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In this post you'll find the last group of logarithmic integrals from the new preprint "The derivation of eighteen special challenging logarithmic integrals" by Cornel Ioan Valean which are evaluated beautifully by fruitfully combining the Fourier series and the integral representation of the Polylogarithm given in Section 1.6, page 4, from the book (Almost) Impossible Integrals, Sums, and Series (it is also stated in the preprint - see Lemma 4), to turn the single integrals into double integrals.

The following equalities hold: \begin{equation*} i) \ \int_0^1 \frac{\log ^2(x)}{x}\operatorname{Li}_2\left(\frac{2 x}{1+x^2}\right) \textrm{d}x=\frac{3}{64}\log (2)\pi ^4 +\frac{5}{48}\pi ^2 \zeta (3)-\frac{31}{128} \zeta (5); \end{equation*} \begin{equation*} ii) \ \int_0^1 \frac{\log ^2(x)}{x}\operatorname{Li}_2\left(-\frac{2 x}{1+x^2}\right) \textrm{d}x=-\frac{7}{192}\log (2)\pi ^4-\frac{1}{12}\pi ^2\zeta (3)-\frac{31}{128} \zeta (5); \end{equation*} \begin{equation*} iii) \ \int_0^1 \frac{\log^4(x)}{x}\operatorname{Li}_2\left(\frac{2 x}{1+x^2}\right) \textrm{d}x=\frac{33}{640} \log (2)\pi ^6+\frac{89}{960} \pi ^4 \zeta (3)+\frac{13}{32} \pi ^2 \zeta (5)-\frac{381}{512} \zeta (7); \end{equation*} \begin{equation*} iv) \ \int_0^1 \frac{\log^4(x)}{x}\operatorname{Li}_2\left(-\frac{2 x}{1+x^2}\right) \textrm{d}x=-\frac{31}{640}\log (2) \pi ^6-\frac{91}{960}\pi ^4 \zeta (3)-\frac{19}{64} \pi ^2 \zeta (5)-\frac{381}{512} \zeta (7); \end{equation*} \begin{equation*} v) \ \int_0^1 \frac{\log^6(x)}{x}\operatorname{Li}_2\left(\frac{2 x}{1+x^2}\right) \textrm{d}x \end{equation*} \begin{equation*} =\frac{2193}{14336}\log(2)\pi ^8+\frac{215}{768}\pi ^6 \zeta (3)+\frac{113}{128} \pi ^4 \zeta (5)+\frac{825}{256}\pi ^2 \zeta (7)-\frac{22995}{4096} \zeta (9); \end{equation*} \begin{equation*} vi) \ \int_0^1 \frac{\log^6(x)}{x}\operatorname{Li}_2\left(-\frac{2 x}{1+x^2}\right) \textrm{d}x \end{equation*} \begin{equation*} =-\frac{2159}{14336}\log (2)\pi ^8-\frac{217}{768}\pi ^6 \zeta (3)-\frac{7}{8}\pi ^4 \zeta (5)-\frac{1185}{512}\pi ^2 \zeta (7)-\frac{22995}{4096}\zeta(9). \end{equation*}

Question: Is it possible to avoid completely the use of Fourier series and calculate it by another route involving harmonic series and generating functions with harmonic series?

A note: Starting from these forms and getting integrals from the previous two posts, A multitude of challenging logarithmic integrals (First part) and A multitude of challenging logarithmic integrals (Second part) might be an easier task. However, starting from the previous two groups of integrals and reducing the calculations to the integrals from this post is not something obvious at all. Good to know where we start from!

This is assured by the identity:

A special dilogarithmic identity. Let $x<1$ be a real number. Then the following equality holds: \begin{equation*} \int_0^x \frac{t \log (1-t)}{1+t^2} \textrm{d}t=\frac{1}{4} \left(\frac{1}{2} \log^2(1+x^2)-2 \operatorname{Li}_2(x)+\frac{1}{2}\operatorname{Li}_2\left(-x^2\right)+\operatorname{Li}_2\left(\frac{2 x}{1+x^2}\right)\right), \end{equation*} where $\displaystyle \operatorname{Li}_2(x)=-\int_0^x\frac{\log(1-t)}{t}\textrm{d}t$ is the Dilogarithm function.

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