# Bernoulli polynomials

Jordan Bell
March 9, 2016

## 1 Bernoulli polynomials

For $k\geq 0$, the Bernoulli polynomial $B_{k}(x)$ is defined by

 $\frac{ze^{xz}}{e^{z}-1}=\sum_{k=0}^{\infty}B_{k}(x)\frac{z^{k}}{k!},\qquad|z|<% 2\pi.$ (1)

The Bernoulli numbers are $B_{k}=B_{k}(0)$, the constant terms of the Bernoulli polynomials. For any $x$, using L’Hospital’s rule the left-hand side of (1) tends to $1$ as $z\to 0$, and the right-hand side tends to $B_{0}(x)$, hence $B_{0}(x)=1$. Differentiating (1) with respect to $x$,

 $\sum_{k=0}^{\infty}B_{k}^{\prime}(x)\frac{z^{k}}{k!}=\frac{z^{2}e^{xz}}{e^{z}-% 1}=\sum_{k=0}^{\infty}B_{k}(x)\frac{z^{k+1}}{k!}=\sum_{k=1}^{\infty}B_{k-1}(x)% \frac{z^{k}}{(k-1)!},$

so $B_{0}^{\prime}(x)=0$ and for $k\geq 1$ we have $\frac{B_{k}^{\prime}(x)}{k!}=\frac{B_{k-1}(x)}{(k-1)!}$, i.e. $B_{k}^{\prime}(x)=kB_{k-1}(x)$. Furthermore, for $k\geq 1$, integrating (1) with respect to $x$ on $[0,1]$ produces

 $1=\sum_{k=0}^{\infty}\left(\int_{0}^{1}B_{k}(x)dx\right)\frac{z^{k}}{k!},% \qquad|z|<2\pi,$

hence $\int_{0}^{1}B_{0}(x)dx=1$ and for $k\geq 1$,

 $\int_{0}^{1}B_{k}(x)dx=0.$

The first few Bernoulli polynomials are

 $B_{0}(x)=1,\quad B_{1}(x)=x-\frac{1}{2},\quad B_{2}(x)=x^{2}-x+\frac{1}{6},% \quad B_{3}(x)=x^{3}-\frac{3}{2}x^{2}+\frac{1}{2}x.$

The Bernoulli polynomials satisfy the following:

 $\displaystyle\sum_{k=0}^{\infty}B_{k}(x+1)\frac{z^{k}}{k!}$ $\displaystyle=\frac{ze^{(x+1)z}}{e^{z}-1}$ $\displaystyle=\frac{ze^{xz}(e^{z}-1+1)}{e^{z}-1}$ $\displaystyle=ze^{xz}+\frac{ze^{xz}}{e^{z}-1}$ $\displaystyle=\sum_{k=0}^{\infty}\frac{x^{k}z^{k+1}}{k!}+\sum_{k=0}^{\infty}B_% {k}(x)\frac{z^{k}}{k!}$ $\displaystyle=\sum_{k=1}^{\infty}\frac{x^{k-1}z^{k}}{(k-1)!}+\sum_{k=0}^{% \infty}B_{k}(x)\frac{z^{k}}{k!},$

hence for $k\geq 1$ it holds that $B_{k}(x+1)=kx^{k-1}+B_{k}(x)$. In particular, for $k\geq 2$, $B_{k}(1)=B_{k}(0)$.

Using (1),

 $\displaystyle\sum_{k=0}^{\infty}B_{k}(1-x)\frac{z^{k}}{k!}$ $\displaystyle=\frac{ze^{(1-x)z}}{e^{z}-1}$ $\displaystyle=\frac{ze^{z}e^{-xz}}{e^{z}-1}$ $\displaystyle=\frac{ze^{-xz}}{1-e^{-z}}$ $\displaystyle=\frac{-ze^{-xz}}{e^{-z}-1}$ $\displaystyle=\sum_{k=0}^{\infty}B_{k}(x)\frac{(-z)^{k}}{k!},$

hence for $k\geq 0$,

 $B_{k}(1-x)=(-1)^{k}B_{k}(x).$

Finally, it is a fact that for $k\geq 2$,

 $\sup_{0\leq x\leq 1}|B_{k}(x)|\leq\frac{2\zeta(k)k!}{(2\pi)^{k}}.$ (2)

