Interesting polynomial: $R_{k+1}(x)=x(2^{-2k-1}-1)\zeta(2k+2)+\int_0^x\int_0^t R_{k}(u)dudt$

Question

I am trying to find the polynomial $R_k(x)$ such that $$R_k(x)=\sum_{n=1}^{\infty}\frac{(-1)^n}{n^{2k+1}}\sin(nx)$$ With the base case $$R_1(x)=\frac{x^3}{12}-\frac{\pi^2x}{12}$$ Which can be found by integrating the Fourier series for $x^2$. I found the recurrence mentioned in the title by starting with $$\sum_{n\geq1}\frac{(-1)^n}{n^{2k+1}}\sin(nu)=R_k(u)$$ $$\sum_{n\geq1}\frac{(-1)^n}{n^{2k+1}}\int_0^t\sin(nu)du=\int_0^tR_k(u)du$$ $$\sum_{n\geq1}\frac{(-1)^n}{n^{2k+2}}(\cos(nt)-1)=\int_0^tR_k(u)du$$ $$\sum_{n\geq1}\frac{(-1)^n}{n^{2k+2}}\cos(nt)=(2^{-2k-1}-1)\zeta(2k+2)+\int_0^tR_k(u)du$$ $$\sum_{n\geq1}\frac{(-1)^n}{n^{2k+2}}\int_0^x\cos(nt)dt=x(2^{-2k-1}-1)\zeta(2k+2)+\int_0^x\int_0^tR_k(u)dudx$$ $$R_{k+1}(x)=x(2^{-2k-1}-1)\zeta(2k+2)+\int_0^x\int_0^tR_k(u)dudx$$ Yet I do not know how to get a closed form in terms of a polynomial for this. Could I have some help? Thanks.

Answer 1

Your integral induction formula yields the coefficients of the polynomials by induction, in term of $\zeta(2k+2)$ (having an expression in term of Bernouili numbers).

Now there is another method. Let $P_k$ be the polynomials such that $$r_k(x)=\sum_{n=1}^{\infty}\frac{\sin(2\pi nx)}{n^{2k+1}}= P_k\left(x+\frac12-\left\lfloor x+\frac12\right\rfloor\right),\deg(P_k)= 2+\deg(P_{k-1}) = 2k+1$$

Look at its Mellin transform $$Q_k(s)= \int_0^\infty r_k(x) x^{s-1}dx= \sum_{n=1}^\infty\int_0^\infty n^{-2k-1-s}\sin(2\pi x) x^{s-1}dx=\zeta(s+2k+1) \int_0^\infty \sin(2\pi x) x^{s-1}dx$$ where $$ \int_0^\infty \sin(2\pi x) x^{s-1}dx = \Gamma(s) (2\pi)^{-s} \frac{e^{-i\pi s/2}-e^{i\pi s/2}}{2i} = \Gamma(s)(2\pi)^{-s} \sin(-\pi s/2)$$

So that by Mellin inversion and the residue theorem, for $x \in (0,1/2)$ $$P_k(x)=r_k(x) = \frac{1}{2i\pi} \int_{\sigma-i\infty}^{\sigma+i\infty} Q_k(s) x^{-s}ds=\sum \operatorname{Res}(Q_k(s) x^{-s})=\sum_{m=0}^{2k+1} \operatorname{Res}_{s=-m}(Q_k(s)) x^m$$

$$P_k(x)= x^{2k} (-\pi / 2)\frac{1}{(2k)!} (2\pi)^{2k}+\sum_{m=1}^k x^{2m+1} (2\pi)^{-2m-1} \frac{-1}{(2m+1)!} (-1)^m \zeta(2k-2m)$$