[Part I] How does the prime connect to Zeta Function?
I looked back to my previous post about Riemann Hypothesis, Defining The Zeta Function.. Well, even It was said there that the theorem plays an important role in Number Theory, but all of the things I put there are likely a complex analysis rather than a Number Theory (of course Complex Number is also a number :D ). Now Let's connect it to the primes.
The connection was established before Riemann consider $\zeta(s)$ in his memoirs. It was Euler who first suggested the following relation between Dirichlet Series and Prime numbers
\[\sum_{n=1}^{\infty} \frac{1}{n^s} = \prod_{p\text{ is prime}} \left(1+\frac{1}{p^s} + \frac{1}{p^{2s}} + \cdots + \frac{1}{p^{ks}} + \cdots\right)\qquad (1)\]
Where the product is understood to range over all prime numbers. The above identity holds for $\Re(s)>1$. Therefore, $\zeta(s)$ would never equals zero for $\Re(s)>1$.
Equation (1) can be thought as the analytic version of Fundamental Theorem of Arithmetic. The intuition is clear here, since whenever we multiply all the product from the Right-Hand Side, ranging over all prime numbers, it would produce any combination of the reciprocal $\frac{1}{p^{2\alpha_1s}p^{2\alpha_2 s}\cdots\,p^{2\alpha_n s}}$, which is some representation of a natural number by fundamental theorem of arithmetic and hence summing to the Left-Hand side. To guarantee the convergence of the series and product, we need the condition $\Re (s) >1$.
I want to reprise Tchebychev exposure in his paper Sur la fonction qui détermine la totalité des nombres premiers inférieurs à une limite donnée, using the modernize notation.
My purpose is to verified some missing details in the paper. The paper can be downloaded here, which is in french.
Tchebychev use (1) to bound the prime counting function $\pi(x)$ (the cardinality of the set of all prime numbers less than $x$). His argument is quite elementary and classical, he considered three expressions
\[\sum_{k=2}^{\infty}\frac{1}{k^{1+\rho}}-\frac{1}{\sigma}\qquad (2)\]
\[\ln \rho-\sum_{p \text{ is prime}}\ln\left(1-\frac{1}{p^{1+\sigma}}\right)\qquad (3)\]
\[\sum_{p \text{ is prime}}\ln\left(1-\frac{1}{p^{1+\sigma}}\right)+\sum_{p \text{ is prime}} \frac{1}{p^{1+\sigma}}\qquad (4)\]
Let $s=\sigma+1$, the above are valid for $\Re(\sigma)>0$, or $\Re(s)>1$.
The first expresssion (2) is equal to
\[T_1(s)=\zeta(s) - 1 - \frac{1}{s-1}\]
and by (1) the second (3) is equal to
\[T_2(s)=\ln(s-1)+\ln \zeta(s),\]
while the third (4) is
\[T_3(s)=\sum_{p \text{ is prime}}\frac{1}{p^s}-\ln\zeta(s).\]
Notice that for
\[\begin{align*}\int_0^{\infty}\frac{e^{-x}}{e^x-1} x^{s-1} \, dx &=\int_0^{\infty} \left(\frac{e^{-x}}{1-e^{-x}}\right) e^{-x} x^{s-1} \, dx \\ &= \int_0^{\infty} \left(\sum_{k=1}^{\infty}e^{-kx}\right)e^{-x} x^{s-1} \, dx\\&=\sum_{k=1}^{\infty} \int_0^{\infty}e^{-(k+1)x}x^{s-1}\, dx\end{align*}\]
Substitute $x=\frac{u}{k+1}$, we have $\int_0^{\infty}e^{-(k+1)x}x^{s-1}\, dx = \int_0^{\infty} e^{-u} \frac{u^{s-1}}{(k+1)^s} \, du = \Gamma(s) \frac{1}{(k+1)^s}$. So denoting integral Gamma, $\Gamma(s)=\int_0^{\infty} e^{-x} x^{s-1} \,dx$ we have
\[\int_0^{\infty}\frac{e^{-x}}{e^x-1} x^{s-1}\, dx=\sum_{k=1}^{\infty}\Gamma(s)\frac{1}{(k+1)^s}=\Gamma(s)\left(\zeta(s)-1\right)\]
Since $\frac{1}{s-1}=\frac{\Gamma(s-1)}{\Gamma(s)}$, therefore (2) is equal to
\[T_1(s)=\zeta(s) - 1 - \frac{1}{s-1} = \frac{\int_0^{\infty}\frac{e^{-x}}{e^x-1} x^{s-1}\, dx - \Gamma(s-1)}{\Gamma(s)} = \frac{\int_0^{\infty}\left(\frac{1}{e^x-1} - \frac{1}{x}\right)x^{s-1}\, dx}{\Gamma(s)}\]
Taking limit $s \rightarrow 1^+$, since $\lim_{x\rightarrow 0^+}\left(\frac{1}{e^x-1}-\frac{1}{x}\right)=-\frac{1}{2}$ then $\int_0^{\infty}\frac{1}{e^x-1}-\frac{1}{x}\, dx$ converges and also $\int_0^{\infty} e^{-x}\,dx=1$. Thus
\[\lim_{s\rightarrow 1^+}T_1(s)=\lim_{s\rightarrow 1^+}\frac{\int_0^{\infty}\left(\frac{1}{e^x-1} - \frac{1}{x}\right)x^{s-1}\, dx}{\Gamma(s)}\] exists.
