Problem
Find the exact value of $$\int_0^1 \cos^2{\pi x}\ln \Gamma(x)dx$$
My Attempt
\begin{align*} I&=\int_0^1 \cos^2{(\pi x)}\ln \Gamma(x)dx\\ 2I&=\int_0^1 \cos^2{(\pi x)}\ln \Gamma(x)dx+\int_0^1 \cos^2{(\pi (1-x))}\ln \Gamma(1-x)dx \tag{Using reflection}\\ 2I&=\int_0^1 \cos^2{(\pi x)}\ln \Gamma(x)+\cos^2{(\pi (1-x))}\ln \Gamma(1-x)dx\\ 2I&=\int_0^1 \cos^2{(\pi x)}\ln \Gamma(x)+\cos^2{\pi x}\ln \Gamma(1-x)dx\\ 2I&=\int_0^1 \cos^2{(\pi x)}[\ln \Gamma(x)+\ln \Gamma(1-x)]dx\\ 2I&=\int_0^1 \cos^2{(\pi x)}\left[\ln\left(\frac{\pi}{\sin(\pi x)}\right)\right]dx\tag{Using Euler's reflection formula} \end{align*}
And here is where I am stuck, I have no idea how to evaluate the integral. Any help is appreciated.
Consider reduce the power of cosine term:
$$ \begin{aligned} \int_0^1\cos^2(\pi x)\log\Gamma(x)\mathrm dx &=\frac12\left[\int_0^1\cos(2\pi x)\log\Gamma(x)\mathrm dx+\int_0^1\log\Gamma(x)\mathrm dx\right] \end{aligned} $$
For the first part, by the property that
$$ \int_a^bf(x)\mathrm dx=\frac12\int_a^b[f(x)+f(a+b-x)]\mathrm dx $$
we have
$$ \begin{aligned} \int_0^1\cos(2\pi x)\log\Gamma(x)\mathrm dx &=\frac12\int_0^1\cos(2\pi x)\log[\Gamma(x)\Gamma(1-x)]\mathrm dx \\ &=\frac12\int_0^1\cos(2\pi x)\log\left[\pi\over\sin(\pi x)\right]\mathrm dx \\ &={\log\pi\over2}\int_0^1\cos(2\pi x)\mathrm dx-\frac12\int_0^1\cos(2\pi x)\log\sin(\pi x)\mathrm dx \\ &=-\frac12\int_0^1\cos(2\pi x)\log\sin(\pi x)\mathrm dx \end{aligned} $$
In fact, by the Fourier expansion of log sine
$$ \log\sin\theta=-\log2-\sum_{k=1}^\infty{\cos(2k\theta)\over k} $$
we can deduce
$$ \begin{aligned} -\frac12\int_0^1\cos(2\pi x)\log\sin(\pi x)\mathrm dx &={\log2\over2}\int_0^1\cos(2\pi x)\mathrm dx+\frac12\sum_{k=1}^\infty\frac1k\int_0^1\cos(2\pi x)\cos(2\pi kx)\mathrm dx \\ &=\frac14 \end{aligned} $$
For the last part, by Raabe's integral we have
$$ \int_0^1\log\Gamma(x)\mathrm dx=\frac12\log(2\pi) $$
All things combined, we obtain
$$ \int_0^1\cos^2(\pi x)\log\Gamma(x)\mathrm dx=\frac18+{\log(2\pi)\over4} $$