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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?

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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).

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  • $\begingroup$ Much thanks, are there any conditions for the convergence of the integral? $\endgroup$ – aleden Nov 28 '17 at 23:39
  • $\begingroup$ What do you mean ? The exact same argument shows $\Gamma(s)$ and all its derivatives are fast decreasing on vertical strips $\endgroup$ – reuns Nov 28 '17 at 23:41
  • $\begingroup$ Doesn't $\zeta(s)$ have singularities for $\Re(s)=1$? The equation on the left implies that the integral is convergent regardless of $\sigma$ and $u$ but I'm confused as to how that would work with $\sigma=1$ due to $\zeta(s)$ being undefined on that line. $\endgroup$ – aleden Nov 28 '17 at 23:46
  • $\begingroup$ @aleden $\sigma > 1$. For $\sigma \in (0,1)$ it becomes $\Gamma(s) \zeta(s) = \int_0^\infty x^{s-2} (\frac{x}{e^x-1}-1)dx$ $\endgroup$ – reuns Nov 29 '17 at 10:30

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