If we differentiate $\frac{\int_0^{\infty}\left(\frac{1}{e^x-1} - \frac{1}{x}\right)x^{s-1}\, dx}{\Gamma(s)}$ with respect to $s$ multiple times, we would have some combination of $\int_0^{\infty}\left(\frac{1}{e^x-1} - \frac{1}{x}\right)x^{s-1} \ln^k x \, dx$ and $\Gamma^{(k)}(s)=\int_0^{\infty} e^{-x} x^{s-1}\ln^k x\,dx$, where for all $k$ these integrals are finite and these integrals also converges when $s \rightarrow 1^+$. So the function $T_1(s)$ along with its $n$-th derivative are defined at $s=1$.
Consider expression (3), which is by (1) equal to $T_2(s)=\ln(s-1)+\ln \zeta(s)$, by our previous result $T_1(s)$ has finite limit for $s \rightarrow 1^+$, thus
\[(s-1)T_1(s)\]
approach zero for $s \rightarrow 1^+$. Notice that
\[(s-1)T_1(s)+s=(s-1)\left(\zeta(s)-1- \frac{1}{s-1}\right) + s = (s-1)\zeta(s)\]
Therefore $\ln((s-1)T_1(s)+s)=T_2(s)$ since $\lim_{s \rightarrow 1^+} (s-1)\zeta(s)=1$ we have $\lim_{s \rightarrow 1^+} T_2 =1$.
Next we prove that $T_2$ has derivative of order $n$, for any $n$. We conclude this from the fact that $(s-1)T_1(s)+s$ would never zero, also since we have proved that $T_1(s)$ has $n$-th derivative, the relation $T_2(s)=\ln ((s-1)T_1(s)+s)$ would implies that $T_2$ also assumes $n$-th derivative.
Now we consider expression (4), which is $T_3(s)$. By the mean value theorem for $x>2$ there exists $c \in [x-1,x]$ such that $\ln(x)-\ln(x-1) = \frac{1}{c} > \frac{1}{x}$, so that
\[\ln(x-1)-\ln x +\frac{1}{x}<0 \qquad \text{for all } x>2\]
Now consider the series
\[\sum_{k=2}^{\infty}\ln(k-1)-\ln k + \frac{1}{k}=\lim_{n \rightarrow \infty}\sum_{k=1}^n \frac{1}{k}-\ln n-1=\gamma -1 \]
where $\gamma$ is Euler-Mascheroni Constant therefore the series converges, and since the terms are always negative
\[\sum_{k=2}^{\infty}\left|\ln(k-1)-\ln k+\frac{1}{k}\right|=-\sum_{k=2}^{\infty}\ln(k-1)-\ln k+ \frac{1}{k}=1-\gamma\]
so the series is absolutely converges therefore some subseries of it let say for those $k$ is prime also converges, we have
\[\sum_{p \text{ is prime} }\ln(p-1)-\ln p + \frac{1}{p}=\sum_{p \text{ is prime}}\ln \left(1-\frac{1}{p}\right) + \frac{1}{p}= T_3(1)\]
converges.
Also note that
\[T_3^{(n)}(s) =\sum_{p \text{ is prime} } \frac{1}{p^{2s}} \left(\frac{\ln^i p }{1-\frac{1}{p^s}}\right) \frac{1}{p^j \left(1-\frac{1}{p^s}\right)^k}\]
where $i,j,k \geq 0$. We can compare $T_3^{(n)}(s)$ with the $p$-series $\sum \frac{1}{n^p}$ where $\Re(p)>1$, thus the series is converges.
we will use $T_1(s)$, $T_2(s)$, and $T_3(s)$ to approximate $|\pi(x) - Li(x)|$ where $Li(x)=\int_2^x \frac{1}{\ln x}\, dx$. We will do this in the next post.
