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A few days ago, I posted the following problems

Prove that \begin{equation} \int_0^{\pi/2}\ln^2(\cos x)\,dx=\frac{\pi}{2}\ln^2 2+\frac{\pi^3}{24}\\[20pt] -\int_0^{\pi/2}\ln^3(\cos x)\,dx=\frac{\pi}{2}\ln^3 2+\frac{\pi^3}{8}\ln 2 +\frac{3\pi}{4}\zeta(3) \end{equation}

and the OP receives some good answers even I then could answer it.


My next question is finding the closed-forms for

\begin{align} \int_0^{\pi/4}\ln^2(\sin x)\,dx\tag1\\[20pt] \int_0^{\pi/4}\ln^2(\cos x)\,dx\tag2\\[20pt] \int_0^1\frac{\ln t~\ln\big(1+t^2\big)}{1+t^2}dt\tag3 \end{align}

I have a strong feeling that the closed-forms exist because we have nice closed-forms for \begin{equation} \int_0^{\pi/4}\ln(\sin x)\ dx=-\frac12\left(C+\frac\pi2\ln2\right)\\ \text{and}\\ \int_0^{\pi/4}\ln(\cos x)\ dx=\frac12\left(C-\frac\pi2\ln2\right). \end{equation} The complete proofs can be found here.

As shown by Mr. Lucian in his answer below, the three integrals are closely related, so finding the closed-form one of them will also find the other closed-forms. How to find the closed-forms of the integrals? Could anyone here please help me to find the closed-form, only one of them, preferably with elementary ways (high school methods)? If possible, please avoiding contour integration and double summation. Any help would be greatly appreciated. Thank you.

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  • $\begingroup$ What type of integration is this, I was taught only basic integrals, can you tell me waht topic which they come under so that I can read about them and learn about them. $\endgroup$
    – RE60K
    Sep 2, 2014 at 16:02
  • $\begingroup$ @Aditya Log-Trig integrals $\endgroup$ Sep 2, 2014 at 16:04
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    $\begingroup$ As I wrote in my answer on this thread, this question is equivalent to finding the closed form expression for $\text{Li}_3\Big(\frac{1+i}2\Big)$. $\endgroup$
    – Lucian
    Sep 4, 2014 at 23:55

7 Answers 7

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Following the same approach as in this answer,

$$ \begin{align} &\int_{0}^{\pi/4} \log^{2} (2 \sin x) \ dx = \int_{0}^{\pi/4} \log^{2}(2) \ dx + 2 \log 2 \int_{0}^{\pi/4}\log(\sin x) \ dx + \int_{0}^{\pi /4}\log^{2}(\sin x) \ dx \\ &= \frac{\pi}{4} \log^{2}(2) - \log (2) \left(G + \frac{\pi}{2} \log (2) \right) + \int_{0}^{\pi/4} \log^{2}(\sin x) \ dx \\ &= \int_{0}^{\pi /4} \left(x- \frac{\pi}{2} \right)^{2} \ dx + \text{Re} \int_{0}^{\pi/4} \log^{2}(1-e^{2ix}) \ dx \\ &= \frac{7 \pi^{3}}{192} + \frac{1}{2} \text{Im} \int_{{\color{red}{1}}}^{i} \frac{\log^{2}(1-z)}{z} \ dz \\ &= \frac{7 \pi^{3}}{192} + \frac{1}{2} \text{Im} \left(\log^{2}(1-i) \log(i) + 2 \log(1-i) \text{Li}_{2}(1-i) - 2 \text{Li}_{3}(1-i) \right) \\ &= \frac{7 \pi^{3}}{192} + \frac{1}{2} \left(\frac{\pi}{8} \log^{2}(2) - \frac{\pi^{3}}{32} + \log(2) \ \text{Im} \ \text{Li}_{2}(1-i) - \frac{\pi}{2} \text{Re} \ \text{Li}_{2}(1-i)- 2 \ \text{Im} \ \text{Li}_{3}(1-i)\right) . \end{align}$$

Therefore,

$$ \begin{align}\int_{0}^{\pi/4} \log^{2}(\sin x) \ dx &= \frac{\pi^{3}}{48} + G \log(2)+ \frac{5 \pi}{16}\log^{2}(2) + \frac{\log(2)}{2} \text{Im} \ \text{Li}_{2}(1-i) - \frac{\pi}{4} \text{Re} \ \text{Li}_{2}(1-i) \\ &- \text{Im} \ \text{Li}_{3}(1-i) \approx 2.0290341368 . \end{align}$$

