calculus - How to evaluate $int_{0}^{pi }theta lntanfrac{theta }{2} , mathrm{d}theta$

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

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

$$\sum_{n=1}^{\infty }\frac{1}{n^{2}}\psi \left ( n+\frac{1}{2} \right )=\frac{7}{2}\zeta \left ( 3 \right )-\left ( \gamma +2\ln 2 \right )\frac{\pi ^{2}}{6}$$