[Part II] The first attempt in the difference of pi(x) and Li(x)

[Part II] The first attempt in the difference of pi(x) and Li(x)

23:43:00   Analytic Number Theory,

Following the post from Part I , where we have

\[T_1(s)=\zeta(s) - 1 - \frac{1}{s-1}\]

\[T_2(s)=\ln(s-1)+\ln \zeta(s),\]

\[T_3(s)=\sum_{p \text{ is prime}}\frac{1}{p^s}-\ln\zeta(s).\]

and $T_i(s)$ has derivative of order $n$ for $s\rightarrow 1^+$. 

Now consider 

\[-\frac{d^n T_3(s)}{d s^n} - \frac{d^n T_2(s)}{ds^n} - \frac{d^{n-1} T_1(s)}{ds^n}\]

by our previous result, the above derivative exist. So the result is finite.  Looking at canceled terms from $-T_1^{(n)}(s)$ and $-T_2^{(n-1)}(s)$ those are $\ln (s-1)$ and $\frac{1}{s-1}$, and the canceled term for $-T_3^{(n)}(s)$ and $-T_2^{(n)}(s)$ is $\ln \zeta(s)$. We have the derivative equals

\[\pm \left(\sum_{p\text{ prime}}\frac{\ln^n p}{p^{s}}-\sum_{k=2}\frac{\ln^{n-1} k}{k^{s}}\right)\]

which is finite.  Now we consider the real valued value of $s$, let say $t=\Re \, s$. Therefore

\[\pm\left(\sum_{p\text{ prime}}\frac{\ln^n p}{p^{t}}-\sum_{k=2}\frac{\ln^{n-1} k}{k^{t}}\right)\]

is finite for $t>1$, or precisely for $t \rightarrow 1^+$.

Taking the positive difference we have

\[S=\left(\sum_{p\text{ prime}}\frac{\ln^n p}{p^{t}}-\sum_{k=2}\frac{\ln^{n-1} k}{k^{t}}\right)\]

Now consider $\pi(x+1)-\pi(x)$ where $\pi(x)$ is the number of prime not exceeding $x$. We see that when $x$ is prime then $\pi(x+1)-\pi(x)$ equals to $1$, while when $x$ is not prime then $\pi(x+1)-\pi(x)$ equals $0$. Thus

\[\sum_{p\text{ prime}}\frac{\ln^n p}{p^{t}}=\sum_{k=2}^{\infty}(\pi(k+1)-\pi(k))\frac{\ln^n k}{k^{t}}\]

Ann expression $S$ can be written as

\[S=\sum_{k=2}^{\infty}\left[(\pi(k+1)-\pi(k))\frac{\ln^n k}{k^{t}}-\frac{\ln^{n-1} k}{k^{t}}\right]\]
\[=\sum_{k=2}^{\infty}\left[\pi(k+1)-\pi(k)-\frac{1}{\ln k}\right]\frac{\ln^n k}{k^{t}}\]

Now consider

\[\frac{1}{\ln x} - \int_x^{x+1} \frac{1}{\ln x}\,dx\]

By the mean value theorem for integral, there exists $c_x \in (x,x+1)$ such that $\int_x^{x+1}\frac{1}{\ln x}\,dx=\frac{1}{\ln c_x}$. Thus $\frac{1}{\ln x}-\int_x^{x+1} \frac{1}{\ln x}\,dx=\frac{1}{\ln x}-\frac{1}{\ln c_x}$ for some $c_x \in (x,x+1)$.
Since $c_x \rightarrow \infty$ whenever $x \rightarrow \infty$, we have

\[\lim_{x \rightarrow \infty}\frac{\frac{1}{\ln x}-\int_x^{x+1} \frac{1}{\ln x}\,dx}{\frac{1}{x}}=\lim_{x\rightarrow\infty}\left(\frac{x}{\ln x}-\frac{x}{\ln c_x}\right)=0\]

Thus there exists $K$ such that for $x>K$ we have $\left|\frac{1}{\ln x}-\int_x^{x+1} \frac{1}{\ln x}\,dx\right|<\frac{1}{x}$. We conclude that

\[T=T^{\prime}+\left|\sum_{k=K}^{\infty}\left(\frac{1}{\ln k}-\int_k^{k+1} \frac{1}{\ln x}\,dx\right)\frac{\ln^{n}k}{k^t}\right| < T^{\prime}+\sum_{k=K}^{\infty}\frac{\ln^{n}k}{k^{t}}\]

which converges to a finite value. Therefore

\[S+T = \sum_{k=1}^{\infty}\left(\pi(k+1)-\pi(k)-\int_k^{k+1} \frac{1}{\ln x}\,dx\right)\frac{\ln^{n}k}{k^{t}}\]

converges to a finite value.