The answer could be further simplified using the dilogarithm reflection formula $$\text{Li}_{2}(x) {\color{red}{+}} \text{Li}_{2}(1-x) = \frac{\pi^{2}}{6} - \log(x) \log(1-x) $$

and the fact that $$ \text{Li}_{2}(i) = - \frac{\pi^{2}}{48} + i G.$$

EDIT:

Specifically, $$\text{Li}_{2}(1-i) = \frac{\pi^{2}}{16} - i G - \frac{i \pi}{4} \log(2). $$

So $$\int_{0}^{\pi /4} \log^{2}(\sin x) \ dx = \frac{\pi^{3}}{192} + G\frac{ \log(2)}{2} + \frac{3 \pi}{16} \log^{2}(2) - \text{Im} \ \text{Li}_{3}(1-i).$$

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  • $\begingroup$ Thank you Mr. Random variable, for now I'll upvote this answer and study it (ô‿ô) $\endgroup$ Sep 5, 2014 at 12:21
  • $\begingroup$ There were a couple of errors that I fixed. $\endgroup$ Sep 5, 2014 at 15:54
  • $\begingroup$ I added a bit more detail to the other answer. $\endgroup$ Sep 6, 2014 at 7:52
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    $\begingroup$ Ack. Apparently I accidentally downvoted, and now it's locked in. I'll fix this if you edit. :( $\endgroup$
    – David H
    Sep 6, 2014 at 10:57
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    $\begingroup$ @DavidH Why did you do that? I consider this answer as a strong candidate winner for my bounty! ᕙ(`皿´)╯ $\endgroup$ Sep 6, 2014 at 11:18
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$$\int_0^\frac\pi4\Big(\ln\sin x\Big)^2~dx~=~\dfrac{23}{384}\cdot\pi^3~+~\dfrac9{32}\cdot\pi\cdot\ln^22~+~\underbrace{\beta(2)}_\text{Catalan}\cdot\dfrac{\ln2}2~-~\Im\bigg[\text{Li}_3\bigg(\dfrac{1+i}2\bigg)\bigg].$$

$$\int_0^\frac\pi4\Big(\ln\cos x\Big)^2~dx~=~\dfrac{-7}{384}\cdot\pi^3~+~\dfrac7{32}\cdot\pi\cdot\ln^22~-~\underbrace{\beta(2)}_\text{Catalan}\cdot\dfrac{\ln2}2~+~\Im\bigg[\text{Li}_3\bigg(\dfrac{1+i}2\bigg)\bigg].$$


$$S=\int_0^\frac\pi4\Big(\ln\sin x\Big)^2~dx~+~\int_0^\frac\pi4\Big(\ln\cos x\Big)^2~dx=I+J.$$

But, by a simple change of variable, $t=\dfrac\pi2-x,~J$ can be shown to equal $\displaystyle\int_\frac\pi4^\frac\pi2\Big(\ln\sin x\Big)^2~dx$,

in which case $I+J=\displaystyle\int_0^\frac\pi2\Big(\ln\sin x\Big)^2~dx=\dfrac{\pi^3}{24}+\dfrac\pi2\ln^22.~$ So we know their sum! Now all

that's left to do is to find out their difference, $D=I-J.~$ Then we'll have $I=\dfrac{S+D}2$ and

$J=\dfrac{S-D}2$.


$$D=I-J=\int_0^\frac\pi4\Big(\ln\sin x\Big)^2~dx-\int_0^\frac\pi4\Big(\ln\cos x\Big)^2~dx=\int_0^\frac\pi4\Big(\ln^2\sin x-\ln^2\cos x\Big)~dx$$

$$=\int_0^\frac\pi4\Big(\ln\sin x-\ln\cos x\Big)~\Big(\ln\sin x+\ln\cos x\Big)~dx=\int_0^\frac\pi4\ln\frac{\sin x}{\cos x}~\ln\big(\sin x~\cos x\big)~dx=$$

$$=\int_0^\frac\pi4\ln\tan x\cdot\ln\frac{\sin2x}2~dx=\frac12\int_0^\frac\pi2\ln\tan\frac x2\cdot\ln\frac{\sin x}2~dx=\int_0^1\ln t\cdot\ln\frac t{1+t^2}\cdot\frac{dt}{1+t^2}$$

