How to evaluate $\int_{0}^{\pi }\theta \ln\tan\frac{\theta }{2} \, \mathrm{d}\theta$

Question

I have some trouble in how to evaluate this integral: $$ \int_{0}^{\pi}\theta\ln\left(\tan\left(\theta \over 2\right)\right) \,\mathrm{d}\theta $$ I think it maybe has another form $$ \int_{0}^{\pi}\theta\ln\left(\tan\left(\theta \over 2\right)\right) \,\mathrm{d}\theta = \sum_{n=1}^{\infty}{1 \over n^{2}} \left[\psi\left(n + {1 \over 2}\right) - \psi\left(1 \over 2\right)\right] $$

Answer 1

Obviously we have $$\int_{0}^{\pi }\theta \ln\tan\frac{\theta }{2}\mathrm{d}\theta =4\int_{0}^{\pi /2}x\ln \tan x\mathrm{d}x$$ then use the definition of Lobachevskiy Function(You can see this in table of integrals,series,and products,Eighth Edition by Ryzhik,page 900) $$\mathrm{L}\left ( x \right )=-\int_{0}^{x}\ln\cos x\mathrm{d}x,~ ~ ~ ~ ~ -\frac{\pi }{2}\leq x\leq \frac{\pi }{2}$$ Hence we have \begin{align*} \int_{0}^{\pi /2}x\ln\tan x\mathrm{d}x &= x\left [ \mathrm{L}\left ( x \right )+\mathrm{L}\left ( \frac{\pi }{2}-x \right ) \right ]_{0}^{\pi /2}-\int_{0}^{\pi /2}\left [ \mathrm{L}\left ( x \right )+\mathrm{L}\left ( \frac{\pi }{2}-x \right ) \right ]\mathrm{d}x\\ &= \left ( \frac{\pi }{2} \right )^{2}\ln 2-2\int_{0}^{\pi /2}\mathrm{L}\left ( x \right )\mathrm{d}x \end{align*} use $$\mathrm{L}\left ( x \right )=x\ln 2-\frac{1}{2}\sum_{k=1}^{\infty }\frac{\left ( -1 \right )^{k-1}}{k^{2}}\sin 2kx$$ (Integrate the fourier series of $\ln\cos x$ from $0$ to $x$.)

we can calculate \begin{align*} \int_{0}^{\pi /2}\mathrm{L}\left ( x \right )\mathrm{d}x&=\frac{1}{2}\left ( \frac{\pi }{2} \right )^{2}\ln 2-\frac{1}{2}\sum_{k=1}^{\infty }\frac{\left ( -1 \right )^{k-1}}{k^{2}}\int_{0}^{\pi /2}\sin 2kx\mathrm{d}x \\ &= \frac{\pi ^{2}}{8}\ln 2-\frac{1}{2}\sum_{k=1}^{\infty }\frac{1}{\left ( 2k-1 \right )^{3}} \end{align*} So \begin{align*} \int_{0}^{\pi/2}x\ln\tan x\mathrm{d}x &=\frac{\pi ^{2}}{4}\ln 2-2\left [ \frac{\pi ^{2}}{8}\ln 2-\frac{1}{2}\sum_{k=1}^{\infty }\frac{1}{\left ( 2k-1 \right )^{3}} \right ] \\ &=\sum_{k=1}^{\infty }\frac{1}{\left ( 2k-1 \right )^{3}}\\ &=\sum_{k=1}^{\infty } \frac{1}{k^{3}}-\sum_{k=1}^{\infty }\frac{1}{\left ( 2k \right )^{3}}=\frac{7}{8}\zeta \left ( 3 \right ) \end{align*} Hence the initial integral is $$\boxed{\Large\color{blue}{\int_{0}^{\pi }\theta \ln\tan\frac{\theta }{2}\mathrm{d}\theta=\frac{7}{2}\zeta \left ( 3 \right )}}$$ in addition,as you mentioned $$\int_{0}^{\pi }\theta \ln\tan\frac{\theta }{2}\mathrm{d}\theta=\color{red}{\sum_{n=1}^{\infty }\frac{1}{n^{2}}\left [ \psi \left ( n+\frac{1}{2} \right )-\psi \left ( \frac{1}{2} \right ) \right ]=\frac{7}{2}\zeta \left ( 3 \right )}$$ or $$\sum_{n=1}^{\infty }\frac{1}{n^{2}}\psi \left ( n+\frac{1}{2} \right )=\frac{7}{2}\zeta \left ( 3 \right )-\left ( \gamma +2\ln 2 \right )\frac{\pi ^{2}}{6}$$