Define the offset Logarithmic Integral Function by $\text{Li}(x)=\int_2^{x}\frac{1}{\ln x}\, dx$. Therefore

\[\pi(k+1)-\pi(k)-\int_k^{k+1} \frac{1}{\ln x}\,dx=[\pi(k+1)-\text{Li}(k+1)]- [\pi(k)-\text{Li}(k)]\qquad (1)\]

Now for the theorem itself:

[Theorem] 
For $x\geq 2$, the inequality $|\pi(x)-\text{Li}(x)| < \frac{\alpha x}{\ln^n x}$ is true for infinitely many $x$, where $\alpha$ is a positive constant, and for a sufficiently large number $n$.


To prove this, suppose for the contrary, that is we can find positive constant $\alpha$, and bound a  number $n$ such that there are only finite value of $x$ such that the inequality is true. Suppose $a$ is the the largest integer value of $x$, such that the inequality is true, and also satisfies $a>e^n$, thus for $x>a$, the inequality is not true and we have

\[|\text{Li}(x)-\pi(x)|\geq \frac{\alpha x}{\ln^n x}\qquad \qquad \ln x>n\]

Now let $U>a$ and consider the sum

\[R=C+\sum_{k=a+1}^{U}\left(\pi(k+1)-\pi(k)-\int_k^{k+1} \frac{1}{\ln x}\,dx\right)\frac{\ln^{n}k}{k^{t}}\]

where

\[C=\sum_{k=2}^{a}\left(\pi(k+1)-\pi(k)-\int_k^{k+1} \frac{1}{\ln x}\,dx\right)\frac{\ln^{n}k}{k^{t}}\]

by equation (1) and then Abel Summation by part we have

\[\begin{align*}&\sum_{k=a+1}^{U}\left(\pi(k+1)-\pi(k)-\int_k^{k+1} \frac{1}{\ln x}\,dx\right)\frac{\ln^{n}k}{k^{t}}\\ &=\sum_{k=a+1}^{t}\left([\pi(k+1)-\text{Li}(k+1)]- [\pi(k)-\text{Li}(k)]\right)\frac{\ln^{n}k}{k^{t+1}}\\ &=(\pi(U+1)-\text{Li}(U))\frac{\ln^n U}{U^{t}}-(\pi(a+1)-\text{Li}(a))\frac{\ln^n a}{a^t}\\ & \qquad\qquad-\sum_{k=a+1}^U [\pi(k)-\text{Li}(k)]\left[\frac{\ln^n (k)}{k^t}-\frac{\ln^n (k-1)}{(k-1)^t}\right]\end{align*}\]


By our hypothesis $(\pi(U+1)-\text{Li}(U))\frac{\ln^n U}{U^{s}}>\left(\frac{\alpha U}{\ln^n U}\right)\frac{\ln^n U}{U^{t}}=\frac{\alpha}{U^{s-1}}>0$.
Moreover by the mean value theorem, there exists $\xi\in[k-1,k]$ such that

\[\left[\frac{\ln^n (k)}{k^t}-\frac{\ln^n (k-1)}{(k-1)^s}\right]=\left(\frac{n}{\ln \xi}-t \right)\frac{\ln^{n}\xi}{\xi^{t+1}}\]

Thus

\[R>C-(\pi(a+1)-\text{Li}(a))\frac{\ln^n a}{a^t}+\sum_{k=a+1}^U[\pi(k)-\text{Li}(k)]\left(t-\frac{n}{\ln \xi}\right)\frac{\ln^{n}\xi}{\xi^{t+1}}\]

Consider $f(x)=\frac{\alpha x}{\ln^n x}$, then $f^{\prime}(x)=\alpha \left(1-\frac{n}{\ln x}\right)\frac{1}{\ln x}>0$ since $\ln x>n$ and $\alpha>0$, thus the function is increasing.

Since for $k\geq a+1$ and $k>\xi$ we have

\[[\pi(k)-\text{Li}(k)]\geq\frac{\alpha k}{\ln^n k}\geq\frac{\alpha \xi}{\ln^n \xi}\]

Writing $D=C-(\pi(a+1)-\text{Li}(a))\frac{\ln^n a}{a^t}$ we have

\[R>D+\sum_{k=a+1}^U\left(\frac{\alpha \xi}{\ln^n \xi}\right)\left(s-\frac{n}{\ln \xi}\right)\frac{\ln^{n}\xi}{\xi^{t+1}}=D+\alpha\sum_{k=a+1}^U\left(t-\frac{n}{\ln \xi}\right)\frac{1}{\xi^t}\]

Also note that $t>1$ and so $t-\frac{n}{\ln \xi} >1-\frac{n}{\ln\xi}$ thus we have

$R>D+\left(1-\frac{n}{\ln\xi}\right)\sum_{k=a+1}^U \frac{1}{\xi^t}$.

