Sorry for asking multiple questions these days about the same topic, but the thing is I was doing a school project about Riemann's zeta function so I kind of suffered when reading Riemann's paper On the Number of Primes Less Than a Given Magnitude. The first half flows pretty well to me but I kind of got stuck after this integral was derived through Fourier Transform:

\begin{align} f(x)=\frac{1}{2\pi i}\int_{a-i\infty}^{a+i\infty} \frac{\log\zeta(s)}{s}x^s ds \end{align}

Since directly plugging in the expression of $\log\zeta(s)$ would cause the integral to diverge, integration by parts is used here:

\begin{align} -\frac{1}{2\pi i\log x}\int_{a-i\infty}^{a+i\infty} \left( \frac{d}{ds}\frac{\log\zeta(s)}{s}\right)x^s ds \end{align}

Everything makes sense right now, but after this, Riemann proposed an identity without giving any proof, and I do not think this is a trivial result (Question 1):

\begin{align} -\log\Gamma\left(\frac{s}{2}+1\right)=\lim_{m\to\infty}\sum_{n=1}^m \log\left(1+\frac{s}{2n}\right)-\frac{s}{2}\log m \end{align}

And then all the terms of the integral have been unified into the form of $\frac{d}{ds}\frac{1}{s}\log\left(1-\frac{s}{\beta}\right)$ after this, which makes sense if we accept the identity, except for the constant term $\log\frac{1}{2}$.

After that, the paper just goes from seemingly nonsense to calculate $\frac{d}{d\beta}\frac{1}{s}\log\left(1-\frac{s}{\beta}\right)=\frac{1}{\beta(\beta-s)}$, I do not see why here (Question 2)

Later on this integral is calculated: \begin{align} \frac{1}{2 \pi i}\int_{a+i\infty}^{a-i\infty} \frac{x^s}{\beta(\beta-s)} ds = \frac{x^\beta}{\beta} = \int_0^x t^{\beta-1} dt \end{align}

which I can see how it comes to appear in this question

And finally, the main formula of this paper, which is evident if we admit the results above: \begin{align} f(x)=Li(x)-\sum_{\rho} Li(x^\rho) + \log\frac{1}{2} + \int_x^\infty \frac{dt}{t(t^2-1)\log t} \end{align}

So my questions are:

  1. How is the expression of the Gamma function deduced
  2. Why is $\frac{d}{d\beta}\frac{1}{s}\log\left(1-\frac{s}{\beta}\right)=\frac{1}{\beta(s-\beta)}$ calculated instead of $\frac{d}{ds}\frac{1}{s}\log\left(1-\frac{s}{\beta}\right)=-\frac{\log\left(1-\frac{s}{\beta}\right)}{s^2}-\frac{1}{s(\beta-s)}$

During my research process, I also found this paper, which gave me a lot of explanation towards some steps skipped by Riemann, but it still did not give any explanation for this process


1 Answer 1

  1. Look for the product formula of the Gamma function, deduced from the Weierstrass factorization theorem or Hadamard factorization theorem.
  2. Riemann introduced auxiliary variable $\beta$ to calculate the integral(first differentiate by $\beta$ and integrating again).
  • $\begingroup$ I do not see the process of integrating by $d\beta$ in the paper, instead everything is regarding $ds$ $\endgroup$
    – Kevin Xu
    Jun 28, 2023 at 3:10
  • $\begingroup$ Integrating $x^\beta/\beta$ with respect to $\beta$ gives ${\rm Li}(x^\beta)$. $\endgroup$
    – Riemann
    Jun 28, 2023 at 3:13
  • $\begingroup$ I might have misread it but I think $Li(x^\beta)$ comes from the fact that $\frac{x^\beta}{\beta}=\int_0^x t^{\beta-1} dt$ and moving in the $\frac{1}{\log x}$, otherwise the $\frac{1}{\log x}$ would be redundant $\endgroup$
    – Kevin Xu
    Jun 28, 2023 at 3:29

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