I am working on this integral: How to evaluate $\int_0^{\pi /2}\frac{u^2\ln{(2\cos u)}}{(u^2+\ln^2{(2\cos u)})^2}du$?
Question
How to evaluate $$\int_{0}^{\frac{\pi}{2}} \frac{x^2 \ln(2 \cos x)}{(x^2 + \ln^2(2 \cos x))^2} \, dx $$
My attempt
\begin{align*} \int_{0}^{\frac{\pi}{2}} \frac{x^2 \ln(2 \cos x)}{(x^2 + \ln^2(2 \cos x))^2} \, dx &= -\frac{1}{4} \mathfrak{J} \cdot \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x}{\ln^2(1 + e^{2ix})} \, dx \\ &= -\frac{1}{16} \mathfrak{J} \cdot \int_{-\pi}^{\pi} \frac{x}{\ln^2(1 + e^{ix})} \, dx \\ &= -\frac{\pi}{8} \mathfrak{R} \int_{-\pi}^{\pi} \int_{0}^{1} \int_{0}^{y} (1+e^{ix})^{y-2} \, dy \, dy \left(\frac{\partial}{\partial \omega} \bigg|_{\omega=1} e^{i \omega x}\right) \, dx \\ &= -\frac{\pi}{8} \mathfrak{R} \int_{0}^{1} \int_{0}^{y} \, dy \, dy \frac{\partial}{\partial \omega} \bigg|_{\omega=1} \int_{-\pi}^{\pi} (1 + e^{ix})^{y-2} e^{-i \omega x} \, dx \\ &= -\frac{\pi}{8} \int_{0}^{1} \int_{0}^{y} \, dy \, dy \frac{\partial}{\partial \omega} \bigg|_{\omega=1} \binom{y-2}{\omega} \\ &= -\frac{\pi}{8} \int_{0}^{1} \int_{0}^{y} (y - 2) \left(\psi(y - 2) - \psi(2)\right) \, dy \, dy \end{align*}
Can someone help me to evaluate
$$\int_{0}^{1} \int_{0}^{y} (y - 2) \left(\psi(y - 2) - \psi(2)\right) \, dy \, dy$$
Don’t mark this as a duplicate. I want to know how to evaluate the last integral where I am stuck. My question is not about evaluating the main integral.
