int_0^\\infty \\frac{t^{2r-1}}{e^{2\\pi t}-1}dt \\\\ \n&= (-1)^n \\sum_{r=1}^{n}\\binom{2n}{2r-1} (-1)^{r}\\frac{(2r-1)! \\zeta(2r)}{(2\\pi)^r }\\\\ \n&= (-1)^n \\sum_{r=1}^{n}\\binom{2n}{2r-1} (-1)^{r} (-1)^{r-1} \\frac{B_{2r}}{4r}\\\\ \n&= \\frac{(-1)^{n-1}}{4}\\sum_{r=1}^{n}\\binom{2n}{2r-1}\\frac{B_{2r}}{r} \\\\\n&= \\frac{(-1)^{n-1}}{2(2n+1)}\\sum_{r=1}^n \\binom{2n+1}{2r} B_{2r} \\\\\n&= \\frac{(-1)^{n-1}}{2(2n+1)} \\left[\\sum_{r=0}^{2n} \\binom{2n+1}{r} B_r – \\binom{2n+1}{0}B_0 – \\binom{2n+1}{1} B_1\\right] \\\\\n&= \\frac{(-1)^{n-1}}{2(2n+1)} \\left[-\\binom{2n+1}{0}B_0 – \\binom{2n+1}{1} B_1\\right] \\\\\n&= \\frac{(-1)^{n-1}}{4}\\cdot \\frac{2n-1}{2n+1}\n\\end{aligned}\n$$\n<\/p>\n\n\n\n
<\/p>\n","protected":false},"excerpt":{"rendered":"
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 … Continue reading