In this post I list another group of integrals from the new preprint "The derivation of eighteen special challenging logarithmic integrals" by Cornel Ioan Valean where they are nicely evaluated.

The following equalities hold: \begin{equation*} i) \ \int_0^1 \frac{x \log ^3(x)\log (1-x) }{1+x^2} \textrm{d}x=\frac{1761}{512} \zeta (5)-\frac{3}{32} \pi ^2 \zeta (3)-\frac{9}{256}\log (2)\pi ^4; \end{equation*} \begin{equation*} ii) \ \int_0^1 \frac{x \log ^3(x)\log (1+x) }{1+x^2} \textrm{d}x=\frac{7}{256}\log (2)\pi ^4+\frac{3}{64} \pi ^2 \zeta (3)-\frac{1215}{512} \zeta (5); \end{equation*} \begin{equation*} iii) \ \int_0^1 \frac{x\log^5(x)\log (1-x) }{1+x^2} \textrm{d}x=\frac{129495}{2048}\zeta (7)-\frac{75 }{128}\pi ^2 \zeta (5)-\frac{\pi ^4}{8} \zeta (3)-\frac{33}{512} \log (2)\pi ^6; \end{equation*} \begin{equation*} iv) \ \int_0^1 \frac{x\log^5(x)\log (1+x) }{1+x^2} \textrm{d}x=\frac{31}{512}\log(2)\pi ^6+\frac{7}{64} \pi ^4 \zeta (3)+\frac{75}{256}\pi^2 \zeta (5)-\frac{114345 }{2048}\zeta (7); \end{equation*} \begin{equation*} v) \ \int_0^1 \frac{x\log^7(x)\log (1-x) }{1+x^2} \textrm{d}x \end{equation*} \begin{equation*} =\frac{42010605}{16384}\zeta (9)-\frac{6615}{1024} \pi ^2 \zeta (7)-\frac{105}{64} \pi ^4 \zeta (5)-\frac{\pi ^6 }{2}\zeta (3)-\frac{2193}{8192} \log (2)\pi ^8; \end{equation*} \begin{equation*} vi) \ \int_0^1 \frac{x\log^7(x)\log (1+x) }{1+x^2} \textrm{d}x \end{equation*} \begin{equation*} =\frac{2159}{8192}\log(2)\pi ^8+\frac{31}{64} \pi ^6 \zeta (3)+\frac{735}{512} \pi ^4 \zeta (5)+\frac{6615}{2048} \pi ^2 \zeta (7)-\frac{40403475}{16384} \zeta (9). \end{equation*}

Question: How would you calculate the integrals by using harmonic series? Are these integrals known in the mathematical literature?

For people that took interest in the Problem 11966 from The American Mathematical Monthly,

\begin{equation*} \int_0^1 \frac{x\log (1+x) }{1+x^2} \textrm{d}x, \end{equation*}

you might want to observe that the integrals from this post contain some additional logs, and in a way can be viewed as more advanced integrals of the one from AMM.

The Problem 11966 may also be found in the book (Almost) Impossible Integrals, Sums, and Series, and in the same book there is the identity 3.62, page 97, (possibly new in the mathematical literature) which has been used for the calculation of these integrals in the present preprint.



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