where the last expression was obtained by using the famous Weierstrass substitution, $t=\tan\dfrac x2$

$$=\int_0^1\frac{\ln t\cdot\Big[\ln t-\ln(1+t^2)\Big]}{1+t^2}dt~=~\int_0^1\frac{\ln^2t}{1+t^2}dt~-~\int_0^1\frac{\ln t~\ln\big(1+t^2\big)}{1+t^2}dt~=~\frac{\pi^3}{16}-K,$$

where $~K=2~\Im\bigg[\text{Li}_3\bigg(\dfrac{1+i}2\bigg)\bigg]-\dfrac{\pi^3}{64}-\dfrac\pi{16}\ln^22-\underbrace{\beta(2)}_\text{Catalan}\ln2.~$ It follows then that our two

definite integrals possess a closed form expression if and only if $~\text{Li}_3\bigg(\dfrac{1+i}2\bigg)$ has one as well. As

an aside, $~\Re\bigg[\text{Li}_3\bigg(\dfrac{1+i}2\bigg)\bigg]=\dfrac{\ln^32}{48}-\dfrac5{192}~\pi^2~\ln2+\dfrac{35}{64}~\zeta(3).~$ Also, $~K=\displaystyle\sum_{n=1}^\infty\frac{(-1)^n~H_n}{(2n+1)^2}$.

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  • $\begingroup$ +1 Mr. Lucian, your method is easy to follow but I have a question, how to evaluate $$\int_0^1\frac{\ln t~\ln\big(1+t^2\big)}{1+t^2}dt$$ $\endgroup$ Sep 3, 2014 at 10:41
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    $\begingroup$ @Lucian Do you mind explaining how you compute $\displaystyle K=\sum^\infty_{n=1}\frac{(-1)^n H_n}{(2n+1)^2}$? Thanks. $\endgroup$ Sep 3, 2014 at 10:53
  • $\begingroup$ @SuperAbound: I didn't. Mathematica did. Basically, I struggled for hours to find an expression which my CAS could finally evaluate, since all other intermediary results available on this page were impenetrable, and, in the end, I actually did. Then I posted the result. As far as the integral is concerned, I tried the two main approaches: series expansion and the Feynman trick, both of which failed miserably, but your Euler sum gives me hope. $($How on earth did you even come up with it, anyway?$)$ My only idea would be to write $H_n$ as a sum, then switch the order of the two summations. $\endgroup$
    – Lucian
    Sep 3, 2014 at 15:03
  • $\begingroup$ @V-Moy: In my opinion, SuperAbound's Euler sum seems the way to go. Indeed, many such similar sums have been solved on this site, so I am rather hopeful. $\endgroup$
    – Lucian
    Sep 3, 2014 at 15:18
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    $\begingroup$ @Anastasiya-Romanova 秀 check this link math.stackexchange.com/q/3449505 $\endgroup$ Aug 29, 2021 at 19:11
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By setting $x=\arctan t$ we have: $$\int_{0}^{\pi/4}\log^2(\cos x)\,dx = \frac{1}{4}\int_{0}^{1}\frac{\log^2(1+t^2)}{1+t^2}.$$ Attack plan: get the Taylor series of $\log^2(1+t^2)$ and integrate it termwise.

Since $$-\log(1-z)=\sum_{n=1}^{+\infty}\frac{z^n}{n}$$ it follows that $$[z^n]\log^2(1-z)=\sum_{k=1}^{n-1}\frac{1}{k(n-k)}=2\frac{H_{n-1}}{n},$$ $$\log^2(1+t^2)=\sum_{n=2}^{+\infty}2\frac{H_{n-1}}{n}(-1)^n t^{2n}.\tag{1}$$ If now we set $$\mathcal{J}_m = \int_{0}^{1}\frac{t^{2m}}{t^2+1}\,dt $$ we have $\mathcal{J}_0=\frac{\pi}{4}$ and $\mathcal{J}_{m+1}+\mathcal{J}_m = \frac{1}{2m+1}$, hence: $$\mathcal{J}_m = (\mathcal{J}_m+\mathcal{J}_{m-1})-(\mathcal{J}_{m-1}+\mathcal{J}_{m-2})+\ldots\pm(\mathcal{J}_1+\mathcal{J}_0)\mp\mathcal{J}_0,$$ $$\mathcal{J}_m = \sum_{j=0}^{m-1}\frac{(-1)^j}{(2m-2j-1)}+(-1)^m\frac{\pi}{4}=(-1)^m \sum_{j\geq m}\frac{(-1)^j}{2j+1}.\tag{2}$$ From $(1)$ and $(2)$ it follows that: $$\int_{0}^{\pi/4}\log^2(\cos x)\,dx=\frac{1}{2}\sum_{n=2}^{+\infty}\frac{H_{n-1}}{n}\sum_{r\geq n}\frac{(-1)^r}{2r+1},\tag{3}$$ and summation by parts gives:

$$\int_{0}^{\pi/4}\log^2(\cos x)\,dx=\frac{1}{4}\sum_{n=2}^{+\infty}(H_n^2-H_n^{(2)})\frac{(-1)^n}{2n+1}.\tag{4}$$

UPDATE: the question is now set in an answer to another question. This site (many thanks to @gammatester) is devoted to the evaluation of sums like the one appearing in the RHS of $(4)$. Through Euler-Landen's identity (see the line below $(608)$ in the linked site) it is not too much difficult to see that the RHS of $(4)$ depends on $\operatorname{Li}_3\left(\frac{1+i}{2}\right)$ as stated in the @Lucian's answer.

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    $\begingroup$ +1 for your answer before I sleep Mr. Jack D'Aurizio (ô‿ô) $\endgroup$ Sep 2, 2014 at 18:00
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    $\begingroup$ another approach: using the following identity: $$\displaystyle \frac{\ln^2(1-x)}{1-x}=\sum_{n=1}^{\infty}\left(H_n^2-H_n^{(2)}\right)x^n$$ replacing $x$ with $-x^2$ and integrating both sides from $0$ to $1$ gives: $$\int_{0}^{1}\frac{\log^2(1+x^2)}{1+x^2}\ dx=\sum_{n=1}^{\infty}(H_n^2-H_n^{(2)})\frac{(-1)^n}{2n+1}$$ $\endgroup$ May 2, 2019 at 1:35
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The strategy in this post will be included in another paper.


A solution (in large steps) by Cornel Ioan Valean

In my opinion, this is a very magical & powerful way that manages to circumvent the necessity of using the already famous method proposed by Random Variable which I think most posts on MSE use it for such integrals. It's time for a new way to come in place and join the existing one!

In this post, we magically prove that $$\int_0^1\frac{\log x\log(1+x^2)}{1+x^2}\textrm{d}x=-\frac{\pi}{16} \log ^2(2) - \log (2)G-\frac{\pi ^3}{64}+2\Im\biggr \{\operatorname{Li}_3\left(\frac{1+i}{2}\right)\biggr \},$$ by wisely combining a result from the book, (Almost) Impossible Integrals, Sums, and Series, namely the special Fourier series (see eq. 3.284, page 244, and eq. 3.288, page 247), \begin{equation} \begin{aligned} \small \sum_{n=1}^{\infty} (-1)^{n-1}\left(\psi\left(\frac{n+1}{2}\right)-\psi\left(\frac{n}{2}\right)-\frac{1}{n}\right)\sin(2nx)&\small=\sum_{n=1}^{\infty} (-1)^{n-1}\left(\int_0^1 t^{n-1}\frac{1-t}{1+t} \textrm{d}t\right)\sin(2nx)\\ &=-\cot(x)\log(\cos(x)), \end{aligned} \end{equation} where $\displaystyle 0< x<\frac{\pi}{2}$, and the Cornel's integral,

$$\int_0^{\pi/2} x\frac{\log(\cos x)}{\sin x}\textrm{d}x=2\log(2)G-\frac{\pi}{8}\log^2(2)-\frac{5}{32}\pi^3+4\Im\left\{\text{Li}_3\left(\frac{1+i}{2}\right)\right\},$$ already calculated in this post How can you approach $\int_0^{\pi/2} x\frac{\ln(\cos x)}{\sin x}dx$.

