About the integral $\int_{0}^{\pi/4}\log^4(\cos\theta)\,d\theta$

The value of logsine integrals $$\int_0^\frac{\pi}4 \log^k(\sin(\theta))d\theta, \ \ \int_0^\frac{\pi}4 \log^k(\cos(\theta))d\theta$$ Where $k=1,2$ are well known. Moreover, this answer by nospoon provides (essentially through the nice structure of the Fourier series of $\log\left(2\left|\sin\frac{x}{2}\right|\right)$) two closed forms for quite non-elementary integrals:

\begin{eqnarray*} \int_{0}^{\pi/4} \log^3(\sin\theta)\,d\theta &=& 3\,\text{Im}\,\text{Li}_4(1-i)-\frac{25\pi^3}{256}\log(2)+\frac{3}{2}\log(2)\,\text{Im}\,\text{Li}_3\left(\tfrac{1+i}{2}\right)\\ && -\frac{3K}{8}\log^2(2)-\frac{17\pi}{64}\log^3(2)-\frac{3\pi}{8}\zeta(3)+\frac{3}{4}\beta(4)\tag{1} \end{eqnarray*} and by differentiating the Euler Beta function $\int_0^\frac{\pi}2 \sin^a(\theta)d\theta$ three times, the logsine integral $\int_0^\frac{\pi}2 \log^3(\sin(\theta))d\theta$ is trivially evaluated, from which and reflection $\theta\to \frac{\pi}2-\theta$ the result below follows: \begin{eqnarray}\label{intlog3}\notag \int_{0}^{\pi/4} \log^3(\cos\theta)\,d\theta &=& -\frac{7\pi^3}{256}\log(2)-\frac{15\pi}{64}\log^3(2)+\frac{3K}{8}\log^2(2)-\frac{3\pi}{8}\zeta(3)\\ && -\frac{3}{4}\beta(4)-\frac{3}{2}\log(2)\,\text{Im}\,\text{Li}_3\left(\tfrac{1+i}{2}\right)-3\,\text{Im}\,\text{Li}_4(1-i).\tag{2} \end{eqnarray}

Q: I am interested in a closed form evaluation (in terms of Euler sums) for the integrals $$\color{blue}{ \int_{0}^{\pi/4}\log^4(\sin\theta)\,d\theta,\qquad \int_{0}^{\pi/4}\log^4(\cos\theta)\,d\theta }$$ whose sum is clearly given by $\frac{19\pi^5}{480}+\frac{\pi^3}{4}\log^2(2)+\frac{\pi}{2}\log^4(2)+3\pi\zeta(3)\log(2)$ thanks to Euler's Beta function again. Are their value already known in the literature? If so, does the evaluation procedure exploit the convolution identity $$ \log^2\left(2\sin\frac{x}{2}\right)\stackrel{L^2(0,\pi)}{=}\frac{\pi^2}{12}+\sum_{n\geq 1}\cos(nx)\frac{H_n+H_{n-1}}{n}\quad?\tag{3} $$

