How to compute $\int_0^1 \cos^2{\pi x}\ln \Gamma(x)dx$ Problem
Find the exact value of
$$\int_0^1 \cos^2{\pi x}\ln \Gamma(x)dx$$
My Attempt
\begin{align*}
I&amp;=\int_0^1 \cos^2{(\pi x)}\ln \Gamma(x)dx\\
2I&amp;=\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&amp;=\int_0^1 \cos^2{(\pi x)}\ln \Gamma(x)+\cos^2{(\pi (1-x))}\ln \Gamma(1-x)dx\\
2I&amp;=\int_0^1 \cos^2{(\pi x)}\ln \Gamma(x)+\cos^2{\pi x}\ln \Gamma(1-x)dx\\
2I&amp;=\int_0^1 \cos^2{(\pi x)}[\ln \Gamma(x)+\ln \Gamma(1-x)]dx\\
2I&amp;=\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:
$$
\begin{aligned}
\int_0^1\cos^2(\pi x)\log\Gamma(x)\mathrm dx
&amp;=\frac12\left[\int_0^1\cos(2\pi x)\log\Gamma(x)\mathrm dx+\int_0^1\log\Gamma(x)\mathrm dx\right]
\end{aligned}
$$
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
$$
\begin{aligned}
\int_0^1\cos(2\pi x)\log\Gamma(x)\mathrm dx
&amp;=\frac12\int_0^1\cos(2\pi x)\log[\Gamma(x)\Gamma(1-x)]\mathrm dx \\
&amp;=\frac12\int_0^1\cos(2\pi x)\log\left[\pi\over\sin(\pi x)\right]\mathrm dx \\
&amp;={\log\pi\over2}\int_0^1\cos(2\pi x)\mathrm dx-\frac12\int_0^1\cos(2\pi x)\log\sin(\pi x)\mathrm dx \\
&amp;=-\frac12\int_0^1\cos(2\pi x)\log\sin(\pi x)\mathrm dx
\end{aligned}
$$
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
$$
\begin{aligned}
-\frac12\int_0^1\cos(2\pi x)\log\sin(\pi x)\mathrm dx
&amp;={\log2\over2}\int_0^1\cos(2\pi x)\mathrm dx+\frac12\sum_{k=1}^\infty\frac1k\int_0^1\cos(2\pi x)\cos(2\pi kx)\mathrm dx \\
&amp;=\frac14
\end{aligned}
$$
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
$$
\int_0^1\cos^2(\pi x)\log\Gamma(x)\mathrm dx=\frac18+{\log(2\pi)\over4}
$$
According to Maple, your last integral is
$$ \frac{1}{4} + \frac{\ln(2\pi)}{2}$$
In fact, $\cos^2(\pi x) \ln(\pi/\sin(\pi x))$ has a rather complicated closed-form antiderivative (involving dilog)