How to calculate




by means of real methods?

This question was suggested by Cornel the author of the book, Almost Impossible Integrals, Sums and Series.

The way I would approach the problem is to use the series property:

$$\sum_{n=1}^\infty (-1)^n f(2n)=\Re \sum_{n=1}^\infty i^n f(n),$$ namely

$$\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{n^3}=8\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{(2n)^3}=8\Re\sum_{n=1}^\infty\frac{i^n H_n}{n^3}$$

then we use the well-known generating function $\sum_{n=1}^\infty\frac{x^nH_n}{n^3}$. But this method is really tedious as we will need to use $\Re\{\text{Li}_2(1+i), \text{Li}_3(1+i),\text{Li}_4(1+i),\text{Li}_4(\frac{1+i}{2})\}.$

By the way, I have not seen a rigorous proof of the following equality:

$$\operatorname{Re} \operatorname{Li}_4 (1 + i)= -\frac{5}{16} \operatorname{Li}_4 \left (\frac{1}{2} \right ) + \frac{97}{9216} \pi^4 + \frac{\pi^2}{48} \ln^2 2 - \frac{5}{384} \ln^4 2\tag1$$

So solving this sum in a different way would be considered a new rigorous proof of $(1)$.

For the second series, I would follow the same approach.

Any idea by real methods? Thanks


1 Answer 1


A (revolutionary) solution by Cornel Ioan Valean

The series in this post are some of the toughest harmonic series ever known in the mathematical literature, and to my best knowledge, up to this moment except a solution involving contour integration by Song (for the first series), there is no known solution that is built on simple real means. Also, the present cocktail of simple ideas opens the road to calculate other very advanced integrals and series by real methods and extract polylogarithmic values involving a complex argument as mentioned in the main post.

Based on the Fourier-like series in the book, (Almost) Impossible Integrals, Sums, and Series, page $248$, eq. $3.292$, we have

$$\sum_{n=1}^{\infty} \left(2H_{2n}-2H_n+\frac{1}{2n}-2\log(2)\right)\frac{\sin^2(2nx)}{n}=\log(\sin(x))\log(\cos(x)), \ 0< x<\frac{\pi}{2}.$$

The BIG magical trick here is to note that

$$\log(\sin(x))\log(\cos(x))=\frac{1}{4} \log ^2\left(\frac{1}{2} \sin (2 x)\right)-\frac{1}{4} \log ^2(\tan (x)),$$ and this simple fact makes a huge difference as you'll see.

Then, the Fourier-like series above can be written as

$$\sum_{n=1}^{\infty} \left(2H_{2n}-2H_n+\frac{1}{2n}-2\log(2)\right)\frac{\sin^2(2nx)}{n}=\frac{1}{4} \log ^2\left(\frac{1}{2} \sin (2 x)\right)-\frac{1}{4} \log ^2(\tan (x)).$$

At this point, we multiply both sides of the Fourier-like series by $x$ and integrate from $x=0$ to $x=\pi/4$ that gives $$\frac{1}{16} \color{blue}{\sum_{n=1}^{\infty}(-1)^{n-1}\frac{ H_{2 n}}{n^3}}-\frac{5}{16}\sum_{n=1}^{\infty}(-1)^{n-1}\frac{ H_n}{n^3}+\frac{3}{16}\sum_{n=1}^{\infty}\frac{H_n}{n^3}$$ $$+\frac{\pi^2}{32}\sum_{n=1}^{\infty}\frac{1}{n}\left(H_{2 n}-H_n-\log(2)\right)-\frac{7}{64} \log (2)\zeta (3)+\frac{75}{512}\zeta(4)$$ $$=\frac{1}{4} \int_0^{\pi/4} x\log ^2\left(\frac{1}{2} \sin (2 x)\right)\textrm{d}x-\frac{1}{4}\int_0^{\pi/4} x\log ^2(\tan (x))\textrm{d}x$$ $$=\frac{1}{16} \int_0^{\pi/2} x\log ^2\left(\frac{1}{2} \sin (x)\right)\textrm{d}x-\frac{1}{4}\int_0^1 \frac{\arctan(x)}{1+x^2}\log ^2(x)\textrm{d}x$$ $$=\frac{1}{16}\log ^2(2)\int_0^{\pi/2}x\textrm{d}x-\frac{1}{8}\log (2) \int_0^{\pi/2} x \log (\sin (x))\textrm{d}x +\frac{1}{16} \int_0^{\pi/2} x \log ^2(\sin (x))\textrm{d}x$$ $$+\frac{1}{32} \sum _{n=1}^{\infty } (-1)^{n-1}\frac{ H_n}{n^3}-\frac{1}{16} \color{blue}{\sum _{n=1}^{\infty } (-1)^{n-1} \frac{H_{2 n}}{n^3}},$$ because $\displaystyle \int_0^1 \frac{\arctan(x)}{1+x^2}\log^2(x) \textrm{d}x=\frac{1}{4} \sum _{n=1}^{\infty } (-1)^{n-1}\frac{ H_{2 n}}{n^3}-\frac{1}{8} \sum _{n=1}^{\infty } (-1)^{n-1}\frac{ H_n}{n^3}$, and then all reduces to $$ \sum_{n=1}^{\infty}(-1)^{n-1}\frac{ H_{2 n}}{n^3}$$ $$=\frac{9}{8} \log ^2(2) \zeta ( 2)-\frac{7}{16} \log (2)\zeta (3) +\frac{1}{2} \int_0^{\pi/2} x \log ^2(\sin (x))\textrm{d}x$$ $$+\frac{11}{4} \sum _{n=1}^{\infty } (-1)^{n-1}\frac{ H_n}{n^3}-\frac{3}{2}\sum_{n=1}^{\infty}\frac{H_n}{n^3}-\frac{\pi^2}{4}\sum_{n=1}^{\infty}\frac{1}{n}\left(H_{2 n}-H_n-\log(2)\right)$$ $$+\frac{7}{8} \log (2)\zeta (3)-\frac{75}{64}\zeta(4),$$