Answer 2

\begin{align} J&=\int_0^\pi \theta\ln\left(\tan\left(\frac{\theta}{2}\right)\right)d\theta\\ &\overset{u=\tan\left(\frac{\theta}{2}\right)}=4\underbrace{\int_0^\infty \frac{\arctan u\ln u}{1+u^2}du}_{=K}\\ K&\overset{\text{IBP}}=\underbrace{\left[\arctan u\left(\int_0^u \frac{\ln t}{1+t^2}dt\right)\right]_0^\infty}_{=0}-\int_0^\infty \frac{1}{1+u^2}\left(\underbrace{\int_0^u \frac{\ln t}{1+t^2}dt}_{y(t)=\frac{t}{u}}\right)du\\ &=-\int_0^\infty \left(\int_0^1 \frac{u\ln(uy)}{(1+u^2)(1+u^2y^2)}dy\right)du\\ &=-\int_0^\infty \left(\int_0^1 \frac{u\ln u}{(1+u^2)(1+u^2y^2)}dy\right)du-\int_0^1 \left(\int_0^\infty \frac{u\ln y}{(1+u^2)(1+u^2y^2)}du\right)dy\\ &=-\int_0^\infty \left[\frac{\arctan(uy)}{1+u^2}\right]_{y=0}^{y=1}\ln udu-\frac{1}{2}\int_0^1 \left[\frac{\ln\left(\frac{1+u^2}{1+u^2y^2}\right)}{1-y^2}\right]_{u=0}^{u=\infty}\ln ydy\\ &=-K+\int_0^1 \frac{\ln^2 y}{1-y^2}dy\\ &=\frac{1}{2}\int_0^1 \frac{\ln^2 y}{1-y}dy-\frac{1}{2}\underbrace{\int_0^1 \frac{y\ln^2 y}{1-y^2}dy}_{z=y^2}\\ &=\frac{7}{16}\int_0^1 \frac{\ln^2 y}{1-y}dy\\ J&=\frac{7}{4}\int_0^1 \frac{\ln^2 y}{1-y}dy\\ J&=\frac{7}{4}\times 2\zeta(3)\\ &=\boxed{\frac{7}{2}\zeta(3)} \end{align}

Answer 3

As an alternative to kishlaya's answer, we have the Fourier series

$$\ln(\cos(x)) = -\ln(2) - \sum_{k=1}^\infty (-1)^k\frac{\cos(2kx)}k \\ \ln(\sin(x)) = -\ln(2) - \sum_{k=1}^\infty \frac{\cos(2kx)}k$$

and hence

$$\int_0^{\frac\pi2} \ln(\cos(x)) \, dx \stackrel{x\mapsto\frac\pi2-x}= \int_0^{\frac\pi2} \ln(\sin(x)) \, dx = -\frac\pi2 \ln(2)$$

Now,

$$\int_0^{\frac\pi2} x \ln(\cos(x)) \, dx \stackrel{x\mapsto\frac\pi2-x}= \int_0^{\frac\pi2} \left(\frac\pi2-x\right) \ln(\sin(x)) \, dx = -\frac{\pi^2}4 \ln(2) - \int_0^{\frac\pi2} x\ln(\sin(x)) \, dx$$

$$\implies \int_0^\pi x \ln\left(\tan\left(\frac x2\right)\right) \, dx = 8\int_0^{\frac\pi2} x\ln(\sin(x)) \, dx + \pi^2 \ln(2)$$