Based on my limited knowledge on logsine integrals, value of these two are not given in literature until this July when "paper $1$" (see link below) was published, and based on whose associated algorithm an equivalent $_pF_q$ representation of these integrals is given in "paper $2$". The solution does not depend on the mentioned convolutional identity but MZV theory instead. Indeed, let $x\to \tan^{-1}(u)$, these integrals are converted to $4$-admissible logarithmic integrals (for its definition, see "paper $1$"). Using the algorithm given by Paper $1$ one immediately get the result: $$\small \int_0^{\frac{\pi }{4}} \log ^4(\sin (x)) \, dx=\frac{1}{4} C \log ^3(2)-3 \beta (4) \log (2)-12 \Im\left(\text{Li}_5\left(\frac{1}{2}+\frac{i}{2}\right)\right)-\frac{3}{2} \log ^2(2) \Im\left(\text{Li}_3\left(\frac{1}{2}+\frac{i}{2}\right)\right)+6 \log (2) \Im\left(\text{Li}_4\left(\frac{1}{2}+\frac{i}{2}\right)\right)-3 S+\frac{3}{2} \pi \zeta (3) \log (2)+\frac{2093 \pi ^5}{30720}+\frac{43}{128} \pi \log ^4(2)+\frac{23}{64} \pi ^3 \log ^2(2)$$ $$\small\int_0^{\frac{\pi }{4}} \log ^4(\cos (x)) \, dx=-\frac{1}{4} C \log ^3(2)+3 \beta (4) \log (2)+12 \Im\left(\text{Li}_5\left(\frac{1}{2}+\frac{i}{2}\right)\right)+\frac{3}{2} \log ^2(2) \Im\left(\text{Li}_3\left(\frac{1}{2}+\frac{i}{2}\right)\right)-6 \log (2) \Im\left(\text{Li}_4\left(\frac{1}{2}+\frac{i}{2}\right)\right)+3 S+\frac{3}{2} \pi \zeta (3) \log (2)-\frac{877 \pi ^5}{30720}+\frac{21}{128} \pi \log ^4(2)-\frac{7}{64} \pi ^3 \log ^2(2)$$ Here $S=\Im L(4,1|i,1)=\sum _{k=1}^{\infty } \frac{(-1)^{k-1} H_{2 k-2}}{(2 k-1)^4}$ is an irreducible colored MZV, which is also expressible via hypergeometric functions by binomial expansion, due to the following identity given by paper $2$: $$\small \sqrt{2} \, _6F_5\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2};\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2};\frac{1}{2}\right)=-\frac{1}{4} S-\frac{1}{8} \beta (4) \log (2)-\Im\left(\text{Li}_5\left(\frac{1}{2}+\frac{i}{2}\right)\right)+\frac{1}{16} \pi \zeta (3) \log (2)+\frac{2093 \pi ^5}{368640}+\frac{1}{512} \pi \log ^4(2)+\frac{23 \pi ^3 \log ^2(2)}{3072}$$ For verification of $2$ integrals above, download the Mathematica package associated to "paper $1$" here and make use of the function MZIntegrate. Note that one may derive an Euler-sum identity using the convolutional identity mentioned and orthogonal relations with help of values of quadratic logsines above. Nevertheless, they themselves are easily transformed into MZVs directly.