Letting $U \rightarrow \infty$ and then $t\rightarrow 1^+$, we have the right hand side is approaching $\infty$, while we already established that $\lim_{t\rightarrow 1^+} R$ is finite. Contradiction.

Continue Reading

[Part I] How does the prime connect to Zeta Function?

[Part I] How does the prime connect to Zeta Function?

20:38:00   Analytic Number Theory,

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. Continue Reading

Mendefinisikan fungsi Zeta Riemann, dan Riemann Hypothesis.

Mendefinisikan fungsi Zeta Riemann, dan Riemann Hypothesis.

04:27:00   Analytic Number Theory,

Walaupun cabang ini berada pada area Teori Bilangan, namun post yang saya buat ini, lebih pekat ke analisa kompleks, pada kenyataannya, pada post ini kita tidak membicarakan hubungan zeta function dengan bilangan prima.


Definisi awal (bukan definisi yang sebenarnya) dari Fungsi Zeta Riemann diperoleh dari Deret Dirichlet

\[\zeta(s)=\sum_{n=1}^{\infty} \frac{1}{n^s}\qquad\Re(s)>1\]

Fungsi ini bukan fungsi yang secara umum dimaksud pada Riemann Hypothesis. Fungsi Zeta Riemann yang sebenarnya akan didefinisikan melalui fungsi ini, dan akan dijelaskan pada post ini.

Fungsi ini analytic di $\Re(s)>1$ (Fungsi nya sendiri diperoleh dari proses analitik pada complex; limit, infinite series; tapi deret diatas bukan power series dari $\zeta(s)$ yang memberikan evidensi bahwa $\zeta(s)$ analytic)
Pernyataan dari Riemann Hypothesis adalah (Mungkin bukan formulasi yang paling sederhana, menurut opini saya, formulasi dengan Liouville's Lambda Function lebih sederhana )

[Riemann Hypothesis] Semua akar tak-trivial dari fungsi $\zeta(s)$ berada pada garis kritis $\Re(s)=\frac{1}{2}$.

Sebuah pertanyaan yang muncul disini adalah:

Bukannya deret tersebut divergen untuk $\Re(s) \leq 1$? Lalu mengapa disebutkan $\Re(s)=\frac{1}{2}$? Lalu akar tak-trivial itu apa?

Pertama-tama kita jawab pertanyaan yang pertama dengan mendefinisikan fungsi zeta yang sebenarnya,kita akan menggunakan argumen yang digunakan Riemann.

Karena fungsi diatas hanya terdefinisi ketika $\Re(s)>1$. Bagaimana caranya kita mencapai (baca:define) di $\Re(s)\leq 1$, yaitu mendefinisikannya untuk seluruh bidang kompleks? Lebih lanjut lagi, karena fungsi awal tersebut sudah analitik di $\Re(s) >1$, kita ingin mendefinisikan fungsi ini sedemikian sehingga fungsi nya juga analitik di $\Re(s) \leq 1$.


Salah satu alat yang bisa kita pakai disini adalah analytic continuation, dimana kita dapat mengisi nilai-nilai dari suatu fungsi analitik diluar natural domain nya sehingga menghasilkan fungsi analitik lain dengan domain yang lebih besar. Yakni, dengan menemukan fungsi lain katakanlah $F(z)$ sedemikian sehingga hasil restriksi dari $F(s)$ di $Re(s)>1$ adalah $\zeta(s)$. Sayangnya tidak ada harapan untuk $s=1$, akan kita liat hal ini nanti, sekarang kita fokus untuk menemukan $F$ sedemikian sehingga $F(s)=\zeta(s)$, untuk $\Re(s)>1$, dan fungsi ini harus analitik di sebuah domain yang lebih besar dari $\Re(s)>1$.

Untuk menemukan fungsi ini kita mulai dengan satu-satunya definisi yang kita punya, yaitu untuk $\Re(s)>1$ diperoleh
\[\begin{align*}\zeta(s)=\sum_{n=1}^{\infty} \frac{1}{n^s}&=\sum_{n=0}^{\infty}\frac{1}{(n+1)^s} \\ &=\sum_{n=1}^{\infty} \frac{1}{(n+1)^s}+1\\ &=\sum_{n=1}^{\infty}\frac{1}{(n+1)^s} + \sum_{n=1}^{\infty} \left(\frac{1}{n^{s-1}} - \frac{1}{(n+1)^{s-1}}\right)\\ &=\sum_{n=1}^{\infty}\left(\frac{1}{(n+1)^s}-\frac{n+1}{(n+1)^s}+\frac{n}{n^{s}}\right)\\ &=\sum_{n=1}^{\infty}n\left(\frac{1}{n^s}-\frac{1}{(n+1)^s}\right)\\ &=\sum_{n=1}^{\infty} n s\int_{n}^{n+1} x^{-s-1}\,dx\end{align*}\]

