I want to proof the series of kummer. Therefore I want to show $$2 \cdot \int_0^1 \log\big(\Gamma(x)\big) \cdot \sin(2 \pi n x) dx = \frac{\gamma + \log(2 \pi) + \log(n)}{n \pi},$$ where $\gamma$ is the Euler-Mascheroni-constant.
Any help would be appreciated. Thanks in advance.
$$ \int_{0}^{1}\log \Gamma (x)\sin(2n\pi x)\,dx = \frac{\gamma + \log(2n\pi)}{2n\pi} \quad \text{for} \quad n \in \mathbb{Z}^{>0} $$
Proof: Multiply the formula $\log \Gamma (x) = \int_{0}^{\infty }\left(\frac{x-1}{t e^{t}} - \frac{1 - e^{-t(x-1)}}{t(e^{t}-1)}\right)dt$ by $\sin(2n\pi x)$ and integrate from $x$ from 0 to 1:
$$ I_{n} := \int_{0}^{1}\log \Gamma (x)\sin(2n\pi x)\,dx \\ = \int_{0}^{1}\int_{0}^{\infty }\left(\frac{x-1}{t e^{t}}\sin(2n\pi x) - \frac{1-e^{-t(x-1)}}{t(e^{t}-1)}\sin(2n\pi x)\right)dt\,dx $$
Exchange the order of integration.
Since $\int_{0}^{1}\sin(2n\pi x)dx=0, \int_{0}^{1}x\sin(2n\pi x)dx=-\frac{1}{2n\pi}$, and $\int_{0}^{1}e^{-t(x-1)}\sin(2n\pi x)dx = \frac{2n\pi}{(2n\pi)^2 + t^2}(e^t-1)$, we have:
$$ I_{n} = \int_{0}^{\infty }\left(-\frac{1}{2n\pi}\frac{1}{t e^{t}} + \frac{1}{t}\frac{2n\pi}{(2n\pi)^2 + t^2}\right)dt \\ 2n\pi I_{n} = \int_{0}^{\infty }\left(\frac{(2n\pi)^2}{(2n\pi)^2 + t^2} - \frac{1}{e^{t}}\right)\frac{dt}{t} \\ = \int_{0}^{\infty }\left(\frac{1}{1+t^2} - \frac{1}{e^{2n\pi t}}\right)\frac{dt}{t} $$
And this is $\gamma + \log(2n\pi)$ due to $\int_{0}^{\infty }\left(\frac{t^{z-1}}{1+t^2} - \frac{t^{z-1}}{e^{2n\pi t}}\right)dt = \frac{\pi}{2}\frac{1}{\sin{\frac{\pi z}{2}}} - \frac{\Gamma(z)}{(2n\pi)^z}$.
$$ = \left(\frac{\pi}{2}\frac{1}{\sin{\frac{\pi z}{2}}} - \frac{1}{z}\right) + \frac{1}{(2n\pi)^z}\left(\frac{1}{z}(2n\pi)^z - \Gamma(z)\right) \to \gamma + \log(2n\pi) \quad \text{for} \quad z \to 0^+. $$