The connection was established before Riemann consider $\zeta(s)$ in his memoirs. It was Euler who first suggested the following relation between Dirichlet Series and Prime numbers
\[\sum_{n=1}^{\infty} \frac{1}{n^s} = \prod_{p\text{ is prime}} \left(1+\frac{1}{p^s} + \frac{1}{p^{2s}} + \cdots + \frac{1}{p^{ks}} + \cdots\right)\qquad (1)\]
Where the product is understood to range over all prime numbers. The above identity holds for $\Re(s)>1$. Therefore, $\zeta(s)$ would never equals zero for $\Re(s)>1$.
Equation (1) can be thought as the analytic version of Fundamental Theorem of Arithmetic. The intuition is clear here, since whenever we multiply all the product from the Right-Hand Side, ranging over all prime numbers, it would produce any combination of the reciprocal $\frac{1}{p^{2\alpha_1s}p^{2\alpha_2 s}\cdots\,p^{2\alpha_n s}}$, which is some representation of a natural number by fundamental theorem of arithmetic and hence summing to the Left-Hand side. To guarantee the convergence of the series and product, we need the condition $\Re (s) >1$.
I want to reprise Tchebychev exposure in his paper Sur la fonction qui détermine la totalité des nombres premiers inférieurs à une limite donnée, using the modernize notation.
My purpose is to verified some missing details in the paper. The paper can be downloaded here, which is in french.
Tchebychev use (1) to bound the prime counting function $\pi(x)$ (the cardinality of the set of all prime numbers less than $x$). His argument is quite elementary and classical, he considered three expressions
\[\sum_{k=2}^{\infty}\frac{1}{k^{1+\rho}}-\frac{1}{\sigma}\qquad (2)\]
\[\ln \rho-\sum_{p \text{ is prime}}\ln\left(1-\frac{1}{p^{1+\sigma}}\right)\qquad (3)\]
\[\sum_{p \text{ is prime}}\ln\left(1-\frac{1}{p^{1+\sigma}}\right)+\sum_{p \text{ is prime}} \frac{1}{p^{1+\sigma}}\qquad (4)\]
Let $s=\sigma+1$, the above are valid for $\Re(\sigma)>0$, or $\Re(s)>1$.
The first expresssion (2) is equal to
\[T_1(s)=\zeta(s) - 1 - \frac{1}{s-1}\]
and by (1) the second (3) is equal to
\[T_2(s)=\ln(s-1)+\ln \zeta(s),\]
while the third (4) is
\[T_3(s)=\sum_{p \text{ is prime}}\frac{1}{p^s}-\ln\zeta(s).\]
Notice that for
\[\begin{align*}\int_0^{\infty}\frac{e^{-x}}{e^x-1} x^{s-1} \, dx &=\int_0^{\infty} \left(\frac{e^{-x}}{1-e^{-x}}\right) e^{-x} x^{s-1} \, dx \\ &= \int_0^{\infty} \left(\sum_{k=1}^{\infty}e^{-kx}\right)e^{-x} x^{s-1} \, dx\\&=\sum_{k=1}^{\infty} \int_0^{\infty}e^{-(k+1)x}x^{s-1}\, dx\end{align*}\]
Substitute $x=\frac{u}{k+1}$, we have $\int_0^{\infty}e^{-(k+1)x}x^{s-1}\, dx = \int_0^{\infty} e^{-u} \frac{u^{s-1}}{(k+1)^s} \, du = \Gamma(s) \frac{1}{(k+1)^s}$. So denoting integral Gamma, $\Gamma(s)=\int_0^{\infty} e^{-x} x^{s-1} \,dx$ we have
\[\int_0^{\infty}\frac{e^{-x}}{e^x-1} x^{s-1}\, dx=\sum_{k=1}^{\infty}\Gamma(s)\frac{1}{(k+1)^s}=\Gamma(s)\left(\zeta(s)-1\right)\]
Since $\frac{1}{s-1}=\frac{\Gamma(s-1)}{\Gamma(s)}$, therefore (2) is equal to
\[T_1(s)=\zeta(s) - 1 - \frac{1}{s-1} = \frac{\int_0^{\infty}\frac{e^{-x}}{e^x-1} x^{s-1}\, dx - \Gamma(s-1)}{\Gamma(s)} = \frac{\int_0^{\infty}\left(\frac{1}{e^x-1} - \frac{1}{x}\right)x^{s-1}\, dx}{\Gamma(s)}\]
Taking limit $s \rightarrow 1^+$, since $\lim_{x\rightarrow 0^+}\left(\frac{1}{e^x-1}-\frac{1}{x}\right)=-\frac{1}{2}$ then $\int_0^{\infty}\frac{1}{e^x-1}-\frac{1}{x}\, dx$ converges and also $\int_0^{\infty} e^{-x}\,dx=1$. Thus
\[\lim_{s\rightarrow 1^+}T_1(s)=\lim_{s\rightarrow 1^+}\frac{\int_0^{\infty}\left(\frac{1}{e^x-1} - \frac{1}{x}\right)x^{s-1}\, dx}{\Gamma(s)}\] exists.
