Integral $\int_0^{\pi/2}\frac{x}{\sin x}\log^2\left(\frac{1+\cos x-\sin x}{1+\cos x+\sin x}\right)dx$ Please help me to evaluate this integral:
$$\large\int_0^{\pi/2}\frac{x}{\sin x}\log^2\left(\frac{1+\cos x-\sin x}{1+\cos x+\sin x}\right)dx$$
 A: I have been dabbling with this one off and on and thought I would post some results.
L.T., may I ask where you found this monster?. 
Using Integrals idea of using $\displaystyle \frac{1+\cos(x)-\sin(x)}{1+\cos(x)+\sin(x)}=\tan(\frac{\pi}{4}-\frac{x}{2})$, one can make the sub $t=\tan(\frac{\pi}{4}-\frac{x}{2})$ and obtain:
$\displaystyle\pi\int_{0}^{1}\frac{log^{2}(t)}{1-t^{2}}dt+4\int_{0}^{1}\frac{\tan^{-1}(t)log^{2}(t)}{1-t^{2}}dt$.
The leftmost integral is rather famous and evaluates to $\displaystyle\frac{7\pi\zeta(3)}{4}$
The other one is the one that is the booger. It evaluates to $\displaystyle\frac{\pi^{2}}{24}G+\beta(4)-\frac{7\pi\zeta(3)}{16}$
So, using the integral form of digamma:  $\displaystyle\int_{0}^{1}\frac{1-t^{a-1}}{1-t}dt=\psi(a)+\gamma$, along with the arctan series, diff this twice w.r.t 'a' in order to introduce the $log^{2}$ term. Make the sub $y=t^{2}$
$\displaystyle1/2\int_{0}^{1}\frac{y^{n+a/2-1/2}}{1-y}\sum_{n=0}^{\infty}\frac{(-1)^{n}}{2n+1}dy$
$\displaystyle=\frac{1}{8}\sum_{n=0}^{\infty}\frac{(-1)^{n}\psi''(n+1)}{2n+1}$,  by noting that a=1.....................................[1]
By using the identity $\psi(x+1)=\psi(x)+1/x$, one may write:
$\displaystyle\frac{1}{8}\sum_{n=1}^{\infty}\frac{(-1)^{n}\psi''(n)}{2n+1}+\frac{1}{4}\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n^{3}(2n+1)}-\frac{\zeta(3)}{4}$
where the n=0 term was brought out of the digamma sum due to the identity not being defined there.
The middle series is equal to $\displaystyle2-\frac{\pi}{2}-log(2)+\frac{\pi^{2}}{24}-\frac{3\zeta(3)}{16}$
Thus, giving $\displaystyle\frac{1}{8}\sum_{n=1}^{\infty}\frac{(-1)^{n}\psi''(n)}{(2n+1)}+2-\frac{\pi}{2}-log(2)+\frac{\pi^{2}}{24}-\frac{7\zeta(3)}{16}$
I tried writing the sum in terms of its odd and even componets. At least, to eliminate the alternation.
$\displaystyle\frac{1}{8}\sum_{n=1}^{\infty}\frac{(-1)^{n}\psi''(n)}{2n+1}=\frac{1}{32}\sum_{n=1}^{\infty}\frac{\psi''(2n)}{n+1/4}-\frac{1}{32}\sum_{n=1}^{\infty}\frac{\psi''(2n+1)}{n+3/4}+\frac{\zeta(3)}{12}$
This is as far as I got.  If anyone knows how to evaluate an alternating tetragamma series, please, I would love to see it. 
I have found little work done on these. May be a good topic for a paper. :)
EDIT:  Due to the identity $\displaystyle \psi''(n+1)=-2\zeta(3)+2H_{n}^{(3)}$, we may write the digamma sum in [1] as:
$\displaystyle \frac{1}{8}\sum_{n=1}^{\infty}\frac{(-1)^{n}\left[-2\zeta(3)+2H_{n}^{(3)}\right]}{2n+1}-\frac{\zeta(3)}{4}$
As a side note, The only other problem I have encountered that has the $\frac{\pi^{2}}{6}G$ term is this one:
A double series yielding Riemann's $\zeta$
Maybe this integral can some how be connected to series like this?.
A: Your integral can be expressed in terms of certain special functions:
$$\begin{align}
\int_0^{\pi/2}\frac{x}{\sin x}\log^2\left(\frac{1+\cos x-\sin x}{1+\cos x+\sin x}\right)dx&=\frac{\,\pi^2}6K+4\,\beta(4)\\&=\frac{\,\pi^2}6K-\frac{\pi^4}{24}+\frac1{192}\psi^{(3)}\left(\frac14\right)\\&=\frac{\,\pi^2}6K-\frac{119\,\pi^4}{2880}-\frac1{32}H^{(4)}_{-3/4},\end{align}$$
where
$K$ is Catalan's constant, $\beta(x)$ is the Dirichlet beta function, $\psi^{(n)}(z)$ is the polygamma function and $H^{(r)}_n$ is the generalized harmonic number.