Denote : \begin{equation} I^{(n)} := \int\limits_0^{\pi/4} [\log(\cos(\theta))]^n d\theta \end{equation} Then by substituting for $z:= \exp(\imath 2 \theta)$ we get: \begin{eqnarray} I^{(4)}&=& \frac{1}{2 \imath}\int\limits_1^{\imath} \left( \log(1+z) - 1/2 \log(z)-\log(2)\right)^4 \frac{d z}{z} \\ &=& \frac{1}{2 \imath} \sum\limits_{0 \le p_1 \le p_2 \le 4} \frac{4!}{p_1!(p_2-p_1)!(4-p_2)!} \int\limits_1^{\imath} \log(1+z)^{p_1} (-1/2 \log(z))^{p_2-p_1} (-\log(2))^{4-p_2} \frac{dz}{z} \end{eqnarray} Now out of the fifteen terms that came about as a result of expanding the integrand there are easy terms and hard terms. Let us write down the terms starting from the easiest ones all the way to the most difficult ones. We have: \begin{eqnarray} &&I^{(4)}= I^{(4)}_{p_1=0} + I^{(4)}_{p_1=1} + I^{(4)}_{p_1=p_2 \& p_1\ge 2} +\\ && 6 \log(2) \underbrace{\int\limits_1^\imath \frac{\log(z)\log(1+z)^2}{z} dz}_{J_1} + \frac{3}{2} \underbrace{\int\limits_1^\imath \frac{\log(z)^2\log(1+z)^2}{z} dz}_{J_2} - 2 \underbrace{\int\limits_1^\imath \frac{\log(z)^1\log(1+z)^3}{z} dz}_{J_3} \end{eqnarray} Here $I^{(4)}_{p_1=0}$ represents five terms having $p_1=0$ then $I^{(4)}_{p_1=1}$ represents four terms having $p_1=1$ and $I^{(4)}_{p_1=p_2 \& p_1\ge 2}$ represents three terms having $p_1=p_2$ and $p_1\ge 2$. We have: \begin{eqnarray} I^{(4)}_{p_1=0} &=& \left( \begin{array}{rrrrr} \log(2)^4, & 2 \log(2)^3, & 3/2 \log(2)^2, & 1/2 \log(2), & 1/16 \end{array}\right)\cdot \left( \frac{\log(\imath)^{q+1}}{q+1}\right)_{q=0}^4 \\ I^{(4)}_{p_1=1} &=& \left( \begin{array}{rrrr} -4 \log(2)^3, && -6 \log(2)^2, && -3 \log(2), && -1/2 \end{array}\right) \cdot \left( - (-1)^{q-1} Li_{2+q}(-1) q!+\sum\limits_{\xi=0}^q (-1)^{\xi-1} Li_{2+\xi}(-\imath) \log(\imath)^{q-\xi} \binom{q}{\xi} \xi! \right)_{q=0}^3 \\ I^{(4)}_{p_1=p_2 \& p_1\ge 2} &=& \left( \begin{array}{rrr} 6 \log(2)^2, && -4 \log(2), && 1 \end{array}\right) \cdot \left( \sum\limits_{\xi=1}^{q+1} \binom{q}{\xi-1} (\xi-1)!(-1)^\xi (Li_\xi(1+\imath) \log(1+\imath)^{q+1-\xi} - Li_\xi(2) \log(2)^{q+1-\xi}) \right)_{q=2}^4 \end{eqnarray} Now the remaining terms are handled via their anti-derivatives. We have: \begin{eqnarray} &&\int \frac{\log(z)\log(1+z)^2}{z} dz = \\ &&\frac{1}{3} \sum\limits_{q=1}^4 Li_q(1+z) \log(1+z)^{4-q} \binom{3}{q-1}(q-1)!(-1)^q-\\ &&\frac{1}{3} \frac{\log(z)^4}{4}+\\ &&\sum\limits_{q=0}^2 Li_{2+q}(-z) \log(z)^{2-q} \binom{2}{q} q! (-1)^{q+1} -\\ &&\frac{1}{3} \sum\limits_{q=1}^4 Li_q(1+\frac{1}{z}) \log(1+\frac{1}{z})^{4-q} \binom{3}{q-1}(q-1)! (-1)^{q-1} \end{eqnarray} and \begin{eqnarray} &&6\int \frac{\log(1+z)^2 \log(z)^2}{z} dz - 4 \int \frac{\log(1+z)^3 \log(z)}{z} dz =\\ && \sum\limits_{q=1}^5 Li_q(1+\frac{1}{z})\log(1+\frac{1}{z})^{5-q} \binom{4}{q-1}(q-1)! (-1)^{q-1}-\\ &&\sum\limits_{q=1}^5 Li_q(1+z) \log(1+z)^{5-q} \binom{4}{q-1}(q-1)!(-1)^q+\\ &&4 \sum\limits_{q=0}^4 Li_{2+q}(-z) \log(z)^{3-q} \binom{3}{q} q! (-1)^{q-1} -\frac{\log(z)^5}{5} \end{eqnarray} where the two identities above have been derived by taking $(u,v):=(\log(1+z),\log(z))$ and then expanding $(u-v)^p$ dividing the result by $z$ and integrating. Here we took $p=3$ and $p=4$ in the first and in the second identities respectively.

In summary we can say that whereas $J_1$ is a function of polylogarithms only this is not the case for the other two terms $J_2$ and $J_3$. Therefore we were not able to express the integral in the original question as a function of polylogs only and we have to leave either $J_2$ or $J_3$ as a parameter in the result.

Update: Below we give another identity related to the integrals in question . Let $z\ge 0$. Then we have: \begin{eqnarray} &&6\int \frac{\log(\frac{z}{1+z})^2 \log(1+z)^2}{1+z} dz + 4 \int \frac{ \log(1+z)^3\log(z)}{z} dz=\\ &&4 \log(z) \sum\limits_{q=1}^4 Li_q(1+z) \log(1+z)^{4-q} \binom{3}{q-1}(q-1)! (-1)^q+\\ &&\frac{1}{15} \left(-18 \log ^5(z+1)+\log (z) \left(45 \log ^4(z+1)-8 \pi ^4\right)-30 \log ^2(z) \log ^3(z+1)\right)+\\ &&12 \text{Li}_2\left(\frac{1}{z+1}\right) \log \left(\frac{z}{z+1}\right) \log ^2(z+1)+\\ &&12 \text{Li}_3\left(\frac{1}{z+1}\right) (2 \log (z)-3 \log (z+1)) \log (z+1)+\\ &&24 \text{Li}_4\left(\frac{1}{z+1}\right) (\log (z)-3 \log (z+1))+\\ &&-72 \text{Li}_5\left(\frac{1}{z+1}\right) \end{eqnarray} That identity comes from integrating by parts the second integral on the left hand side and then exploiting the inversion formula for the polylogarithm http://functions.wolfram.com/ZetaFunctionsandPolylogarithms/PolyLog/17/02/01/01/0005/ .