I'm a famous constant

This is a very nice analysis problem as it's been also solved by Sir Ali Jaffal from Lebanon in complete analysis way.

Here is my try for solution

For convenience we write $\sin^2x = S(x)$ and $\cos^2 x = C(x)$ then we observe that $\small{\psi(n)= \sum_{i=1}^{n } \gamma_i^{S(x)} \cdot \gamma_{n+1-i}^{C(x)} =\sum_{i=1}^{n}\sum_{k=n+1-i}^n\gamma_i^{S(x)} \gamma_k^{C(x)}=\sum_{i=1}^n\gamma_i^{S(x)}\sum_{k=1}^{n} \gamma_k^{C(x)}\\ =\sum_{m=1}^n \gamma_m^{S(x)+C(x)} =\sum_{m=1}^{n } \gamma_{m} \ \ \ (\text{for } i=k=m )}$ since $S(x)+C(x)=1$ Now $\lim_{n\to \infty} \dfrac{\psi(n) } {n}=\lim_{n\to \infty} \dfrac{1}{n} \sum_{m=1}^{n} \gamma_m=\lim_{n\to \infty} \dfrac{1}{n} \sum_{m=1}^{n} (H_m -\log m ) \\ = \lim_{n\to \infty} \dfrac{1}{n}\left(\sum_{m=1}^{n} \sum_{j=1}^{m} \dfrac{1}{j} -\log\left(\prod_{m=1}^{n} m\right)\right)=\lim_{n\to \infty} \dfrac{1}{n}\left(\sum_{m=1}^n\dfrac{n-m+1}{m}-\log(n!) \right)\\ = \lim_{n\to \infty}\left(\sum_{m=1}^n\dfrac{1}{m} -\dfrac{n\log n-n +O\log(n) }{n} - 1 +\dfrac{1}{n}\sum_{m=1}^n\dfrac{1}{n} \right) \\ \therefore \Omega(x) =\lim_{n\to \infty}\left(\sum_{m=1}^n\dfrac{1}{m}-\log n\right)+0=\gamma$ and hence $\log(\Omega(x))=\log(\gamma) = -0.238661\cdots$ .