Proof: We differentiate both sides of the Fourier series that leads to $$2 \sum_{n=1}^{\infty} (-1)^{n-1}\left(\int_0^1 t^{n-1}\frac{1-t}{1+t} \textrm{d}t\right)n\cos(2nx)=1+\frac{\log(\cos(x))}{\sin^2(x)},$$ and if we multiply both side by $x \sin(x)$ and integrate from $x=0$ to $x=\pi/2$, we arrive at $$\int_0^{\pi/2} x\sin(x)\textrm{d}x+\int_0^{\pi/2}x\frac{\log(\cos(x))}{\sin(x)}\textrm{d}x$$

$$=2 \log (2)-1+2 \log (2)\underbrace{\int_0^1 \frac{\log (x)}{1+x^2}\textrm{d}x}_{\displaystyle \text{Trivial}}+\frac{1}{2}\underbrace{\int_0^1 \log (x) \log \left(1-x^2\right)\textrm{d}x}_{\displaystyle \text{Trivial}}$$ $$+\frac{1}{2}\underbrace{\int_0^1\frac{\log (x) \log \left(1-x^2\right)}{x^2}\textrm{d}x}_{\displaystyle \text{Trivial}}-2\underbrace{\int_0^1\frac{ \log (x) \log \left(1-x^4\right)}{1-x^4}\textrm{d}x}_{\displaystyle \text{Beta function in disguise}}$$ $$+2\underbrace{\int_0^1\frac{x^2 \log (x) \log \left(1-x^4\right)}{1-x^4}\textrm{d}x}_{\displaystyle \text{Beta function in disguise}}+2\color{blue}{\int_0 ^1 \frac{\log (x) \log(1+x^2)}{1+x^2}\textrm{d}x},$$ from which the desired result follows.

Note the following values of the Beta function forms in disguise:

$$\int_0^1 \frac{\log (x) \log \left(1-x^4\right)}{1-x^4} \textrm{d}x=\frac{1}{16}\int_0^1 \frac{\log(x)\log (1-x)}{ x^{3/4}(1-x) } \textrm{d}x$$ $$=\frac{7 }{4}\zeta (3)+\frac{\pi ^3}{32}-\frac{3}{16}\log (2)\pi ^2-\frac{\pi }{4}G-\frac{3}{2}\log(2)G,$$ and $$\int_0^1 \frac{x^2\log (x) \log \left(1-x^4\right)}{1-x^4} \textrm{d}x=\frac{1}{16}\int_0^1 \frac{\log(x)\log (1-x)}{x^{1/4}(1-x)} \textrm{d}x$$ $$=\frac{7}{4} \zeta (3)+\frac{3}{2} \log (2)G-\frac{1}{4} \pi G-\frac{3}{16}\log(2)\pi^2-\frac{\pi ^3}{32}.$$

A note: this method can also be adjusted to extract other very difficult integrals, which is possible by further exploiting and developing ideas like the ones in the paper A symmetry-related treatment of two fascinating sums of integrals by C.I. Valean.

End of story

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    $\begingroup$ (+1) hats off to Cornel. His magic never stop. $\endgroup$ Aug 20, 2020 at 22:47
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we can prove, using the same strategy of Random Variable, the following equality:

$$\int_0^{\pi/4}\ln^2(\cos x)\ dx=\frac7{192}\pi^3+\frac5{16}\pi\ln^22-\frac12\ln2G-\text{Im}\operatorname{Li_3}(1+i)$$ proof :
\begin{align*} \ln(1+e^{2ix}) &= \ln (e^{-ix}+e^{ix}) + \ln(e^{ix}) \\ &= \ln(2\cos x) + ix \end{align*}

squaring both sides and integrating, we get

$$\int_0^{\pi/4}\ln^2(1+e^{2ix})\ dx=\int_0^{\pi/4}(\ln(2\cos x)+ix)^2\ dx$$ equating the real parts on both sides and rearranging the terms, we have:

$$ \int_0^{\pi/4}\ln^2(\cos x)\ dx=\int_0^{\pi/4}(x^2-\ln^22)\ dx-2\ln2\int_0^{\pi/4}\ln(\cos x)\ dx$$ $$+\text{Re}\int_0^{\pi/4}\ln^2(1+e^{2ix})\ dx$$ $$=\frac{\pi^3}{192}-\frac{\pi}{4}\ln^22-2\ln2\left(\frac12G-\frac{\pi}{4}\ln2\right)+\text{Re}\int_0^{\pi/4}\ln^2(1+e^{2ix})\ dx$$ $$=\frac{\pi^3}{192}+\frac{\pi}{4}\ln^22-\ln2G+\text{Re}\int_0^{\pi/4}\ln^2(1+e^{2ix})\ dx \tag{1}$$

