I have been messing around with the Fourier Transform and wanted to see if I could manipulate this equation: $$\Gamma(s)\zeta(s)=\int_{0}^\infty\frac{x^{s-1}}{e^x-1}\rm dx$$
into an integral over the Gamma and Zeta Functions.
My Work: $$\Gamma(s)\zeta(s)=\int_{0}^\infty\frac{e^{s\log(x)}}{e^x-1}\frac{\rm dx}{x}$$
Let $2\pi k=\log(x)$
$$\Gamma(s)\zeta(s)=\int_{-\infty}^\infty \frac{e^{2\pi ks}}{e^{e^{2\pi k}}-1}2\pi\rm dk$$
Let $s=u+iv$
$$\Gamma(u+iv)\zeta(u+iv)=\int_{-\infty}^\infty \frac{e^{2\pi ku}e^{2\pi kiv}}{e^{e^{2\pi k}}-1}2\pi\rm dk$$
Through this, I now apply the Fourier Inversion Theorem:$$\frac{e^{2\pi ku}}{e^{e^{2\pi k}}-1}2\pi=\int_{-\infty}^\infty\Gamma(u+iv)\zeta(u+iv)e^{-2\pi ikv}\rm dv$$
I'm sure that I messed up somewhere since this does not look very feasable/pretty but can anyone confirm?
For $\sigma > 1$, with $x = e^{-u}$ $$\Gamma(\sigma+2i\pi\xi)\zeta(\sigma+2i\pi\xi) = \int_0^\infty \frac{x^{(\sigma+2i\pi\xi)-1}}{e^x-1}dx = \int_{-\infty}^\infty \frac{e^{-(\sigma+2i\pi\xi) u}}{e^{e^{-u}}-1} du= \mathcal{F}[\frac{e^{-\sigma u}}{e^{e^{-u}}-1}](\xi)$$
$$\frac{e^{-\sigma u}}{e^{e^{-u}}-1}= \int_{-\infty}^\infty \Gamma(\sigma+2i\pi\xi)\zeta(\sigma+2i\pi\xi) e^{2i \pi \xi u}d\xi$$ Everything converges nicely because $\frac{e^{-\sigma u}}{e^{e^{-u}}-1}$ is Schwartz, thus so is $\Gamma(\sigma+2i\pi\xi)\zeta(\sigma+2i\pi\xi)$ (it is fast decreasing).