## 2 Periodic Bernoulli functions

For $x\in\mathbb{R}$, let $[x]$ be the greatest integer $\leq x$, and let $R(x)=x-[x]$, called the fractional part of $x$. Write $\mathbb{T}=\mathbb{R}/\mathbb{Z}$ and define the periodic Bernoulli functions $P_{k}:\mathbb{T}\to\mathbb{R}$ by

 $P_{k}(t)=B_{k}(R(t)),\qquad t\in\mathbb{T}.$

For $k\geq 2$, because $B_{k}(1)=B_{k}(0)$, the function $P_{k}$ is continuous. For $f:\mathbb{T}\to\mathbb{C}$ define its Fourier transform $\widehat{f}:\mathbb{Z}\to\mathbb{C}$ by

 $\widehat{f}(n)=\int_{\mathbb{T}}f(t)e^{-2\pi int}dt,\qquad n\in\mathbb{Z}.$

For $k\geq 1$, one calculates $\widehat{P}_{k}(0)=0$ and using integration by parts,

 $\widehat{P}_{k}(n)=-\frac{1}{(2\pi in)^{k}}$

for $n\neq 0$. Thus for $k\geq 1$, the Fourier series of $P_{k}$ is1

 $P_{k}(t)\sim\sum_{n\in\mathbb{Z}}\widehat{P}_{k}(n)e^{2\pi int}=-\frac{1}{(2% \pi i)^{k}}\sum_{n\neq 0}n^{-k}e^{2\pi int}.$

For $k\geq 2$, $\sum_{n\in\mathbb{Z}}|\widehat{P}_{k}(n)|<\infty$, from which it follows that $\sum_{|n|\leq N}\widehat{P}_{k}(n)e^{2\pi int}$ converges to $P_{k}(t)$ uniformly for $t\in\mathbb{T}$. Furthermore, for $t\not\in\mathbb{Z}$,22 2 Hugh L. Montgomery and Robert C. Vaughan, Multiplicative Number Theory I: Classical Theory, p. 499, Theorem B.2.

 $P_{1}(t)=-\frac{1}{\pi}\sum_{n=1}^{\infty}\frac{1}{n}\sin 2\pi nt.$

For $f,g\in L^{1}(\mathbb{T})$ and $n\in\mathbb{Z}$,

 $\displaystyle\widehat{f*g}(n)$ $\displaystyle=\int_{\mathbb{T}}\left(\int_{\mathbb{T}}f(x-y)g(y)dy\right)e^{-2% \pi inx}dx$ $\displaystyle=\int_{\mathbb{T}}g(y)\left(\int_{\mathbb{T}}f(x-y)e^{-2\pi inx}% dx\right)dy$ $\displaystyle=\int_{\mathbb{T}}g(y)\left(\int_{\mathbb{T}}f(x)e^{-2\pi inx}e^{% -2\pi iny}dx\right)dy$ $\displaystyle=\widehat{f}(n)\widehat{g}(n).$

For $k,l\geq 1$ and for $n\neq 0$,

 $\displaystyle\widehat{P_{k}*P_{l}}(n)$ $\displaystyle=\widehat{P_{k}}(n)\widehat{P_{l}}(n)$ $\displaystyle=-(2\pi in)^{-k}\cdot-(2\pi in)^{-l}$ $\displaystyle=(2\pi in)^{-k-l}$ $\displaystyle=-\widehat{P_{k+l}}(n),$

and $\widehat{P_{k}*P_{l}}(0)=0=-\widehat{P_{k+l}}(0)$, so $P_{k}*P_{l}=-P_{k+l}$.

## 3 Euler-Maclaurin summation formula

The Euler-Maclaurin summation formula is the following.33 3 Hugh L. Montgomery and Robert C. Vaughan, Multiplicative Number Theory I: Classical Theory, p. 500, Theorem B.5. If $a are real numbers, $K$ is a positive integer, and $f$ is a $C^{K}$ function on an open set that contains $[a,b]$, then

 $\displaystyle\sum_{a $\displaystyle=\int_{a}^{b}f(x)dx+\sum_{k=1}^{K}\frac{(-1)^{k}}{k!}(P_{k}(b)f^{% (k-1)}(b)-P_{k}(a)f^{(k-1)}(a))$ $\displaystyle-\frac{(-1)^{K}}{K!}\int_{a}^{b}P_{K}(x)f^{(K)}(x)dx.$

Applying the Euler-Maclaurin summation formula with $a=1,b=n,K=2,f(x)=\log x$ yields44 4 Hugh L. Montgomery and Robert C. Vaughan, Multiplicative Number Theory I: Classical Theory, p. 503, Eq. B.25.