Karena $\lfloor x \rfloor = n$ ketika $x \in [n,n+1)$, kita peroleh $\int_{n}^{n+1}\lfloor x \rfloor x^{-s-1}\,dx=n\int_{n}^{n+1} x^{-s-1}\,dx$. Kita peroleh

\[\begin{align*}\zeta(s)&=s\sum_{n=1}^{\infty} n \int_{n}^{n+1} x^{-s-1}\,dx\\ &=s\sum_{n=1}^{\infty}\int_{n}^{n+1}\lfloor x \rfloor x^{-s-1}\,dx\\ &=s\int_1^{\infty}\lfloor x \rfloor x^{-s-1}\,dx\end{align*}\]

Tulis $\lfloor x \rfloor = x - \{x\}$ dimana $\{x\}$ adalah bagian pecahan dari $x$. Jadi

\[\begin{align*}\zeta(s)&=s \int_1^{\infty} x^{-s-1}\lfloor x \rfloor \,dx\\ &=s \int_1^{\infty}(x-\{x\})x^{-s-1}\,dx\\ &=s \int_1^{\infty} x^{-s}\, dx - s\int_1^{\infty} x^{-s-1}\{x\}\, dx\end{align*}\]

Karena $s \int_1^{\infty} x^{-s} \,dx = s \frac{-x^{-s+1}}{-s+1}\mid_1^{\infty}=\frac{s}{s-1}$, kita dapatkan

\[\zeta(s)=\frac{s}{s-1} - \int_1^{\infty}\{x\}x^{-s-1}\,dx\]

Dengan representasi dari $\zeta(s)$ yang ini, dapat dilihat bahwa $\zeta(s)$ mempunyai simple pole di $s=1$, dan juga karena $\{x\} \in [0,1)$ integral $\int_1^{\infty}\{x\}x^{-s-1}\,dx$ konvergen untuk $\Re(s)>0$, berdasarkan uji banding dengan $\int_1^{\infty} x^{-\Re(s)-1} \, dx$. Jadi, dengan menggunakan representasi ini , $\zeta(s)$ terdefinisi untuk $\Re(s)>0$. Kita telah memperbesar domain dari $\zeta(s)$ (dengan menggunakan analytic continuation) dari $\Re(s)>1$ ke $\Re(s)>0$, dan memperoleh informasi tentang simple pole di $s=1$ dengan residue 1 (koefisien dari $(s-1)^{-1}$). Tetapi pekerjaan untuk mendefinsikan $\zeta(s)$ untuk (hampir) setiap bidang kompleks belum selesai.


Untuk melanjutkan, kita pinjam sebuah fungsi dari kuliah Kalkulus Tingkat Lanjut (di Universitas Indonesia kami menyebutnya Matematika Dasar IV), yaitu Fungsi Gamma (Gamma Function).
Fungsi Gamma, awalnya didefinisikan sebagai $\Gamma(s)=\int_0^{\infty}t^{s-1}e^{-t}\,dt$ untuk $s$ bilangan real positif, karena integral tersebut juga konvergen untuk $s$ bilangan kompleks dengan $\Re(s)>0$, maka fungsi Gamma juga terdinisi pada bilangan kompleks , fungsi gamma analitik juga untuk $\Re(s)>0$ (beberapa kali kita memakai fakta bahwa fungsi yang menggunakan integral atau deret dalam representasinya itu analitik). Bagaimanapun juga hanya dengan menggunakan hal ini, kita tidak bisa lanjut lebih jauh melebihi $\Re(s)>0$, tapi untungnya, dengan menggunakan analytic continuation, Fungsi Gamma juga bisa diperluas menjadi fungsi analitik di seluruh bidang kompleks kecuali untuk $s=0, -1,-2, \cdots, -n, \cdots$ (liat link, atau baca lagi buku Kalkulus Tingkat Lanjut)

Menggunakan hal diatas fungsi $\Gamma(s)$ dapat didefinsikan di $\mathbb{C}/\{0,-1,-2,\cdots\}$, dan terlebih lagi analitik disana. Hubungan antara $\zeta$ dan $\Gamma$ dapat diperoleh sebagai berikut
\[\Gamma\left(\frac{s}{2}\right)=\int_0^{\infty}t^{s/2-1}e^{-t}\, dt\]

Substitusikan $t=n^2\pi u$ kita dapatkan

\[\Gamma\left(\frac{s}{2}\right)=\int_0^{\infty}\pi^{\frac{s-2}{2}} n^{s-2} u^{\frac{s-2}{2}} e^{-n^2 \pi u} (n^2 \pi)\,du=n^s \pi^{\frac{s}{/2}}\int_0^{\infty} u^{\frac{s}{2}-1} e^{-n^2 \pi u} \, du\]

