Jtdylktuy

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{3pi}{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),dxtag1\[20pt]
int_0^{pi/4}ln^2(cos x),dxtag2\[20pt]
int_0^1frac{ln t~lnbig(1+t^2big)}{1+t^2}dttag3
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=-frac12left(C+fracpi2ln2right)\
text{and}\
int_0^{pi/4}ln(cos x) dx=frac12left(C-fracpi2ln2right).
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.

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

$$int_0^fracpi4Big(lnsin xBig)^2~dx~=~dfrac{23}{384}cdotpi^3~+~dfrac9{32}cdotpicdotln^22~+~underbrace{beta(2)}_text{Catalan}cdotdfrac{ln2}2~-~Imbigg[text{Li}_3bigg(dfrac{1+i}2bigg)bigg].$$

$$int_0^fracpi4Big(lncos xBig)^2~dx~=~dfrac{-7}{384}cdotpi^3~+~dfrac7{32}cdotpicdotln^22~-~underbrace{beta(2)}_text{Catalan}cdotdfrac{ln2}2~+~Imbigg[text{Li}_3bigg(dfrac{1+i}2bigg)bigg].$$

$$S=int_0^fracpi4Big(lnsin xBig)^2~dx~+~int_0^fracpi4Big(lncos xBig)^2~dx=I+J.$$

But, by a simple change of variable, $t=dfracpi2-x,~J$ can be shown to equal $displaystyleint_fracpi4^fracpi2Big(lnsin xBig)^2~dx$,

in which case $I+J=displaystyleint_0^fracpi2Big(lnsin xBig)^2~dx=dfrac{pi^3}{24}+dfracpi2ln^22.~$ 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

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

$$D=I-J=int_0^fracpi4Big(lnsin xBig)^2~dx-int_0^fracpi4Big(lncos xBig)^2~dx=int_0^fracpi4Big(ln^2sin x-ln^2cos xBig)~dx$$

$$=int_0^fracpi4Big(lnsin x-lncos xBig)~Big(lnsin x+lncos xBig)~dx=int_0^fracpi4lnfrac{sin x}{cos x}~lnbig(sin x~cos xbig)~dx=$$

$$=int_0^fracpi4lntan xcdotlnfrac{sin2x}2~dx=frac12int_0^fracpi2lntanfrac x2cdotlnfrac{sin x}2~dx=int_0^1ln tcdotlnfrac t{1+t^2}cdotfrac{dt}{1+t^2}$$

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

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

where $~K=2~Imbigg[text{Li}_3bigg(dfrac{1+i}2bigg)bigg]-dfrac{pi^3}{64}-dfracpi{16}ln^22-underbrace{beta(2)}_text{Catalan}ln2.~$ It follows then that our two

definite integrals possess a closed form expression if and only if $~text{Li}_3bigg(dfrac{1+i}2bigg)$ has one as well. As

an aside, $~Rebigg[text{Li}_3bigg(dfrac{1+i}2bigg)bigg]=dfrac{ln^32}{48}-dfrac5{192}~pi^2~ln2+dfrac{35}{64}~zeta(3).~$ Also, $~K=displaystylesum_{n=1}^inftyfrac{(-1)^n~H_n}{(2n+1)^2}$.

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
$$[z^n]log^2(1-z)=sum_{k=1}^{n-1}frac{1}{k(n-k)}=2frac{H_{n-1}}{n},$$
$$log^2(1+t^2)=sum_{n=2}^{+infty}2frac{H_{n-1}}{n}(-1)^n t^{2n}.tag{1}$$
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:
$$mathcal{J}_m = (mathcal{J}_m+mathcal{J}_{m-1})-(mathcal{J}_{m-1}+mathcal{J}_{m-2})+ldotspm(mathcal{J}_1+mathcal{J}_0)mpmathcal{J}_0,$$
$$mathcal{J}_m = sum_{j=0}^{m-1}frac{(-1)^j}{(2m-2j-1)}+(-1)^mfrac{pi}{4}=(-1)^m sum_{jgeq m}frac{(-1)^j}{2j+1}.tag{2}$$
From $(1)$ and $(2)$ it follows that:
$$int_{0}^{pi/4}log^2(cos x),dx=frac{1}{2}sum_{n=2}^{+infty}frac{H_{n-1}}{n}sum_{rgeq n}frac{(-1)^r}{2r+1},tag{3}$$
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 $operatorname{Li}_3left(frac{1+i}{2}right)$ as stated in the @Lucian's answer.