Log Trig $\int_0^{\pi/2}\log^4 \tan \frac{x}{2}dx=\frac{5\pi^5}{32}$ Hi I am trying to integrate a log trigonometric integral given by
$$
I:=\int_0^{\pi/2}\log^4 \tan \frac{x}{2}dx=\frac{5\pi^5}{32}.
$$
This is very similar to a previous integral posted except the power of the logarithm.
Note this integral is also equal to
$$
\int_0^{\pi/2}\log^4 \tan x \, dx=\frac{5\pi^5}{32}.
$$
I have wrote I as
$$
I=\int_0^{\pi/2} \left(\log \sin \frac{x}{2}-\log \cos \frac{x}{2}\right)^4 dx
$$
but got stuck here since factoring this out seems like a mess.  Having seen how David H solved a similar integral I posted, I tried another method starting with I and using $t=\tan x/2$, and obtained
$$
I=2\int_0^{1}\log^4 t \frac{dt}{1+t^2}.
$$
Following this I tried $u=-\log t$ but got stuck after this.  Thanks, it would be nice to see a solution that doesn't reduce the integral to a difficult sum to evaluate
 A: This post and related comments and answers are very interesting and allow the generalization of the problem to $$I(n)=\int_0^{\pi/2}\log ^n\left[\tan \left(\frac{x}{2}\right)\right]dx$$ for which the result is $$I(n)=(-1)^n 2^{-2 n-1} \left[\zeta \left(n+1,\frac{1}{4}\right)-\zeta
   \left(n+1,\frac{3}{4}\right)\right] \Gamma (n+1).$$ When $n$ is even, one can find a simple formula in terms of the Euler numbers: $$I(2n)=(-1)^nE_{2n}\left(\dfrac\pi2\right)^{2n+1},$$ e.g. $$I(0)=\frac{\pi}{2}$$ $$I(2)=\frac{\pi ^3}{8}$$ $$I(4)=\frac{5 \pi ^5}{32}$$ $$I(6)=\frac{61 \pi ^7}{128}$$ $$I(8)=\frac{1385 \pi ^9}{512}$$ $$I(10)=\frac{50521 \pi ^{11}}{2048}$$ and so on.
When $n$ is odd, all values are negative and cannot be expressed in any form simpler than the $\zeta$ function.
A: \begin{align}
J&=\int_0^{\frac{\pi}{2}}\log^4 \left(\tan x\right) \, dx\\
&\overset{y=\tan x}=\int_0^\infty \frac{\ln^4 x}{1+x^2}\,dx\\
J_n&=\int_0^\infty \frac{\ln^n x }{1+x^2}\,dx\\
J_0&=\frac{\pi}{2}\\
K_n&=\int_0^\infty\int_0^\infty \frac{\ln^n(xy) }{(1+x^2)(1+y^2)}\,dx\,dy\\
&\overset{u(x)=xy}=\int_0^\infty \int_0^\infty \frac{\ln^n u}{(u^2+y^2)(1+y^2)}\,du\,dy\\
&=\int_0^\infty \frac{\ln^{n+1}u }{u^2-1}\,du\\
&=\int_0^1 \frac{\ln^{n+1}u }{u^2-1}\,du+\int_1^\infty \frac{\ln^{n+1}u }{u^2-1}\,du\\
&=\Big(1+(-1)^n\Big)\int_0^1 \frac{\ln^{n+1}u }{u^2-1}\,du\\
&=\Big(1+(-1)^n\Big)\left(\int_0^1 \frac{\ln^{n+1}u }{u-1}\,du-\int_0^1 \frac{u\ln^{n+1}u }{u^2-1}\,du\right)\\
&=\Big(1+(-1)^n\Big)\left(1-\frac{1}{2^{n+2}}\right)\int_0^1 \frac{\ln^{n+1}u }{u-1}\,du\\
&=\Big(1+(-1)^n\Big)\left(1-\frac{1}{2^{n+2}}\right)(-1)^{n+2}(n+1)!\zeta(n+2)\\
&=\Big(1+(-1)^n\Big)\left(1-\frac{1}{2^{n+2}}\right)(n+1)!\zeta(n+2)\\
K_n&=\sum_{k=0}^n \binom{n}{k} J_kJ_{n-k}\\
K_2&=\pi J_2\\
K_2&=\frac{45}{4}\zeta(4)\\
K_4&=\pi J_4+6J_2^2\\
K_4&=\frac{945}{4}\zeta(6)\\
\end{align}Therefore,
\begin{align}
J_4&=\frac{K_4-6{J_2}^2}{\pi}\\
&=\frac{K_4-6\left(\frac{K_2}{\pi}\right)^2}{\pi}\\
&=\frac{K_4-6{J_2}^2}{\pi}\\
&=\frac{\frac{945}{4}\zeta(6)-6\left(\frac{45}{4}\zeta(4)\times\frac{1}{\pi}\right)^2}{\pi}\\
&=\frac{945\zeta(6)}{4\pi}-\frac{6075\zeta(4)^2}{8\pi^3}\\
\end{align}
Moreover, if you assume,
\begin{align}
\zeta(4)&=\frac{\pi^4}{90}\\
\zeta(6)&=\frac{\pi^6}{945}
 \end{align}
Therefore,
\begin{align}
J&=J(4)\\
&=\frac{945}{4\pi}\times \frac{\pi^6}{945}-\frac{6075}{8\pi^3}\times \left(\frac{\pi^4}{90}\right)^2\\
&=\boxed{\frac{5}{32}\pi^5}
\end{align}
A: $\newcommand{\+}{^{\dagger}}
 \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\down}{\downarrow}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\isdiv}{\,\left.\right\vert\,}
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
 \newcommand{\wt}[1]{\widetilde{#1}}$
By following my previous answer you'll arrive to
\begin{align}
&\half\,\lim_{\mu \to 0}\partiald[4]{\sec\pars{\pi\mu/2}}{\mu}={\pi \over 2}\times
\\[3mm]&\lim_{\mu \to 0}\bracks{%
{5 \over 16}\,\pi^{4}\sec^{5}\left(\frac{\pi\mu}{2}\right)+\frac{9}{8} \pi ^4 \tan ^2\left(\frac{\pi\mu}{2}\right) \sec^3\left(\frac{\pi\mu}{2}\right)
+\frac{1}{16}\pi^{4}\tan^{4}\left(\frac{\pi\mu}{2}\right)\sec\left(\frac{\pi\mu}{2}\right)}
\\[3mm]&={\pi \over 2}\,{5\pi^{4} \over 16}
=\color{#00f}{\large{5\pi^{5} \over 32}}
\end{align}
A: You've established that $$\int_0^{\pi/2}\log^4\tan\frac{x}{2}\,dx=\int_0^{\pi/2}\log^4\tan x\,dx,$$
now use $\log\tan x=\log\sin x-\log\cos x$ as well as
$$\int_0^{\pi/2}\log^m(\sin x)\log^n(\cos x)dx=\left[\left(\frac{\partial}{\partial a}\right)^m\left(\frac{\partial}{\partial b}\right)^n\frac{\Gamma(\tfrac{a+1}2)\Gamma(\tfrac{b+1}2)}{2\Gamma(\tfrac{a+b}2+1)}\right]_{(a,b)=(0,0)},$$
which comes from repeated differentiation of
$$\int_0^{\pi/2}\sin(x)^a\cos(x)^bdx=\frac{\Gamma(\tfrac{a+1}2)\Gamma(\tfrac{b+1}2)}{2\Gamma(\tfrac{a+b}2+1)}.$$
This should yield the result.
