Malmsten-like integrals for $\zeta(n) $

Question

Context

Let us define a Malmsten integral when it is of the form $$M_{P,Q} := \int_{0}^{1} \frac{P(x)}{Q(x)} \ln \ln \left( \frac{1}{x} \right) dx. \tag{1} $$ In a paper by Blagouchine [1], the author traces the origins of this type of integral to the Swedish mathematician Carl Johan Malmsten. For instance, he attributes the evaluation of the following integral to him: $$\int_{0}^{1} \frac{\ln \ln \frac{1}{x}}{1+x^2} dx = \frac{\pi}{2} \ln \left( \frac{\Gamma(3/4)}{\Gamma(1/4)} \sqrt{2 \pi} \right) , \tag{2}$$ which can also be found on p. 570 of Gradshteyn and Ryzhik's book [2] on Integrals and Series (entry 4.325.4). Blagouchine deduces a Malmsten integral identity for Apéry's constant on p. 61 of his article:

$$ \label{3} \zeta(3) = \frac{8 \pi^{2}}{7} \int_{0}^{1} \frac{x(x^{4} -4x^{2} +1 ) \ln \ln \frac{1}{x} }{(1+x^{2})^{4}} dx. \tag{3} $$ This was done by defining and evaluating Malmsten integrals of the form $$I_{n} := \int_{0}^{1} \frac{x^{n-1} \ln \ln \frac{1}{x} }{(1+x^{2})^{n}} dx $$ and simplifying $C_{1} :=I_{4} - \left( \frac{1}{6} \right) I_{2} $. Although the author does not do so in the paper, one could proceed to obtain a similar integral expression for $\zeta(5)$ by incorporating his results for $I_{6}$ and $I_{4}$ by defining \begin{align} C_{2} &:= I_{6} - \left( \frac{1}{5} \right) I_{4} \\ &= - \int_{0}^{1} \frac{x^{3}(x^{4} -3x^{2}+1 ) \ln \ln \frac{1}{x} }{5(1+x^{2})^{6}} dx,\end{align} which by the identities on p. 60 implies that $$\frac{31 \zeta(5) }{20 \pi^{4}} = \int_{0}^{1} \frac{x^{3}(x^{4} -3x^{2}+1 ) \ln \ln \frac{1}{x} }{5(1+x^{2})^{6}} - \frac{7 \zeta(3)}{960 \pi^{2}}. \tag{4} $$ When we proceed by substituting the expression for $\zeta(3)$ obtained in $\eqref{3} $ and simplify, we obtain $$\zeta(5) = - \frac{8 \pi^{4}}{93} \int_{0}^{1} \frac{x(x^8-26x^6+ 66x^4-26x^2+1) \ln \ln \frac{1}{x}}{(1+x^{2})^{6}} dx. \tag{5} $$

Likewise, it is possible to obtain integral expressions for $\zeta(2n+1)$ in a similar manner.

Questions

1. As the author indicates, another Malmsten integral for Catalan's constant was found by Adamchik [3]. However, I haven't been able to find such integrals for the zeta values at even integers yet. Do these exist and are they known?
2. Is there a general identity that expresses all zeta values as a Malmsten integral family, i.e. $$\zeta(n) = A_{n} \int_{0}^{1} \frac{P_{n}(x) \ln \ln \frac{1}{x} }{Q_{n} (x)} dx $$ for $n \in \mathbb{Z}_{\geq 2}$, whereby $A_{n}, $ $P_{n}(\cdot) $, and $Q_{n}(\cdot) $ adhere to some pattern?

Sources

[1] Blagouchine, I. V. (2014). Rediscovery of Malmsten’s integrals, their evaluation by contour integration methods and some related results. The Ramanujan Journal, 35(1), 21–110. doi:10.1007/s11139-013-9528-5 , url

[2] I. S. Gradshteyn, I. M. Ryzhik (2007). Table of Integrals, Series, and Products, seventh edition. Academic Press, url

[3] V. Adamchik (1997). A Class of Logarithmic Integrals. Conference paper, url

Answer 1

Three remarks:

1. If you change $x$ into $\exp(-t)$ your integrals are linear combinations of $\int_0^\infty \log(t)\exp(-mt)/(1+\exp(-2t))^n\,dt$.
2. In turn, these integrals are the derivative with respect to $s$ at $s=1$ of $\int_0^\infty t^{s-1}\exp(-mt)/(1+\exp(-2t))^n\,dt$, which are easily related to $\Gamma(s)\zeta(s)$ by expanding $(1+z)^{-n}$.
3. The reason you do not get even values of zeta is that your values are in fact linear combinations of $\zeta'(-k)$, and only for $k$ even is this a rational multiple of $\zeta(k+1)/\pi^k$.