I wanted to share an approach for proving the functional equation of the Riemann zeta function, did you ever see it, what is the most elementary and transparent way to fill the details, what are the pro/cons compared to the other approaches ?
For $\Re(s) \in (0,1)$
\begin{align} 2\frac{\sin(\pi s/2)}{s} \Gamma(1-s)&(2\pi)^{s-1}(1-2^{1-s})\zeta(1-s) \\&=\int_{-\pi/2}^{\pi/2} e^{it}e^{it(s-1)}(2\pi)^{s-1} \int_0^\infty x^{-s}(\frac{1}{e^ x-1}-\frac{1}{e^{x/2}-1}-\frac1{x/2})dx dt \\ & =\int_{-\pi/2}^{\pi/2} e^{it}\int_0^\infty x^{-s}(\frac{1}{e^{2\pi e^{it} x}-1}-\frac{1}{e^{\pi e^{it} x}-1}-\frac1{\pi e^{it}x})dx dt \\ & =\int_0^\infty x^{-s}\int_{-\pi/2}^{\pi/2} (\frac{1}{e^{2\pi e^{it} x}-1}-\frac{1}{e^{\pi e^{it} x}-1}-\frac1{\pi e^{it}x})e^{it}dtdx \\ &=\int_0^\infty x^{-s-1} \frac1{4i\pi}\int_{|z|=2\pi x} (\frac{1}{e^z-1}-\frac{1}{e^{z/2}-1}-\frac1{z/2})dzdx \\ & = \int_0^\infty x^{-s-1} \frac{1-(-1)^{\lfloor x \rfloor}}{2}dx\\&=\frac{(1-2^{1-s})\zeta(s)}{s}. \end{align}