 $\sum_{1\leq m\leq n}\log n=n\log n-n+\frac{1}{2}\log n+\frac{1}{2}\log 2\pi+O(% n^{-1}).$

Since $e^{1+O(n^{-1})}=1+O(n^{-1})$,

 $n!=n^{n}e^{-n}\sqrt{2\pi n}(1+O(n^{-1})),$

Stirling’s approximation.

Write $a_{n}=-\log n+\sum_{1\leq m\leq n}\frac{1}{m}$. Because $\log(1-x)$ is concave,

 $a_{n}-a_{n-1}=\frac{1}{n}+\log\left(1-\frac{1}{n}\right)\leq 1+1-\frac{1}{n}=0,$

which means that the sequence $a_{n}$ is nonincreasing. For $f(x)=\frac{1}{x}$, because $f$ is positive and nonincreasing,

 $\sum_{1\leq m\leq n}f(m)\geq\int_{1}^{n+1}f(x)dx=\log(n+1)>\log n,$

hence $a_{n}>0$. Because $a_{n}$ is positive and nonincreasing, there exists some nonnegative limit, $\gamma$, called Euler’s constant. Using the Euler-Maclaurin summation formula with $a=1,b=n,K=1,f(x)=\frac{1}{x}$, as $P_{1}(x)=[x]-\frac{1}{2}$,

 $\sum_{1

which is

 $\sum_{1

as $0\leq R(x)x^{-2}\leq x^{-2}$, the function $x\mapsto R(x)x^{-2}$ is integrable on $[1,\infty)$. Since $0\leq\int_{n}^{\infty}R(x)x^{-2}dx\leq\int_{n}^{\infty}x^{-2}dx=n^{-1}$,

 $\sum_{1\leq m\leq n}\frac{1}{m}=\log n+C+O(n^{-1})$

for $C=1-\int_{1}^{\infty}R(x)x^{-2}$. But $-\log n+\sum_{1\leq m\leq n}\frac{1}{m}\to\gamma$ as $n\to\infty$, from which it follows that $C=\gamma$, and thus

 $\sum_{1\leq m\leq n}\frac{1}{m}=\log n+\gamma+O(n^{-1}).$

## 4 Hurwitz zeta function

For $0<\alpha\leq 1$ and $\mathrm{Re}\,s>1$, define the Hurwitz zeta function by

 $\zeta(s,\alpha)=\sum_{n\geq 0}(n+\alpha)^{-s}.$

For $\mathrm{Re}\,s>0$,

 $\Gamma(s)=\int_{0}^{\infty}t^{s-1}e^{-t}dt,$

and for $n\geq 0$ do the change of variable $t=(n+\alpha)u$,

 $\displaystyle\Gamma(s)$ $\displaystyle=\int_{0}^{\infty}(n+\alpha)^{s-1}u^{s-1}e^{-(n+\alpha)u}(n+% \alpha)du$ $\displaystyle=(n+\alpha)^{s}\int_{0}^{\infty}u^{s-1}e^{-nu}e^{-\alpha u}du.$

For real $s>1$,

 $(n+\alpha)^{-s}\Gamma(s)=\int_{0}^{\infty}u^{s-1}e^{-nu}e^{-\alpha u}du.$

Then

 $\sum_{0\leq n\leq N}(n+\alpha)^{-s}\Gamma(s)=\sum_{0\leq n\leq N}\int_{0}^{% \infty}u^{s-1}e^{-nu}e^{-\alpha u}du=\int_{0}^{\infty}f_{N}(s,u)du,$

where

 $f_{N}(s,u)=\begin{cases}u^{s-1}e^{-\alpha u}\frac{1-e^{-(N+1)u}}{1-e^{-u}}&u>0% \\ 0&u=0.\end{cases}$

$f_{N}(s,u)\geq 0$ and the sequence $f_{N}(s,u)$ is pointwise nondecreasing, and

 $\lim_{N\to\infty}f_{N}(s,u)=f(s,u)=\begin{cases}u^{s-1}e^{-\alpha u}\frac{1}{1% -e^{-u}}&u>0\\ 0&u=0.\end{cases}$

