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).

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.