Setara dengan

\[\pi^{-\frac{s}{2}}\Gamma\left(\frac{s}{2}\right)\left(\frac{1}{n^s}\right)=\int_0^{\infty} u^{\frac{s}{2}-1} e^{-n^2 \pi u} \, du\]

Jumlahkan dari $n=1$ ke $\infty$ dan karena integral di sisi kiri konvergen uniform, kita bisa menukar integral dan sumasi dan menghasilkan

\[\pi^{-\frac{s}{2}}\Gamma\left(\frac{s}{2}\right) \zeta(s) = \int_0^{\infty} u^{\frac{s}{2}-1} \left(\sum_{n=1}^{\infty}e^{-n^2 \pi u}\right) \, du\]

Kita pinjam lagi fungsi dari kalkulus tingkat lanjut yaitu, Jacobi-Theta Function, yang didefinisikan sebagai $\theta(t)=\sum_{n=-\infty}^{\infty} e^{n^2 i\pi t}$. Oleh karena itu $\theta(it)=\sum_{n=-\infty}^{\infty} e^{-n^2 \pi t}$. Karena "$\sum_{n=-\infty}^{0} e^{-n^2 \pi t} = \sum_{n=0}^{\infty} e^{-n^2\pi t}$", jadi

\[\theta(it)=\sum_{n=0}^{\infty} e^{-n^2 \pi t}+ \sum_{n=1}^{\infty}e^{-n^2 \pi t}=2 \sum_{n=1}^{\infty} e^{-n^2 \pi t}+1.\]

sehingga $\sum_{n=1}^{\infty}e^{-n^2 \pi t} = \frac{\theta(it)-1}{2}$. Sehingga fungsi zeta dapat dinyatakan sebagai

\[\zeta(s)=\frac{\pi^{\frac{s}{2}}}{\Gamma\left(\frac{s}{2}\right)}\int_0^{\infty} u^{\frac{s}{2}-1} \left(\frac{\theta(ui)-1}{2}\right)\, du\]

Dari sini kita akan menggunakan persamaan fungsi dari $\theta$, yaitu $(-ix)^{\frac{1}{2}}\theta(x)=\theta\left(-\frac{1}{x}\right)$, yang buktinya cukup susah (liat di buku tentang Special Function), pesamaan ini dapat diperoleh melalui transformasi modular dari jacobi function, biasanya juga disebut Identitas Jacobi (Liat link diatas, tapi tidak ada buktinya). Substitusikan $x \mapsto ix$ ke persamaan fungsi diatas diperoleh $x^{\frac{1}{2}} \theta(ix)=\theta\left(\frac{-1}{ix}\right)=\theta\left(\frac{i}{x}\right)$.

\[\begin{align*}&\int_0^{\infty}x^{s/2-1}\left(\frac{\theta(xi)-1}{2}\right)\,dx=\\ &\int_0^{1}x^{s/2-1}\left(\frac{\theta(xi)-1}{2}\right)\,dx+\int_1^{\infty}x^{s/2-1}\left(\frac{\theta(xi)-1}{2}\right)\,dx\end{align*}\]

Dengan substitusi $x \mapsto \frac{1}{x}$ ke Integral pertama diperoleh

\[\begin{align*}\int_0^1x^{s/2-1}\left(\frac{\theta(xi)-1}{2}\right)\, dx&= \int_{\infty}^{1}x^{-s/2+1}\left(\frac{\theta\left(\frac{i}{x}\right)-1}{2}\right)\left(\frac{-1}{x^2}\right)\, dx \\ &= \int_1^{\infty} x^{-s/2-1}\left(\frac{x^{1/2}\theta(ix)-1}{2}\right)\\ &= \int_1^{\infty} \frac{x^{-s/2-1/2}\theta(ix)-x^{-s/2-1/2}}{2}\\ &+ \int_1^{\infty} \frac{x^{-s/2-1/2}-x^{-s/2-1}}{2} \, dx\\ &=\int_1^{\infty} \frac{x^{-s/2-1/2}\theta(ix)-x^{-s/2-1/2}}{2}+\frac{1}{s(s-1)}\end{align*}\]

Jadi

\[\int_0^{\infty}x^{s/2-1}\left(\frac{\theta(xi)-1}{2}\right)\,dx\]
\[=\frac{1}{s(s-1)}+\int_1^{\infty}(x^{-s/2-1/2}+x^{s/2-1})\left(\frac{\theta(xi)-1}{2}\right)\,dx\]

Dan Fungsi Zeta Riemann terdefinisi sebagai
\[\zeta(s)=\frac{\pi^{\frac{s}{2}}}{\Gamma\left(\frac{s}{2}\right)}\left(\frac{1}{s(s-1)}+\int_1^{\infty}(x^{-s/2-1/2}+x^{s/2-1})\left(\frac{\theta(xi)-1}{2}\right)\,dx\right)\]

