Successive derivative of $\tanh u$ can be expressed as polynomial functions of $\tanh u$: \begin{align} \frac{d}{du}\tanh u&=1-\tanh^2u\\ \frac{d^2}{du^2}\tanh u&=-2\tanh u\left(1-\tanh^2u\right)\\ \frac{d^3}{du^3}\tanh u&=2\left(1-\tanh^2u\right)\left(3\tanh^2u-1\right)\\ \frac{d^4}{du^4}\tanh u&=-8\tanh u\left(1-\tanh^2u\right)\left(3\tanh^2u-2\right)\\ \ldots \end{align} I am specially interested in the derivatives of even order. Their expressions can be found in this article by Boyadzhiev: Denoting $z=\tanh u$ and $C_{2n}\left( \tanh u \right)=\tfrac{d^{2n}}{du^{2n}}\left( \tanh u \right) $, \begin{equation} C_{2n}\left(z \right)=2^{2n}\left( 1+ z\right)\sum_{k=0}^{2n}\frac{k!}{2^k}S\left( 2n,k \right)\left( z-1 \right)^k \end{equation} where $S(n,k)$ are the Stirling numbers of the second kind. We have also $C_{2n}(0)=C_{2n}(\pm1)=0$. As $\tanh u$ is an odd function of $u$, they can also be expressed as \begin{equation} C_{2n}\left(z \right)=z\left( 1-z^2 \right)Q_{2n-2}(z^2) \tag{A}\label{eqA} \end{equation} for $n=1,2,\ldots$. Then, $Q_0(z)=-2$, $Q_2(z)=-8\left( 3z^2-2 \right)$... One may establish the recurrence relation (with $Z=z^2 $): \begin{equation} Q_{2n}=4Z\left( 1-Z \right)^2Q''_{2n-2}-6\left( 1-Z \right)\left( 3Z-1 \right)Q'_{2n-2}+4\left( 3Z-2 \right)Q_{2n-2} \end{equation} What else can be found about the family $Q_{2n}$? Are they related to a well-known family of polynomials?
Context: Answering a MO question, an integral representation for $\zeta(3)$ appeared: \begin{equation} \int_{0}^\infty \frac{u-\tanh u}{u^3}\,du=\frac{7}{\pi^2}\zeta(3) \end{equation} It can easily be checked using the residue theorem, taking advantage of the parity of the integrand. This result can be generalized for $n=1,2,3,\ldots$ as: \begin{equation} \int_{0}^\infty \frac{T_{2n-1}(u)-\tanh u}{u^{2n+1}}\,du=(-1)^{n+1}\frac{2^{2n+1}-1}{\pi^{2n}}\zeta(2n+1) \end{equation} where $T_{2n-1}(u)$ is the odd polynomial identical to the Maclaurin expansion for $\tanh u$: \begin{equation} \tanh u=u-\frac{1}{3}u^3+\frac{2}{15}u^5-\frac{17}{315}u^7+\ldots \end{equation} up to the $u^{2n-1}$ term. This identity can also be checked using the residue theorem.
Now, integrating $2n$ times by parts, one obtains the following representation \begin{equation} \int_0^\infty C_{2n}\left(\tanh u \right)\frac{du}{u}=(-1)^{n}\frac{2^{2n+1}-1}{\pi^{2n}}\left( 2n \right)!\zeta(2n+1) \end{equation} With the representation \eqref{eqA} and the substitution $u=e^{-2v}$, one obtains \begin{equation} \int_0^1 \frac{v-1}{\left( v+ 1\right)^3}Q_{2n-2}\left( \left( \frac{1-v}{1+v} \right)^2 \right)\frac{dv}{\ln v}=(-1)^{n}\frac{2^{2n+1}-1}{4\pi^{2n}}\left( 2n \right)!\zeta(2n+1) \end{equation} which provides another integral representation for $\zeta(2n+1)$.
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