Can You Believe A Closed Form Exists?

In 1847 Kummer gave a Fourier series regarding Log Gamma which states ,

$\displaystyle\color{#3D99F6}{ \ln\Gamma(x)-\frac{1}{2}\ln(2\pi) = -\frac{1}{2}\ln|2\sin(\pi x)| + \frac{1}{2}(1-2x)(\gamma + \ln(2\pi)) + \frac{1}{\pi}\sum_{k=2}^{\infty}\frac{\ln k}{k}\sin(2\pi kx)}$

Considering the symmetry of $\displaystyle \sin(2\pi kx)$ & a substitution of variable $\displaystyle x\to (1-x)$ for $\displaystyle x\in(0,1)$ we have ,

$\displaystyle\color{#3D99F6}{ \ln\Gamma(1-x)-\frac{1}{2}\ln(2\pi) = -\frac{1}{2}\ln|2\sin(\pi x)| - \frac{1}{2}(1-2x)(\gamma + \ln(2\pi)) - \frac{1}{\pi}\sum_{k=2}^{\infty}\frac{\ln k}{k}\sin(2\pi kx)}$

The above two results may be used to compute $\displaystyle \mathfrak{LG}_2 = \int_{0}^{1} (\ln|\Gamma(x)|)^2 dx$

Adding the two $\color{#3D99F6}{\text{Blue}}$ results we get,

$\displaystyle \ln|\Gamma(x)\Gamma(1-x)| = \ln(2\pi)-\ln|2\sin(\pi x)|$

Squaring and integrating both sides we get ,

$\displaystyle \int_{0}^{1} (\ln\Gamma(x)\Gamma(1-x)))^2 dx = \ln^2(2\pi) - \frac{1}{\pi}Ls_{3}(\pi)$

$\displaystyle \color{#D61F06}{Ls_3(\pi)}$ denotes the Log-Sine Integrals .

As a standard result we may use $\displaystyle Ls_{3}(\pi) = -\frac{\pi^3}{12}$ which yields ,

$\displaystyle \color{#20A900} {\int_{0}^{1} (\ln\Gamma(x)\Gamma(1-x)))^2 dx = \ln^2(2\pi) + \frac{\pi^2}{12}}$

Again by subtracting the $\color{#3D99F6}{\text{Blue}}$ results we get ,

$\displaystyle \ln\frac{\Gamma(x)}{\Gamma(1-x)} = (1-2x)(\gamma+\ln(2\pi))+\frac{2}{\pi}\sum_{k=2}^{\infty}\frac{\ln k}{k}\sin(2\pi kx)$

Which on squaring followed by integration and obviously some basic reductions gives us,

$\displaystyle \int_{0}^{1} (\ln\frac{\Gamma(x)}{\Gamma(1-x)})^2 dx = \frac{1}{3}(\gamma+\ln(2\pi))^2 + \frac{4}{\pi^2}(\gamma+\ln(2\pi)) \sum_{k=2}^{\infty}\frac{\ln k}{k^2} + \frac{2}{\pi^2}\sum_{k=2}^{\infty}\frac{\ln^2 k}{k^2}$

Using $\displaystyle \zeta'(2)=-\sum_{k=2}^{\infty} \frac{\ln k}{k^2}$ & $\displaystyle \zeta''(2)=\sum_{k=2}^{\infty} \frac{\ln^2 k}{k^2}$

$\displaystyle \color{#20A900}{ \int_{0}^{1} (\ln\frac{\Gamma(x)}{\Gamma(1-x)})^2 dx = \frac{1}{3}(\gamma+\ln(2\pi))^2 - \frac{4}{\pi^2}(\gamma+\ln(2\pi)) \zeta'(2) + \frac{2}{\pi^2}\zeta''(2)}$

Now simply adding the two $\color{#20A900}{\text{Green}}$ results and using the fact $\displaystyle \int_{0}^{1}\ln\Gamma(x) dx = \int_{0}^{1}\ln\Gamma(1-x)dx$ we may deduce that ,

$\displaystyle \color{#333333}{ \int_{0}^{1} (\ln\Gamma(x))^2 dx = \frac{\ln^2(2\pi)}{3}+\frac{\pi^2}{48}+\frac{\gamma^2}{12} + \frac{\gamma\ln(\sqrt{2\pi})}{3} - \frac{\ln(2\pi)}{\pi^2}\zeta'(2) - \frac{\gamma}{\pi^2}\zeta'(2) + \frac{\zeta''(2)}{2\pi^2}}$

So the answer is $\boxed{97}$