and since we have that

$$\sum_{n=1}^{\infty}(-1)^{n-1} \frac{H_n}{n^3}$$ $$=\frac{11}{4}\zeta(4)-\frac{7}{4}\log(2)\zeta(3)+\frac{1}{2}\log^2(2)\zeta(2)-\frac{1}{12}\log^4(2)-2 \operatorname{Li}_4\left(\frac{1}{2}\right);$$ $$\sum_{n=1}^{\infty}\frac{H_n}{n^3}=\frac{5}{4}\zeta(4);$$ $$\sum_{n=1}^{\infty}\frac{1}{n}\left(H_{2 n}-H_n-\log(2)\right)=\log^2(2)-\frac{1}{2}\zeta(2);$$ $$\int_0^{\pi/2} x \log ^2(\sin (x))\textrm{d}x=\operatorname{Li}_4\left(\frac{1}{2}\right)+\frac{1}{24}\log^4(2)+\frac{1}{2}\log^2(2)\zeta(2)-\frac{19}{32}\zeta(4),$$ where the first three result are found in the book, (Almost) Impossible Integrals, Sums, and Series (see pages $309$-$310$, $87$, eq. $3.45$, $250$, eq. $3.294$), and the fourth one is already calculated here (and also other ways are possible), we conclude that

$$\sum_{n=1}^{\infty}(-1)^{n-1}\frac{ H_{2 n}}{n^3}$$ $$=\frac{195}{32}\zeta(4)+\frac{5}{4} \log ^2(2)\zeta(2)-\frac{35}{8}\log (2)\zeta (3) -\frac{5}{24} \log ^4(2)-5 \operatorname{Li}_4\left(\frac{1}{2}\right).$$

Finally, to extract the second series, we use the relation already established in the book, (Almost) Impossible Integrals, Sums, and Series, Sect. $4.58$, page $313$, $$\sum _{n=1}^{\infty}(-1)^{n-1}\frac{ H_{2 n}}{n^3}+\sum _{n=1}^{\infty}(-1)^{n-1}\frac{ H_{2 n}^{(2)}}{n^2}=2G^2+\frac{37}{64}\zeta(4),$$ that if we combine with the previous result, we obtain

$$\sum _{n=1}^{\infty}(-1)^{n-1}\frac{ H_{2 n}^{(2)}}{n^2}$$ $$=2G^2-\frac{353}{64}\zeta(4)-\frac{5}{4} \log ^2(2)\zeta(2)+\frac{35}{8}\log (2)\zeta (3) +\frac{5}{24} \log ^4(2)+5 \operatorname{Li}_4\left(\frac{1}{2}\right).$$

  • $\begingroup$ @User Thank you! I was in the process of editing. All fixed now. $\endgroup$ Aug 26, 2020 at 10:53
  • 3
    $\begingroup$ (+1) Revolutionary indeed. This work deserves publishing for sure. Congrats to Cornel. $\endgroup$ Aug 26, 2020 at 13:12

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