By the monotone convergence theorem,

 $\int_{0}^{\infty}f_{N}(s,u)du\to\int_{0}^{\infty}f(s,u)du,$

which means that, for real $s>1$,

 $\zeta(s,\alpha)\Gamma(s)=\int_{0}^{\infty}f(s,u)du.$

Write

 $\int_{0}^{\infty}f(s,u)du=\int_{0}^{1}f(s,u)du+\int_{1}^{\infty}f(s,u)du.$

Now, by (1), for $0,

 $\displaystyle f(s,u)$ $\displaystyle=u^{s-1}e^{-\alpha u}\frac{1}{1-e^{-u}}$ $\displaystyle=u^{s-2}\cdot\frac{-ue^{-\alpha u}}{e^{-u}-1}$ $\displaystyle=u^{s-2}\sum_{k=0}^{\infty}B_{k}(\alpha)\frac{(-u)^{k}}{k!}$ $\displaystyle=\sum_{k=0}^{\infty}(-1)^{k}B_{k}(\alpha)\frac{u^{k+s-2}}{k!}.$

For $k\geq 2$, real $s>1$, and $0, by (2),

 $\left|B_{k}(\alpha)\frac{u^{k+s-2}}{k!}\right|\leq\frac{2\zeta(k)k!}{(2\pi)^{k% }}\cdot u^{k+s-2}\cdot\frac{1}{k!}=2\zeta(k)\left(\frac{u}{2\pi}\right)^{k}u^{% s-2},$

which is summable, and thus by the dominated convergence theorem,

 $\displaystyle\int_{0}^{1}f(s,u)du$ $\displaystyle=\int_{0}^{1}\sum_{k=0}^{\infty}(-1)^{k}B_{k}(\alpha)\frac{u^{k+s% -2}}{k!}du$ $\displaystyle=\sum_{k=0}^{\infty}(-1)^{k}B_{k}(\alpha)\frac{1}{k!}\frac{1}{k+s% -1}.$

Check that $s\mapsto\sum_{k=0}^{\infty}(-1)^{k}B_{k}(\alpha)\frac{1}{k!}\frac{1}{k+s-1}$ is meromorphic on $\mathbb{C}$, with poles of order 0 or 1 at $s=-k+1$, $k\geq 0$ (the order of the pole is $0$ if $B_{k}(\alpha)=0$), at which the residue is $(-1)^{k}B_{k}(\alpha)\frac{1}{k!}$.55 5 Kazuya Kato, Nobushige Kurokawa, and Takeshi Saito, Number Theory 1: Fermat’s Dream, p. 96. On the other hand, check that $s\mapsto\int_{1}^{\infty}f(s,u)du$ is entire. Therefore $\zeta(s,\alpha)\Gamma(s)$ is meromorphic on $\mathbb{C}$, with poles of order 0 or 1 at $s=-k+1$, $k\geq 0$ and the residue of $\zeta(s,\alpha)\Gamma(s)$ at $s=-k+1$ is $(-1)^{k}B_{k}(\alpha)\frac{1}{k!}$. But it is a fact that $\Gamma(s)$ has poles of order $1$ at $s=-n$, $n\geq 0$, with residue $\frac{(-1)^{n}}{n!}$. Hence the only pole of $\zeta(s,\alpha)$ is at $s=1$, at which the residue is $1$.

###### Theorem 1.

For $n\geq 1$ and for $0<\alpha\leq 1$,

 $\zeta(1-n,\alpha)=-\frac{B_{n}(\alpha)}{n}.$
###### Proof.

For $n\geq 1$, because $\zeta(s,\alpha)$ does not have a pole at $s=1-n$ and because $\Gamma(s)$ has a pole of order $1$ at $s=1-n$ with residue $\frac{(-1)^{n-1}}{(n-1)!}$,

 $\displaystyle\lim_{s\to 1-n}(s-(1-n))\Gamma(s)\zeta(s,\alpha)$ $\displaystyle=\zeta(1-n,\alpha)\cdot\lim_{s\to 1-n}(s-(1-n))\Gamma(s)$ $\displaystyle=\zeta(1-n,\alpha)\cdot\mathrm{Res}_{s=1-n}\Gamma(s)$ $\displaystyle=\zeta(1-n,\alpha)\cdot\frac{(-1)^{n-1}}{(n-1)!}.$

