$\int\limits_0^{1/2} \ln(Γ(x)) \cos⁡(\pi x)\ dx$

Question

Evaluate: $$\int\limits_0^{1/2} \ln(Γ(x)) \cos⁡(\pi x)\ dx$$

I tried to find this integral using different methods including integration by parts, without success.Question from Jalil Hajimir

Answer 1

Borrowing the Malmsten-Kummer Fourier series from M.N.C.E post $$\color{blue}{\ln{\Gamma(x)}=\frac{1}{2}\ln(2\pi)+\sum^\infty_{n=1}\left\{\frac{1}{2n}\cos(2\pi nx)+\frac{\gamma+\ln(2\pi n)}{n\pi}\sin(2\pi nx)\right\}}$$ we get $$ \int_{0}^{1/2}\cos(\pi x)\log\Gamma(x)\,dx = \frac{\log(2\pi)}{2\pi}+\sum_{n\geq 1}\left\{\frac{(-1)^{n+1}}{2\pi n(4n^2-1)}+\frac{\gamma+\ln(2\pi n)}{n\pi}\cdot\frac{2n}{\pi(4n^2-1)}\right\} $$ and the RHS can be simplified into $$ \frac{\log(2\pi)}{2\pi}+\frac{1-\log(2)}{2\pi}+\frac{2}{\pi^2}\sum_{n\geq 1}\frac{\gamma+\log(2\pi)+\log(n)}{(4n^2-1)} $$ then into $$ \frac{1+\log(\pi)}{2\pi}+\frac{2}{\pi^2}\left[\frac{\gamma+\log(2\pi)}{2}+\sum_{n\geq 1}\frac{\log(n)}{(2n+1)(2n-1)}\right]$$ which, except for known constants, depends on the derivative of the Dirichlet $\beta$-function at $s=1$.
Malmsten showed that $$ \beta'(1) = \frac{\pi}{4}(\gamma-\log\pi)+\pi\log\Gamma\left(\tfrac{3}{4}\right) $$ and this gives a closed form in terms of $\pi,\log\pi,\gamma$ and $\log\Gamma\left(\frac{3}{4}\right).$