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How to show the validity of the following integral inside the critical strip:

$$ \Gamma(s)\zeta(s)-\Gamma(s-1) = \int_{0}^{\infty}x^{s-2}\left(\frac{x}{e^x-1}-\frac{1}{e^x}\right)\,dx \qquad\colon\space Re\{s\}\gt0 \tag{1} $$


By definition: $$ \begin{align} \Gamma(s) &= \int_{0}^{\infty}\frac{x^{s-1}}{e^x}\,dx \space\colon\space Re\{s\}\gt0 \space\Rightarrow\space \Gamma(s-1) = \int_{0}^{\infty}x^{s-2}\left(\frac{1}{e^x}\right)\,dx \space\colon\space Re\{s\}\gt1 \\[4mm] \Gamma(s)\zeta(s) &= \int_{0}^{\infty}\frac{x^{s-1}}{e^x-1}\,dx = \int_{0}^{\infty}x^{s-2}\left(\frac{x}{e^x-1}\right)\,dx \quad\colon\space Re\{s\}\gt1 \end{align} $$ Thus: $$ \begin{align} \Gamma(s)\zeta(s)-\Gamma(s-1) &= \int_{0}^{\infty}x^{s-2}\left(\frac{x}{e^x-1}-\frac{1}{e^x}\right)\,dx \qquad\colon\space Re\{s\}\gt1 \\[4mm] &= \int_{0}^{\infty}x^{s-2}\left(\frac{x}{e^x-1}\color{red}{-1}-\frac{1}{e^x}\color{red}{+1}\right)\,dx \\[4mm] &= \int_{0}^{\infty}x^{s-2}\left(\frac{x}{e^x-1}-1\right)\,dx - \int_{0}^{\infty}x^{s-2}\left(\frac{1}{e^x}-1\right)\,dx \space\colon\space \color{red}{Re\{s\}\gt0} \end{align} $$ And, if I am not mistaken, this would imply:

$$ \begin{align} \Gamma(s)\zeta(s) &= \int_{0}^{\infty}x^{s-1}\left(\frac{1}{e^x-1}-\frac{1}{x}\right)\,dx \quad\colon\space 0\lt Re\{s\}\le1 \tag{2} \\[4mm] \Gamma(s) &= \int_{0}^{\infty}x^{s-1}\left(\frac{1}{e^x}-1\right)\,dx \space\quad\colon\space -1\lt Re\{s\}\le0 \tag{3} \end{align} $$ Is this correct? How can I show these results in another way? Appreciating.

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  • $\begingroup$ anything unclear ? $\endgroup$ – reuns Jan 3 '17 at 12:54
  • $\begingroup$ @user1952009: Many thank for the answer. May I seek your patient, let us see other opinions/answers/votes. $\endgroup$ – Hazem Orabi Jan 3 '17 at 13:02
  • $\begingroup$ What do you want more ? If you are not sure of what is happening here, you need to study a complex analysis and a Fourier/Laplace transform course. The first thing is to see why $\frac{z}{e^z-1}$ is analytic at $z= 0$ $\endgroup$ – reuns Jan 3 '17 at 13:16
  • $\begingroup$ @user1952009: Thanks for been always interested in my posts, you definitely deserve a bounty for your skills. Regarding this question, appreciating if you could respond to my comment below, I think the answer is still uncompleted! $\endgroup$ – Hazem Orabi Jan 10 '17 at 20:40
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Yes (1),(2),(3) are correct. A way to show (2) is with the Taylor expansion $\frac{x}{e^x-1}=1+\mathcal{O}(x)$ as $x \to 0$ so that $$\Gamma(s)\zeta(s) - \frac{1}{s-1} = \int_{0}^{\infty}x^{s-2}\left(\frac{x}{e^x-1}-1_{x < 1}\right)dx$$ (which is obvious for $Re(s)> 1$) converges and is analytic for $Re(s) > 0$.

Then, use $$\frac{1}{s} = \begin{cases} \ \int_0^1 x^{s-1}dx \ \ \text{ for } Re(s) > 0 \\ \ -\int_1^\infty x^{s-1}dx \text{ for } Re(s) < 0 \end{cases}$$ this is why the domain of convergence is so important when talking of a Laplace/Mellin transform.

Note the inverse Laplace/Mellin transforms $$\frac{1}{e^x-1} = \frac{1}{2i\pi}\int_{\sigma-i\infty}^{\sigma+i\infty} \zeta(s) \Gamma(s) x^{-s}dx, \quad \sigma > 1$$ $$ \quad \frac{1}{e^x-1}-\frac{1}{x} = \frac{1}{2i\pi}\int_{\sigma-i\infty}^{\sigma+i\infty} \zeta(s) \Gamma(s) x^{-s}dx, \quad \sigma \in (0,1)$$ the convergence being ensured by $\zeta(s) = \mathcal{O}(s)$ as $ Im(s) \to \infty$ while $\Gamma(s)$ decreases faster than any negative power of $s$ as $Im(s) \to \infty$, since $\Gamma(s) = \int_{-\infty}^\infty e^{-e^{-u}}e^{-su}du$ is the Laplace/Fourier transform of a Schwartz function for $Re(s) > 0$.

And see Riemann's proof of the functional equation, starting from $E(s) = \pi^{-s/2} \Gamma(s/2)\zeta(s) = \int_0^\infty x^{s/2-1}\sum_{n=1}^\infty e^{-\pi n^2 x}dx$ and using that $\theta(x) = 1+2\sum_{n=1}^\infty e^{-\pi n^2 x}$ is modular, that is $\theta(1/x) = x \theta(x)$ revealing that $E(s) = E(1-s)$.

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  • $\begingroup$ I meant $\theta(1/x) = \sqrt{x}\theta(x)$ so it is modular but with a non-integral weight $\endgroup$ – reuns Jan 4 '17 at 23:52
  • $\begingroup$ Up to my knowledge, analytic continuation of $\,\zeta\,$ by Integral $(2)$ is known {i.e. books of E.C. Titchmarsh & Aleksandar Ivic}, while analytic continuation of $\,\Gamma\,$ by Integral $(3)$ is not. Would you please explain Integral $(3)$, at least in similar way of explaining Integral $(2)$. Thanks. $\endgroup$ – Hazem Orabi Jan 10 '17 at 20:41
  • $\begingroup$ @HazemOrabi For $(2)$ or $(3)$ it really works the same way. $e^{-x}-1_{x < 1} = \mathcal{O}(x)$ as $x \to 0$ so that $\Gamma(s) - \frac{1}{s} = \int_0^\infty x^{s-1} (e^{-x}-1_{x < 1})dx$ for $Re(s) > 0$ stays convergent and analytic for $Re(s) > -1$. Then add $\frac{1}{s} = -\int_0^\infty x^{s-1} 1_{x > 1} dx \ (Re(s) < 0)$ for obtaining $(2)$ $\endgroup$ – reuns Jan 10 '17 at 21:19
  • $\begingroup$ Meanwhile, for the benefit of public readers, whenever you have time, kindly review/revise your answer to include everything about the question on a bit more comprehensive/clear way. Do so, and I shall have no excuse not to accept it! $\endgroup$ – Hazem Orabi Jan 10 '17 at 22:05

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