If we differentiate $\frac{\int_0^{\infty}\left(\frac{1}{e^x-1} - \frac{1}{x}\right)x^{s-1}\, dx}{\Gamma(s)}$ with respect to $s$ multiple times, we would have some combination of $\int_0^{\infty}\left(\frac{1}{e^x-1} - \frac{1}{x}\right)x^{s-1} \ln^k x \, dx$ and $\Gamma^{(k)}(s)=\int_0^{\infty} e^{-x} x^{s-1}\ln^k x\,dx$, where for all $k$ these integrals are finite and these integrals also converges when $s \rightarrow 1^+$. So the function $T_1(s)$ along with its $n$-th derivative are defined at $s=1$.
Consider expression (3), which is by (1) equal to $T_2(s)=\ln(s-1)+\ln \zeta(s)$, by our previous result $T_1(s)$ has finite limit for $s \rightarrow 1^+$, thus
\[(s-1)T_1(s)\]
approach zero for $s \rightarrow 1^+$. Notice that
\[(s-1)T_1(s)+s=(s-1)\left(\zeta(s)-1- \frac{1}{s-1}\right) + s = (s-1)\zeta(s)\]
Therefore $\ln((s-1)T_1(s)+s)=T_2(s)$ since $\lim_{s \rightarrow 1^+} (s-1)\zeta(s)=1$ we have $\lim_{s \rightarrow 1^+} T_2 =1$.
Next we prove that $T_2$ has derivative of order $n$, for any $n$. We conclude this from the fact that $(s-1)T_1(s)+s$ would never zero, also since we have proved that $T_1(s)$ has $n$-th derivative, the relation $T_2(s)=\ln ((s-1)T_1(s)+s)$ would implies that $T_2$ also assumes $n$-th derivative.
Now we consider expression (4), which is $T_3(s)$. By the mean value theorem for $x>2$ there exists $c \in [x-1,x]$ such that $\ln(x)-\ln(x-1) = \frac{1}{c} > \frac{1}{x}$, so that
\[\ln(x-1)-\ln x +\frac{1}{x}<0 \qquad \text{for all } x>2\]
Now consider the series
\[\sum_{k=2}^{\infty}\ln(k-1)-\ln k + \frac{1}{k}=\lim_{n \rightarrow \infty}\sum_{k=1}^n \frac{1}{k}-\ln n-1=\gamma -1 \]
where $\gamma$ is Euler-Mascheroni Constant therefore the series converges, and since the terms are always negative
\[\sum_{k=2}^{\infty}\left|\ln(k-1)-\ln k+\frac{1}{k}\right|=-\sum_{k=2}^{\infty}\ln(k-1)-\ln k+ \frac{1}{k}=1-\gamma\]
so the series is absolutely converges therefore some subseries of it let say for those $k$ is prime also converges, we have
\[\sum_{p \text{ is prime} }\ln(p-1)-\ln p + \frac{1}{p}=\sum_{p \text{ is prime}}\ln \left(1-\frac{1}{p}\right) + \frac{1}{p}= T_3(1)\]
converges.
Also note that
\[T_3^{(n)}(s) =\sum_{p \text{ is prime} } \frac{1}{p^{2s}} \left(\frac{\ln^i p }{1-\frac{1}{p^s}}\right) \frac{1}{p^j \left(1-\frac{1}{p^s}\right)^k}\]
where $i,j,k \geq 0$. We can compare $T_3^{(n)}(s)$ with the $p$-series $\sum \frac{1}{n^p}$ where $\Re(p)>1$, thus the series is converges.
we will use $T_1(s)$, $T_2(s)$, and $T_3(s)$ to approximate $|\pi(x) - Li(x)|$ where $Li(x)=\int_2^x \frac{1}{\ln x}\, dx$. We will do this in the next post.
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