Evaluating the last integral: \begin{align*} I&=\text{Re}\int_0^{\pi/4}\ln^2(1+e^{2ix})\ dx=\frac12\text{Im}\int_1^i\frac{\ln^2(1+x)}{x}\ dx\\ &=\frac12\text{Im}\left(\ln(-i)\ln^2(1+i)+2\ln(1+i)\operatorname{Li_2}(1+i)-2\operatorname{Li_3}(1+i)\right)\\ &=\frac{\pi^3}{32}+\frac{\pi}{16}\ln^22+\frac12\ln2G-\text{Im}\operatorname{Li_3}(1+i)\tag{2} \end{align*} Plugging $(2)$ in $(1)$ we get our result.

note that we used: $$\ln(-i)=-\frac{\pi}{2}i$$ $$\ln(1+i)=\frac12\ln2+\frac{\pi}{4}i$$ $$\operatorname{Li_2}(1+i)=\frac{\pi^2}{16}+\left(\frac{\pi}{4}\ln2+G\right)i$$ which give us: $$\ln(-i)\ln^2(1+i)=\frac{\pi^2}{8}\ln2+\left(\frac{\pi^3}{32}-\frac{\pi}{8}\ln^22\right)i$$ $$\ln(1+i)\operatorname{Li_2}(1+i) =-\frac{\pi}{4}G-\frac{\pi^2}{32}\ln2+\left(\frac12\ln2G+\frac{\pi^3}{64}+\frac{\pi}{8}\ln^22\right)i$$

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my approach to problem $(3)$: \begin{align} I&=\int_0^1\frac{\ln x\ln(1+x^2)}{1+x^2}\ dx=-2\int_0^{\pi/4}\ln(\tan x)\ln(\cos x)\ dx\\ &=-2\int_0^{\pi/4}\ln(\sin x)\ln(\cos x)\ dx+2\int_0^{\pi/4}\ln^2(\cos x)\ dx\\ &=-\int_0^{\pi/2}\ln(\sin x)\ln(\cos x)\ dx+2\int_0^{\pi/4}\ln^2(\cos x)\ dx\\ &=-\left(\frac{\pi}{2}\ln^22-\frac{\pi^3}{48}\right)+2\left(\frac7{192}\pi^3+\frac5{16}\pi\ln^22-\frac12\ln2~G-\text{Im}\operatorname{Li_3}(1+i)\right)\\ &=\frac3{32}\pi^3+\frac{\pi}8\ln^22-\ln2~G-2\text{Im}\operatorname{Li_3}(1+i) \end{align}

note that we evaluated the first integral using the derivative of beta function and as follows: \begin{align} J&=\int_0^{\pi/2}\ln(\sin x)\ln(\cos x)\ dx=\frac18\frac{\partial^2}{\partial{a}\partial{b}}\beta(a,b)\Bigg\rvert_{a\to1/2,~b\to1/2}\\ &=\frac18\beta(a,b)\left(\left(\psi(a)-\psi(a+b)\right)\left(\psi(b)-\psi(a+b)\right)-\psi^{(1)}(a+b)\right)\Bigg\rvert_{a\to1/2,~b\to1/2}\\ &=\frac18\beta(1/2,1/2)\left((\psi(1/2)-\psi(1))^2-\psi^{(1)}(1)\right)\\ &=\frac{\pi}8\left(4\ln^22-\zeta(2)\right)\\ &=\frac{\pi}2\ln^22-\frac{\pi^3}{48} \end{align}

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Evaluation of $\int_0^1\frac{\ln(x)\ln(1+x^2)}{1+x^2}dx:$

First note that

$$\int_0^1\frac{\ln(x)\ln(1+x^2)}{1+x^2}dx=\sum_{n=1}^\infty (-1)^n H_n\int_0^1 \ln(x) x^{2n}dx$$ $$=\sum_{n=1}^\infty\frac{(-1)^nH_n}{(2n+1)^2}=\sum_{n=0}^\infty\frac{(-1)^nH_n}{(2n+1)^2}.$$

