Here (https://arxiv.org/abs/math/0305191) you can find Báez-Duarte's proof of the functional equation for $\zeta (s)$. However, I'm stuck at this central identity used in the proof: $$\sum_{n=1}^\infty \frac{1}{n\pi}\int_0^\infty x^{-s-1}(\sin 4n\pi x-\sin 2n\pi x)\, dx=-(2^s-1)2^s \pi ^{s-1}\Gamma (-s)\sin\frac{\pi s}{2}\, \zeta (1-s)$$ where $-1\lt \operatorname{Re}s\lt 0$.
How could one show the equivalence between them? Trying to split the integral and calculate $$\int_0^\infty x^{-s-1}\sin 4n\pi x\, dx$$ does not seem to help.
With the change of variable $y =2\pi n x$ then $$\int_0^\infty x^{-s-1}\sin 4n\pi x\, dx=(4\pi n)^s \int_0^\infty y^{-s-1}\sin y\, dy$$ Next, it is a complex analysis (analytic continuation) theorem that $a^{s}\Gamma(-s)=\int_0^\infty x^{-s-1}e^{-ax}dx$ for $a > 0$ extends to $\int_0^\infty x^{-s-1}\sin x\, dx=\sin(\pi s/2)\Gamma(-s)$