On the other hand, $\zeta(s,\alpha)\Gamma(s)$ has a pole of order $1$ at $s=1-n$ with residue $(-1)^{n}B_{n}(\alpha)\frac{1}{n!}$. Therefore

 $\zeta(1-n,\alpha)\cdot\frac{(-1)^{n-1}}{(n-1)!}=(-1)^{n}B_{n}(\alpha)\frac{1}{% n!},$

i.e. for $n\geq 1$ and $0<\alpha\leq 1$,

 $\zeta(1-n,\alpha)=-\frac{B_{n}(\alpha)}{n}.$

## 5 Sobolev spaces

For real $s\geq 0$, we define the Sobolev space $H^{s}(\mathbb{T})$ as the set of those $f\in L^{2}(\mathbb{T})$ such that

 $|\widehat{f}(0)|^{2}+\sum_{n\in\mathbb{Z}\setminus\{0\}}|\widehat{f}(n)|^{2}|n% |^{2s}<\infty.$

For $f,g\in H^{s}(\mathbb{T})$, define

 $\left\langle f,g\right\rangle_{H^{s}(\mathbb{T})}=\widehat{f}(0)\overline{% \widehat{g}(0)}+\sum_{n\in\mathbb{Z}\setminus\{0\}}\widehat{f}(n)\overline{% \widehat{g}(n)}|n|^{2s}.$

This is an inner product, with which $H^{s}(\mathbb{T})$ is a Hilbert space.6

For $c\in\mathbb{C}^{\mathbb{Z}}$, if $s>r+\frac{1}{2}$,

 $\begin{split}&\displaystyle\left\|\sum_{|n|\leq N}c_{n}e^{2\pi inx}\right\|_{C% ^{r}(\mathbb{T})}\\ \displaystyle=&\displaystyle\sup_{0\leq j\leq r}\sup_{x\in\mathbb{T}}\left|% \sum_{|n|\leq N}c_{n}(2\pi in)^{j}e^{2\pi inx}\right|\\ \displaystyle\leq&\displaystyle|c_{0}|^{2}+\sup_{0\leq j\leq r}\sup_{x\in% \mathbb{T}}\left|\sum_{1\leq|n|\leq N}c_{n}(2\pi in)^{j}e^{2\pi inx}\right|\\ \displaystyle\leq&\displaystyle|c_{0}|^{2}+(2\pi)^{r}\sum_{1\leq|n|\leq N}|c_{% n}||n|^{r}\\ \displaystyle=&\displaystyle|c_{0}|^{2}+(2\pi)^{r}\sum_{1\leq|n|\leq N}|c_{n}|% |n|^{s}|n|^{-(r-s)}\\ \displaystyle\leq&\displaystyle|c_{0}|^{2}+(2\pi)^{r}\left(\sum_{1\leq|n|\leq N% }|c_{n}|^{2}|n|^{2s}\right)^{1/2}\left(\sum_{1\leq|n|\leq N}|n|^{-(2s-2r)}% \right)^{1/2}\\ \displaystyle\leq&\displaystyle|c_{0}|^{2}+(2\pi)^{r}\cdot(2\cdot\zeta(2s-2r))% ^{1/2}\cdot\left(\sum_{1\leq|n|\leq N}|c_{n}|^{2}|n|^{2s}\right)^{1/2}.\end{split}$

For $f\in H^{s}(\mathbb{T})$, the partial sums $\sum_{|n|\leq N}\widehat{f}(n)e^{2\pi inx}$ are a Cauchy sequence in $H^{s}(\mathbb{T})$ and by the above are a Cauchy sequence in the Banach space $C^{r}(\mathbb{T})$ and so converge to some $g\in C^{r}(\mathbb{T})$. Then $\widehat{g}=\widehat{f}$, which implies that $g=f$ almost everywhere.

For $k\geq 1$, $\widehat{P}_{k}(0)=0$ and $\widehat{P}_{k}(n)=-(2\pi in)^{-k}$ for $n\neq 0$. For $k,l>s+\frac{1}{2}$,

 $\displaystyle\left\langle P_{k},P_{l}\right\rangle_{H^{s}(\mathbb{T})}$ $\displaystyle=\sum_{n\in\mathbb{Z}\setminus\{0\}}-(2\pi in)^{-k}\overline{-(2% \pi in)^{-l}}$ $\displaystyle=\sum_{n\in\mathbb{Z}\setminus\{0\}}i^{-k+l}(2\pi n)^{-k-l}$ $\displaystyle=i^{-k+l}(2\pi)^{-k-l}\cdot 2\cdot\zeta(k+l).$

Thus if $k>s+\frac{1}{2}$ then $P_{k}\in H^{s}(\mathbb{T})$, and in particular $P_{k}\in H^{k-1}(\mathbb{T})$ for $k\geq 1$.