We have here

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

$$=\ln(2)+H_n-H_{2n+1}+\frac1{2n+1}.$$

Multiply both sides by $\frac{(-1)^n}{(2n+1)^2}$ then $\sum_{n=0}^\infty$ we get

$$\text{G}\ln(2)+\sum_{n=0}^\infty\frac{(-1)^nH_n}{(2n+1)^2}-\sum_{n=0}^\infty\frac{(-1)^nH_{2n+1}}{(2n+1)^2}+\underbrace{\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^3}}_{\pi^3/32}$$

$$=\int_0^1\frac{1}{1+x}\left(\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n+1)^2}\right)dx=\int_0^1\frac{1}{1+x}\left(\Im\frac{\text{Li}_2(ix)}{x}\right)dx$$

$$\int_0^1\frac{1}{1+x}\left(\Im\int_0^1-\frac{i\ln(y)}{1-ixy}dy\right)dx=\int_0^1\frac{1}{1+x}\left(\int_0^1-\frac{\ln(y)}{1+x^2y^2}dy\right)dx$$

$$\overset{xy=t}{=}\int_0^1\int_0^x\frac{\ln(x/t)}{x(1+x)(1+t^2)}dtdx=\int_0^1\frac{1}{1+t^2}\left(\int_t^1\frac{\ln(x/t)}{x(1+x)}dx\right)dt$$

$$=\int_0^1\frac{1}{1+t^2}\left(\text{Li}_2(-t)+\frac12\ln^2(t)+\ln(2)\ln(t)+\frac12\zeta(2)\right)dt$$

$$=\int_0^1\frac{\text{Li}_2(-t)}{1+t^2}dt+\frac{\pi^3}{32}-\text{G}\ln(2)+\frac{\pi^3}{48}$$

Therefore

$$\sum_{n=0}^\infty\frac{(-1)^nH_n}{(2n+1)^2}=\sum_{n=0}^\infty\frac{(-1)^nH_{2n+1}}{(2n+1)^2}+\int_0^1\frac{\text{Li}_2(-t)}{1+t^2}dt-2\text{G}\ln(2)+\frac{\pi^3}{48}\tag1$$

where

$$\sum_{n=0}^\infty\frac{(-1)^nH_{2n+1}}{(2n+1)^2}=\Im\sum_{n=1}^\infty\frac{i^nH_{n}}{n^2}=-\frac{\pi}{16}\ln^2(2)-\frac12\text{G}\ln(2)+\Im\operatorname{Li}_3(1+i)\tag2$$

and

$$\int_0^1\frac{\text{Li}_2(-t)}{1+t^2}dt=\int_0^1\frac{1}{1+t^2}\left(\int_0^1\frac{t\ln(x)}{1+tx}dx\right)dt$$

$$=\int_0^1\ln x\left(\int_0^1\frac{t}{(1+t^2)(1+tx)}dt\right)dx$$

$$=\int_0^1\ln(x)\left(\frac{\pi}{4}\frac{x}{1+x^2}+\frac{\ln(2)}{2}\frac{1}{1+x^2}-\frac{\ln(1+x)}{1+x^2}\right)dx$$

$$=-\frac{\pi^3}{192}-\frac12\text{G}\ln(2)-\int_0^1\frac{\ln(x)\ln(1+x)}{1+x^2}dx$$

Substitute

$$\int_0^1\frac{\ln(x)\ln(1+x)}{1+x^2}dx=3\ \text{Im}\operatorname{Li}_3(1+i)-\frac{5\pi^3}{64}-\frac{3\pi}{16}\ln^2(2)-2\ln(2)\ G,$$

we get

$$\int_0^1\frac{\text{Li}_2(-t)}{1+t^2}dt=-3\ \text{Im}\operatorname{Li}_3(1+i)+\frac{7\pi^3}{96}+\frac{3\pi}{16}\ln^2(2)+\frac32\ln(2)\ G.\tag3$$

Substitute $(2)$ and $(3)$ in $(1)$, we get

$$\sum_{n=0}^\infty\frac{(-1)^nH_n}{(2n+1)^2}:=\int_0^1\frac{\ln(x)\ln(1+x^2)}{1+x^2}dx=\frac{3\pi^3}{32}+\frac{\pi}8\ln^2(2)-\text{G}\ln(2)-2\Im\operatorname{Li_3}(1+i).$$

$\endgroup$
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  • 1
    $\begingroup$ (+1) Ingenious work. $\endgroup$ Aug 21, 2020 at 11:28
  • $\begingroup$ @user97357329 Thank you :) $\endgroup$ Aug 21, 2020 at 11:30

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