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How to prove (any of) the following

  • $\small \sum _{n=1}^{\infty } \frac{1}{n^4 2^n \binom{3 n}{n}}=-2 \pi \Im(\text{Li}_3(1+i))-\frac{21 \text{Li}_4\left(\frac{1}{2}\right)}{2}-\frac{57}{8} \zeta (3) \log (2)+\frac{83 \pi ^4}{480}-\frac{23}{48} \log ^4(2)+\frac{7}{12} \pi ^2 \log ^2(2)$

  • $\int_0^1 \frac{x \text{Li}_2(x) \log (1-x)}{x^2+1}dx=\frac{C^2}{2}-\frac{1}{8} \pi C \log (2)+\frac{15 \text{Li}_4\left(\frac{1}{2}\right)}{16}-\frac{511 \pi ^4}{46080}+\frac{5 \log ^4(2)}{128}-\frac{7}{384} \pi ^2 \log ^2(2)$

  • $\small \int_0^1 \frac{\text{Li}_2(x) \log \left(x^2-2 x+2\right)}{x} dx=\frac{1}{2} \pi \Im(\text{Li}_3(1+i))+\frac{5 \text{Li}_4\left(\frac{1}{2}\right)}{8}+\frac{35}{64} \zeta (3) \log (2)-\frac{577 \pi ^4}{23040}+\frac{5 \log ^4(2)}{192}-\frac{1}{96} \pi ^2 \log ^2(2)$

The first identity is an unproved conjecture of J. M. Borwein found by PSLQ, originally written in terms of multiple-polylogarithm $L_{3,1}$. I have already established the equivalence of $3$ identities but neither of them is trivial. Any help will be appreciated.


Update: Ali Shadhar had established the indeterminate integral. See here for proof of the first one using his result. The last one is direct via IBP.

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    $\begingroup$ Very nice question +1 $\endgroup$ Mar 31, 2020 at 8:11
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    $\begingroup$ I think " harmonic number " should be tagged $\endgroup$ Mar 31, 2020 at 8:13
  • $\begingroup$ @AliShather I think there's a maximum of $5$ tags allowed. $\endgroup$
    – saulspatz
    Mar 31, 2020 at 8:18
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    $\begingroup$ "The main purpose of this article is the evaluation of 85 specific logarithmic integrals, 89 alternating Euler sums and 263 polylogarithmic generalizations with their weights not exceed 5. By establishing linear relations between 3 kinds of values, we discover the common pattern on their closed-forms and present a systematic proof. Based on previous results, we solved series of problems on related integrals and series, among which 193 quadratic logarithmic integrals with weights no more than 4 are calculated analytically." Well, no one can say you haven't put any work into this! $\endgroup$ Mar 31, 2020 at 9:04

1 Answer 1

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I am going to establish two relations involving the following two integrals

$$I=\int_0^1\frac{x\text{Li}_2(x)\ln(1-x)}{1+x^2}\ dx$$

$$K=\int_0^1\frac{x\ln(x)\ln^2(1-x)}{1+x^2}\ dx$$

and solve for $I$ and $K$ by elimination.


The first relation

Start with using $\text{Li}_2(x)=-\int_0^1\frac{x\ln(y)}{1-xy}\ dy$

$$I=\int_0^1\frac{x\text{Li}_2(x)\ln(1-x)}{1+x^2}\ dx=-\int_0^1\ln(y)\left(\int_0^1\frac{x^2\ln(1-x)}{(1+x^2)(1-xy)}dx\right)dy$$

$$=-\int_0^1\ln(y)\left(\frac{1}{1+y^2}\left(G-\frac{\pi\ln(2)}{8}\right)+\frac{y}{1+y^2}\left(\frac{5\pi^2}{96}-\frac{\ln^2(2)}{8}\right)+\frac{\text{Li}_2\left(\frac{y}{y-1}\right)}{y(1+y^2)}\right)\ dy$$

$$=G^2-\frac{\pi\ln(2)}{8}G+\frac{5\pi^4}{4608}-\frac{\pi^2\ln^2(2)}{384}-\underbrace{\int_0^1\frac{\ln(y)\text{Li}_2\left(\frac{y}{y-1}\right)}{y(1+y^2)}\ dy}_{A}$$

$$A=\underbrace{\int_0^1\frac{\ln(y)\text{Li}_2\left(\frac{y}{y-1}\right)}{y}\ dy}_{A_1}-\underbrace{\int_0^1\frac{y\ln(y)\text{Li}_2\left(\frac{y}{y-1}\right)}{1+y^2}\ dy}_{A_2}$$

By integration by parts we have

$$A_1=-\frac12\int_0^1\frac{\ln^2(y)\ln(1-y)}{y(1-y)}\ dy=\frac12\sum_{n=1}^\infty H_n\int_0^1y^{n-1}\ln^2(y)\ dy=\sum_{n=1}^\infty\frac{H_n}{n^3}=\frac54\zeta(4)$$