Perhatikan bahwa fungsi Gamma mempunyai simple pole di $s=0$, jadi $\frac{1}{\Gamma\left(\frac{s}{2}\right)}$ saling menghilangkan simple pole yang juga di $s=0$ yang diperoleh dari $\frac{1}{s(s-1)}$.
Karena $\theta(xi)$ tumbuh secara eksponensial (baca buku Persamaan Differensial, Pengantar transformasi Laplace), integrak diatas konvergen untuk $s\in \mathbb{C}$, jadi ekspresi yang berada di kanan merupakan fungsi analitik di $s \in \mathbb{C}$ kecuali untuk for $s=1$ (dimana penyebutnya nol) dan $s=-2,-4,-6, \cdots$ (dimana $\Gamma\left(\frac{s}{2}\right)$ tidak analitik, yaitu mempunyai pole) jadi kita telah memperluas fungsi zeta menjadi fungsi analitik pada domain $\mathbb{C}/\{1,-2,-4,\cdots,-2k,\cdots\}$. Pertanyaan pertama selesai.

Untuk pertanyaan kedua, apa itu akar tak-trivial dari $\zeta(s)$, singkatnya, kita sebut $s$ sebagai akar trivial jika dan hanya jika $s$ merupakan pole dari $\Gamma\left(\frac{s}{2}\right)$ yang tidak ter-cancel ($s\neq 0$), yaitu $s=-2,-4,\cdots$. Akar lain yang diketahui berada pada $0\leq \Re(z) \leq 1$, disebut akar tak-trivial. Yup, Riemann Hypothesis menanyakan apakah akar ini selalu berada pada garis $\Re(s)=\frac{1}{2}$, atau berbentuk $s=\frac{1}{2} + bi$

Telah diketahui bahwa beberapa milyar akar pertama dari akar tak-trivial dari $\zeta(s)$ memenuhi prediksi Riemann Hypothesis. Jadi pernyataan nya mungkin benar. Tapi pembuktiannya sangatlah susah, dan diketahui, teknik yang dipakai pada area ini sangat beragam, dan belum ada yang berani mengklaim teknik mana yang akan membawa ke bukti yang diinginkan (silahkan liat di arXiv.org)

Post selanjutnya (sedang di draft) membicarakan tentang hubungan nya dengan bilangan prima, dan Li function.

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Riemann Hypothesis for Dummies?

Riemann Hypothesis for Dummies?

05:23:00   Analytic Number Theory,

First, i'm sorry if the title is a little bit offensive. I just wonder if there's such a thing as "Riemann Hypothesis For Dummies", while the original statement of the conjecture itself would require at least advance undergraduate complex analysis, and (no math undergraduate is dummy right?).  I'm going to write down some analysis preliminary that will lead us to understand the statement of Riemann Hypothesis.

Recall that the Initial Riemann zeta function is defined via the Dirichlet series by

\[\zeta(s)=\sum_{n=1}^{\infty} \frac{1}{n^s}\qquad\Re(s)>1\]

Note that the function is clearly analytic on $\Re(s)>1$.

The statement of Riemann Hypothesis is (it is not the most simple form, perhaps the most simple form i know is the thing related to Liouville's Lambda Function )

[Riemann Hypothesis] All nontrivial zeros of the Riemann Zeta Function  lie on the critical line $\Re(s)=\frac{1}{2}$.


The question immediately arises here, isn't the series diverges for $\Re(s) \leq 1$? Then why does it
mention $\Re(s)=\frac{1}{2}$? Also what is the nontrivial zero?

First we resolve the problem on defining the Riemann Zeta Function (without initial), we will use Riemann's Argument, since so far this function only defined for $\Re(s)>1$.  How could we reach (read:define) it on $\Re(s)\leq 1$, that is to define it for all complex plane? Moreover, since the function is already analytic on $\Re(s) >1$, we want the function to be also analytic on the other region $\Re(s) \leq 1$.
One tool we can use here is the analytic continuation, for which we can fill the the value of the function outside its analytic domain resulting the other analytic function with a larger domain,  this can be done if  we can find another analytic function $F(s)$ such that the restriction of $F$ on $Re(s)>1$ is $\zeta(s)$ itself.  Unfortunately, there is no hope for  $s=1$, we will see this later, now we focus on finding an analytic function $F$ such that $F(s)=\zeta(s)$, for $\Re(s)>1$, this function must be analytic on the domain larger than just $\Re(s)>1$.