For $s>r+\frac{1}{2}$, if $f\in H^{s}(\mathbb{T})$ then there is some $g\in C^{r}(\mathbb{T})$ such that $g=f$ almost everywhere. Thus if $r+\frac{1}{2}, i.e. $k>r+1$, then there is some $g\in C^{r}(\mathbb{T})$ such that $g=P_{k}$ almost everywhere. But for $k\neq 1$, $P_{k}$ is continuous, so in fact $g=P_{k}$. In particular, $P_{k}\in C^{k-2}(\mathbb{T})$ for $k\geq 2$.

## 6 Reproducing kernel Hilbert spaces

For $x\in\mathbb{T}$ and $f:\mathbb{T}\to\mathbb{C}$, define $(\tau_{x}f)(y)=f(y-x)$. We calculate

 $\displaystyle\widehat{\tau_{x}f}(n)$ $\displaystyle=\int_{\mathbb{T}}f(y-x)e^{-2\pi iny}dy$ $\displaystyle=e^{-2\pi inx}\int_{\mathbb{T}}f(y)e^{-2\pi iny}dy$ $\displaystyle=e^{-2\pi inx}\widehat{f}(n).$

Let $r\geq 1$. For $x\in\mathbb{T}$, define $F_{x}:\mathbb{T}\to\mathbb{R}$ by

 $F_{x}=1+(-1)^{r-1}(2\pi)^{2r}\tau_{x}P_{2r}.$

For $n\in\mathbb{Z}$,

 $\displaystyle\widehat{F_{x}}(n)$ $\displaystyle=\delta_{0}(n)+(-1)^{r-1}(2\pi)^{2r}\cdot e^{-2\pi inx}\widehat{P% }_{2r}(n).$

$\widehat{F_{x}}(0)=1$, and for $n\neq 0$,

 $\widehat{F_{x}}(n)=(-1)^{r-1}(2\pi)^{2r}\cdot e^{-2\pi inx}\cdot-(2\pi in)^{-2% r}=|n|^{-2r}e^{-2\pi inx}.$

For $f\in H^{r}(\mathbb{T})$,

 $\displaystyle\left\langle f,F_{x}\right\rangle_{H^{r}(\mathbb{T})}$ $\displaystyle=\widehat{f}(0)\overline{\widehat{F_{x}}(0)}+\sum_{n\in\mathbb{Z}% \setminus\{0\}}\widehat{f}(n)\overline{\widehat{F_{x}}(n)}|n|^{2r}$ $\displaystyle=\widehat{f}(0)+\sum_{n\in\mathbb{Z}\setminus\{0\}}\widehat{f}(n)% |n|^{-2r}e^{2\pi inx}|n|^{2r}$ $\displaystyle=\widehat{f}(0)+\sum_{n\in\mathbb{Z}\setminus\{0\}}\widehat{f}(n)% e^{2\pi inx}$ $\displaystyle=f(x).$

This shows that $H^{r}(\mathbb{T})$ is a reproducing kernel Hilbert space.

Define $F:\mathbb{T}\times\mathbb{T}\to\mathbb{R}$ by

 $\displaystyle F(x,y)$ $\displaystyle=\left\langle F_{x},F_{y}\right\rangle_{H^{r}(\mathbb{T})}$ $\displaystyle=F_{x}(y)$ $\displaystyle=1+(-1)^{r-1}(2\pi)^{2r}P_{2r}(y-x).$

Thus the reproducing kernel of $H^{r}(\mathbb{T})$ is77 7 cf. Alain Berlinet and Christine Thomas-Agnan, Reproducing Kernel Hilbert Spaces in Probability and Statistics, p. 318, who use a different inner product on $H^{r}(\mathbb{T})$ and consequently have a different expression for the reproducing kernel.

 $F(x,y)=1+(-1)^{r-1}(2\pi)^{2r}P_{2r}(y-x).$