Log Trig $\int_0^{\pi/2}\log^4 \tan \frac{x}{2}dx=\frac{5\pi^5}{32}$

Question

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

Answer 1

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ 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}

Answer 2

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.

Answer 3

\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}

Answer 4

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.

Answer 5

As OP did, $$ \int_0^{\frac{\pi}{2}} \log ^4\left(\tan \frac{x}{2}\right) d x=2 \int_0^1 \frac{\ln ^4 t}{1+t^2} d t=2 I^4(0), $$ where $\displaystyle I(a)=\int_0^1 \frac{t^a}{1+t^2} d t$.

By the expansion $\frac{1}{1+t^2} =\sum_{n=0}^\infty (-1)^nt^{2n}$ for $|t|<1$, we have $$ I(a) =\int_0^1 \frac{t^a}{1+t^2} d t=\sum_{n=0}^{\infty}(-1)^n \int_0^1 t^a t^{2 n} d t =\sum_{n=0}^{\infty} \frac{(-1)^n}{2 n+1+a} $$ Differentiating $I(a)$ by $4$ times w.r.t. $a$ yields

$$ I^{(4)}(a) =\sum_{n=0}^{\infty} \frac{(-1)^n4!}{(2 n+1+a)^4} =4!\left[\sum_{n=0}^{\infty} \frac{1}{(4 n+1+a)^5}-\sum_{n=0}^{\infty} \frac{1}{(4 n+3+a)^5}\right] $$

Putting $a=0$ gives $$ \begin{aligned} I^{(4)}(0) & =4!\left[\sum_{n=0}^{\infty} \frac{1}{(4 n+1)^5}-\sum_{n=0}^{\infty} \frac{1}{(4 n+3)^5} \right]\\ & =\frac{3}{128} \left[ \zeta\left(5, \frac{1}{4}\right)-\zeta\left(5, \frac{3}{4}\right)\right] \end{aligned} $$ Hence we can conclude that $$\int_0^{\frac{\pi}{2}} \log ^4\left(\tan \frac{x}{2}\right) d x =2 I^4(0) = \frac{3}{64}\left[\zeta\left(5, \frac{1}{4}\right)-\zeta\left(5, \frac{3}{4}\right) \right]= \frac{5 \pi^5}{32} $$

In general, $$ \int_0^{\frac{\pi}{2}} \log ^n\left(\tan \frac{x}{2}\right) d x=\frac{ n !}{2^{2n+1}}\left[\zeta\left(n+1, \frac{1}{4}\right)-\zeta\left(n+1, \frac{3}{4}\right)\right] $$