closed form for $\int_{0}^{1/2}\frac{x\cos\left(x\pi\right)^{2}\cos\left(2\pi kx\right)}{\sin\left(x\pi\right)}dx,k\in \mathbb{N}$

Question

I would know if exists a closed form for $$\int_{0}^{1/2}\frac{x\cos^{2}\left(\pi x\right)\cos\left(2\pi kx\right)}{\sin\left(\pi x\right)}dx,k\in\mathbb{N}.$$I tried integration by parts without success.

Answer 1

We have: $$I_k=\int_{0}^{1/2}\frac{x\cos\left(\pi x\right)^{2}\cos\left(2\pi kx\right)}{\sin\left(\pi x\right)}dx=\frac{1}{\pi^2}\int_{0}^{\pi/2}\frac{x\cos^2 x\cos(2kx)}{\sin x}\,dx$$ where $$ I_0 = \frac{2K-1}{\pi^2} $$ ($K$ is the Catalan constant) and: $$ I_{k+1}-I_k = -\frac{2}{\pi^2}\int_{0}^{\pi/2}x\cos^2 x\sin((2k+1)x)\,dx=-\frac{2}{\pi^2}\frac{(2-24 k (1+k))(-1)^k}{\left(3-2 k \left(-1+6 k+4 k^2\right)\right)^2} $$ hence:

$$ I_k = \frac{2K-1}{\pi^2}-\frac{4}{\pi^2}\sum_{n=1}^{k}\frac{(-1)^n(12n^2-12n-1)}{(8n^3-12n^2-2n+3)^2}.$$