For the remaining integral, we have

$$\begin{align*} \int_0^{\frac\pi2} x \ln(\sin(x)) \, dx &= -\ln(2) \int_0^{\frac\pi2} x \, dx - \int_0^{\frac\pi2} \sum_{k=1}^\infty \frac{x\cos(2kx)}k \, dx \\[1ex] &= -\frac{\pi^2}8\ln(2) - \sum_{k=1}^\infty \frac1k \int_0^{\frac\pi2} x \cos(2kx) \, dx \\[1ex] &= -\frac{\pi^2}8 \ln(2) - \frac14 \sum_{k=1}^\infty \frac{(-1)^k-1}{k^3} \\[1ex] &= -\frac{\pi^2}8 \ln(2) - \frac14 \left(-\frac34\zeta(3)-\zeta(3)\right) \\[1ex] &= -\frac{\pi^2}8 \ln(2) + \frac7{16} \zeta(3) \end{align*}$$

and it follows that

$$\int_0^\pi x \ln\left(\tan\left(\frac x2\right)\right) \, dx = \boxed{\frac72\zeta(3)}$$

Answer 4

I ran across this question while researching a duplicate question.

I had written the following answer, which seems to be a different approach to the answers I've seen so far.

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As shown in this answer $$ \log(\sin(x))=-\log(2)-\sum_{k=1}^\infty\frac{\cos(2kx)}{k}\tag1 $$ and $$ \log(\cos(x))=-\log(2)+\sum_{k=1}^\infty(-1)^{k-1}\frac{\cos(2kx)}{k}\tag2 $$ Subtraction yields $$ \log(\tan(x))=-2\sum_{k=0}^\infty\frac{\cos((4k+2)x)}{2k+1}\tag3 $$ Therefore, $$ \begin{align} \int_0^{\pi/2}x\log(\tan(x))\,\mathrm{d}x &=-2\sum_{k=0}^\infty\int_0^{\pi/2}\frac{x\cos((4k+2)x)}{2k+1}\,\mathrm{d}x\tag{4a}\\ &=-4\sum_{k=0}^\infty\frac1{(4k+2)^3}\int_0^{(2k+1)\pi}x\cos(x)\,\mathrm{d}x\tag{4b}\\ &=-\frac12\sum_{k=0}^\infty\frac1{(2k+1)^3}\int_0^{(2k+1)\pi}x\,\mathrm{d}\sin(x)\tag{4c}\\ &=\frac12\sum_{k=0}^\infty\frac1{(2k+1)^3}\int_0^{(2k+1)\pi}\sin(x)\,\mathrm{d}x\tag{4d}\\ &=\sum_{k=0}^\infty\frac1{(2k+1)^3}\tag{4e}\\[3pt] &=\frac78\zeta(3)\tag{4f} \end{align} $$ Explanation:
$\text{(4a):}$ apply $(3)$
$\text{(4b):}$ substitute $x\mapsto\frac{x}{4k+2}$
$\text{(4c):}$ prepare to integrate by parts
$\text{(4d):}$ integrate by parts
$\text{(4e):}$ evaluate the integral
$\text{(4f):}$ $\underbrace{\sum\frac1{(2k+1)^3}}_\text{odds}=\underbrace{\sum\frac1{(k+1)^3}}_\text{all}-\underbrace{\sum\frac1{(2k+2)^3}}_\text{evens}=\frac78\sum\frac1{(k+1)^3}$

Thus, $$ \begin{align} \int_0^\pi\theta\log(\tan(\theta/2))\,\mathrm{d}\theta &=4\int_0^{\pi/2}x\log(\tan(x))\,\mathrm{d}x\tag{5a}\\[3pt] &=\frac72\zeta(3)\tag{5b} \end{align} $$ Explanation:
$\text{(5a):}$ $\theta=2x$
$\text{(5b):}$ apply $(4)$