To find the extended function, we start with what we only have, for $\Re(s)>1$ we have

\[\begin{align*}\zeta(s)=\sum_{n=1}^{\infty} \frac{1}{n^s}&=\sum_{n=0}^{\infty}\frac{1}{(n+1)^s} \\ &=\sum_{n=1}^{\infty} \frac{1}{(n+1)^s}+1\\ &=\sum_{n=1}^{\infty}\frac{1}{(n+1)^s} + \sum_{n=1}^{\infty} \left(\frac{1}{n^{s-1}} - \frac{1}{(n+1)^{s-1}}\right)\\ &=\sum_{n=1}^{\infty}\left(\frac{1}{(n+1)^s}-\frac{n+1}{(n+1)^s}+\frac{n}{n^{s}}\right)\\ &=\sum_{n=1}^{\infty}n\left(\frac{1}{n^s}-\frac{1}{(n+1)^s}\right)\\ &=\sum_{n=1}^{\infty} n s\int_{n}^{n+1} x^{-s-1}\,dx\end{align*}\]

Since $\lfloor x \rfloor = n$ when $x \in [n,n+1)$, we have $\int_{n}^{n+1}\lfloor x \rfloor x^{-s-1}\,dx=n\int_{n}^{n+1} x^{-s-1}\,dx$. We have

\[\begin{align*}\zeta(s)&=s\sum_{n=1}^{\infty} n \int_{n}^{n+1} x^{-s-1}\,dx\\ &=s\sum_{n=1}^{\infty}\int_{n}^{n+1}\lfloor x \rfloor x^{-s-1}\,dx\\ &=s\int_1^{\infty}\lfloor x \rfloor x^{-s-1}\,dx\end{align*}\]

Write $\lfloor x \rfloor = x - \{x\}$ where $\{x\}$ is fractional part of $x$. We have

\[\begin{align*}\zeta(s)&=s \int_1^{\infty} x^{-s-1}\lfloor x \rfloor \,dx\\ &=s \int_1^{\infty}(x-\{x\})x^{-s-1}\,dx\\ &=s \int_1^{\infty} x^{-s}\, dx - s\int_1^{\infty} x^{-s-1}\{x\}\, dx\end{align*}\]

Since $s \int_1^{\infty} x^{-s} \,dx = s \frac{-x^{-s+1}}{-s+1}\mid_1^{\infty}=\frac{s}{s-1}$, we have

\[\zeta(s)=\frac{s}{s-1} - \int_1^{\infty}\{x\}x^{-s-1}\,dx\]

Via this representation of $\zeta(s)$,  we see that $\zeta(s)$ has simple pole for $s=1$, and also since $\{x\} \in [0,1)$ the integral $\int_1^{\infty}\{x\}x^{-s-1}\,dx$ converges for $\Re(s)>0$, by comparison test with $\int_1^{\infty} x^{-\Re(s)-1} \, dx$. Thus, on this representation, $\zeta(s)$ is defined for $\Re(s)>0$. We actually have extend the domain of $\zeta(s)$ (applying analytic continuation) from $\Re(s)>1$ to $\Re(s)>0$, and resulting the simple pole in $s=1$ with residue 1. Still, our works to define $\zeta(s)$ for the whole complex plane, not finish yet.

To continue, we borrow a function from Advanced Calculus Course (in Universitas Indonesia we call it Calculus IV), Gamma Function.

Gamma function, initially defined by $\Gamma(s)=\int_0^{\infty}t^{s-1}e^{-t}\,dt$ for $s$ is real positive, since the integral also converges whenever $s$ is a complex number with $\Re(s)>0$, therefore the Gamma function is defined as a complex function, this function is analytic on where it is defined that is $\Re(s)>0$. However using this definition, we can't go anywhere other than $\Re(s)>0$, but luckily, using also analytic continuation, the Gamma Function can also extendable to an analytic function in the whole complex plane except for $s=0, -1,-2, \cdots, -n, \cdots$ (see the link above)

Using it then, Gamma Function is extendable to $\mathbb{C}/\{0,-1,-2,\cdots\}$.  In the relation between $\zeta$ and $\Gamma$, consider

\[\Gamma\left(\frac{s}{2}\right)=\int_0^{\infty}t^{s/2-1}e^{-t}\, dt\]

Substitue $t=n^2\pi u$ we have

\[\Gamma\left(\frac{s}{2}\right)=\int_0^{\infty}\pi^{\frac{s-2}{2}} n^{s-2} u^{\frac{s-2}{2}} e^{-n^2 \pi u} (n^2 \pi)\,du=n^s \pi^{\frac{s}{/2}}\int_0^{\infty} u^{\frac{s}{2}-1} e^{-n^2 \pi u} \, du\]

Equivalent to

\[\pi^{-\frac{s}{2}}\Gamma\left(\frac{s}{2}\right)\left(\frac{1}{n^s}\right)=\int_0^{\infty} u^{\frac{s}{2}-1} e^{-n^2 \pi u} \, du\]



Taking the summation from $n=1$ to $\infty$ and since the integral on the left converges uniformly, we can exchange the integral and summation thus resulting

\[\pi^{-\frac{s}{2}}\Gamma\left(\frac{s}{2}\right) \zeta(s) = \int_0^{\infty} u^{\frac{s}{2}-1} \left(\sum_{n=1}^{\infty}e^{-n^2 \pi u}\right) \, du\]