For $A_2$, use Landen's identity

$$A_2=-\underbrace{\int_0^1\frac{y\ln(y)\text{Li}_2(y)}{1+y^2}\ dy}_{J}-\frac12\underbrace{\int_0^1\frac{y\ln(y)\ln^2(1-y)}{1+y^2}\ dy}_{K}$$

For $J$, use $\text{Li}_2(y)=-\int_0^1\frac{y\ln(t)}{1-ty}\ dt$ again

$$J=-\int_0^1\ln(t)\left(\int_0^1\frac{y^2\ln(y)}{(1+y^2)(1-yt)}dy\right)dt$$

$$=-\int_0^1\ln(t)\left(\frac{G}{1+t^2}+\frac{\pi^2 t}{48(1+t^2)}-\frac{\text{Li}_2(t)}{t}+\frac{t\text{Li}_2(t)}{1+t^2}\right)dt$$

$$=G^2+\frac{\pi^4}{2304}-\frac{\pi^4}{90}-J\Longrightarrow J=\frac12G^2-\frac{41\pi^4}{7680}$$

Collect all the pieces, we get

$$2I+K=G^2-\frac{\pi\ln(2)}{4}G-\frac{\pi^2\ln^2(2)}{192}-\frac{43\pi^4}{2880}\tag1$$


The second relation

We have the reflection identity $$\text{Li}_2(x)+\ln(x)\ln(1-x)=\zeta(2)-\text{Li}_2(1-x)$$

multiply both sides by $\frac{x\ln(1-x)}{1+x^2}$ then $\int_0^1$ we get

$$I+K=\zeta(2)\underbrace{\int_0^1\frac{x\ln(1-x)}{1+x^2}\ dx}_{B}-\underbrace{\int_0^1\frac{x\ln(1-x)\text{Li}_2(1-x)}{1+x^2}\ dx}_{C}$$

For both of $B$ and $C$, use $\Im \frac{1}{1-ix}=\frac{x}{1+x^2}$

$$B=\Im \int_0^1\frac{\ln(1-x)}{1-ix}\ dx\overset{1-x=t}{=}\Im\int_0^1\frac{\ln(t)}{1-i+it}\ dt$$

$$=-\Re\int_0^1\frac{i\ln(t)}{1-i+it}\ dt=-\Re\text{Li}_2\left(\frac{i}{i-1}\right)=\frac{\ln^2(2)}{8}-\frac{5\pi^2}{96}$$

where the last result follows from using the generalzaition

$$\int_0^1\frac{y\ln^{n}(x)}{1-y+yx}dx=(-1)^{n-1}n!\operatorname{Li}_{n+1}\left(\frac{y}{y-1}\right)$$ which can be found in the book Almost Impossible Integrals, sums and series in page 5.

$$C=\Im\int_0^1\frac{\ln(1-x)\text{Li}_2(1-x)}{1-ix}\ dx\overset{1-x=t}{=}\Im\frac{1}{1-i}\int_0^1\frac{\ln(t)\text{Li}_2(t)}{1-at}\ dt,\quad a=\frac{i}{i-1}$$

By using Cornel's identity

$$\int_0^1\frac{\ln(t)\text{Li}_2(t)}{1-at}\ dt=\frac{\text{Li}_2^2(a)}{2a}+3\frac{\text{Li}_4(a)}{a}-2\zeta(2)\frac{\text{Li}_2(a)}{a}$$

we have

$$C=-\frac12G^2+\frac{\pi\ln(2)}{8}G+\frac{\pi^2\ln^2(2)}{128}-\frac{223\pi^4}{46080}+\frac{5}{128}\ln^4(2)+\frac{15}{16}\text{Li}_4\left(\frac12\right)$$

Combine the results of $B$ and $C$ we get

$$I+K=\frac12G^2-\frac{\pi\ln(2)}{8}G+\frac{5\pi^2\ln^2(2)}{384}-\frac{59\pi^4}{15360}-\frac{5}{128}\ln^4(2)-\frac{15}{16}\text{Li}_4\left(\frac12\right)\tag2$$

Solving $(1)$ and $(2)$ we get

$$I=\frac12G^2-\frac{\pi\ln(2)}{8}G-\frac{7\pi^2\ln^2(2)}{384}-\frac{511\pi^4}{46080}+\frac{5}{128}\ln^4(2)+\frac{15}{16}\text{Li}_4\left(\frac12\right)$$

and

$$K=\frac{\pi^2\ln^2(2)}{32}-\frac{167\pi^4}{23040}-\frac{5}{64}\ln^4(2)-\frac{15}{16}\text{Li}_4\left(\frac12\right)$$


The interesting thing about this solution is that I didn't use any harmonic series. Big thanks to Cornel for his magical identities I used in my solution.

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    $\begingroup$ By the way, the integral representation for your sum is $2\int_0^1\frac1x \text{Li}_3\left(\frac{x^2-x^3}{2}\right)\ dx$. Still tough to crack I guess. $\endgroup$ Mar 31, 2020 at 10:14

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