Question

$\displaystyle\int^1_0\log\Gamma(x)\cos(2n\pi x)dx$

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Answer 1

$&=\int^1_0\log\Gamma(x)\cos(2n\pi x)dx\\ &=-\frac12\int^1_0\log(\sin(\pi x))\cos(2n\pi x)dx\\ &=-\frac1{4n\pi}\int^1_0\log(\sin(\pi x))d(\sin(2n\pi x))\\ &=\frac1{4n\pi}\int^1_0\sin(2n\pi x)d(\log(\sin(\pi x)))\\ &=\frac1{4n}\int^1_0\sin(2n\pi x)\cot(\pi x)dx\\ &=\frac1{4n}\int^1_0\frac{\sin(2n\pi x)}{\sin (\pi x)}\cos(\pi x)dx\\ &=\frac1{2n}\int^1_0\left(\sum_{k=1}^{n}\cos((2k-1)\pi x)\right)\cos(\pi x)dx\\ &=\frac1{2n}\int^1_0\cos^2(\pi x)dx\\ &=\frac1{4n}$