We borrow again a function from advanced calculus,  Jacobi-Theta Function, defined by $\theta(t)=\sum_{n=-\infty}^{\infty} e^{n^2 i\pi t}$. Therefore $\theta(it)=\sum_{n=-\infty}^{\infty} e^{-n^2 \pi t}$. Since "$\sum_{n=-\infty}^{0} e^{-n^2 \pi t} = \sum_{n=0}^{\infty} e^{-n^2\pi t}$", using this we have

\[\theta(it)=\sum_{n=0}^{\infty} e^{-n^2 \pi t}+ \sum_{n=1}^{\infty}e^{-n^2 \pi t}=2 \sum_{n=1}^{\infty} e^{-n^2 \pi t}+1.\]

Thus $\sum_{n=1}^{\infty}e^{-n^2 \pi t} = \frac{\theta(it)-1}{2}$. Therefore the zeta function is represented as

\[\zeta(s)=\frac{\pi^{\frac{s}{2}}}{\Gamma\left(\frac{s}{2}\right)}\int_0^{\infty} u^{\frac{s}{2}-1} \left(\frac{\theta(ui)-1}{2}\right)\, du\]

From this point we shall use the functional equation for $\theta$, that is $(-ix)^{\frac{1}{2}}\theta(x)=\theta\left(-\frac{1}{x}\right)$,  which is quite hard to prove, this functional equation can be obtained from modular transform of theta function. Usually given as Jacobi Identities for theta function (See the Link above, but no proof there).  Substituing $x \mapsto ix$ to the functional equation we have $x^{\frac{1}{2}} \theta(ix)=\theta\left(\frac{-1}{ix}\right)=\theta\left(\frac{i}{x}\right)$.

\[\begin{align*}&\int_0^{\infty}x^{s/2-1}\left(\frac{\theta(xi)-1}{2}\right)\,dx=\\ &\int_0^{1}x^{s/2-1}\left(\frac{\theta(xi)-1}{2}\right)\,dx+\int_1^{\infty}x^{s/2-1}\left(\frac{\theta(xi)-1}{2}\right)\,dx\end{align*}\]

Using substitution $x \mapsto \frac{1}{x}$ on the first integral, we have

\[\begin{align*}\int_0^1x^{s/2-1}\left(\frac{\theta(xi)-1}{2}\right)\, dx&= \int_{\infty}^{1}x^{-s/2+1}\left(\frac{\theta\left(\frac{i}{x}\right)-1}{2}\right)\left(\frac{-1}{x^2}\right)\, dx \\ &= \int_1^{\infty} x^{-s/2-1}\left(\frac{x^{1/2}\theta(ix)-1}{2}\right)\\ &= \int_1^{\infty} \frac{x^{-s/2-1/2}\theta(ix)-x^{-s/2-1/2}}{2}\\ &+ \int_1^{\infty} \frac{x^{-s/2-1/2}-x^{-s/2-1}}{2} \, dx\\ &=\int_1^{\infty} \frac{x^{-s/2-1/2}\theta(ix)-x^{-s/2-1/2}}{2}+\frac{1}{s(s-1)}\end{align*}\]

Thus we have

\[\int_0^{\infty}x^{s/2-1}\left(\frac{\theta(xi)-1}{2}\right)\,dx\]
\[=\frac{1}{s(s-1)}+\int_1^{\infty}(x^{-s/2-1/2}+x^{s/2-1})\left(\frac{\theta(xi)-1}{2}\right)\,dx\]

Hence the Riemann Zeta Function would takes form

\[\zeta(s)=\frac{\pi^{\frac{s}{2}}}{\Gamma\left(\frac{s}{2}\right)}\left(\frac{1}{s(s-1)}+\int_1^{\infty}(x^{-s/2-1/2}+x^{s/2-1})\left(\frac{\theta(xi)-1}{2}\right)\,dx\right)\]

Note that Gamma function has simple pole on $s=0$, thus $\frac{1}{\Gamma\left(\frac{s}{2}\right)}$ cancel the simple pole also at $s=0$ obtained from $\frac{1}{s(s-1)}$.
Since $\theta(xi)$ has exponential order, the above integral converges for every $s\in \mathbb{C}$, therefore the expression on the right performs an analytic function for all $s \in \mathbb{C}$ except for $s=1$ (in which the denominator is zero) and $s=-2,-4,-6, \cdots$ (in which $\Gamma\left(\frac{s}{2}\right)$ is not analytic) thus we have extend the zeta function to an analytic function defined on $\mathbb{C}/\{1,-2,-4,\cdots,-2k,\cdots\}$.

We actually have any other question to resolve, that is a nontrivial zeros of $\zeta(s)$, to make it short, we call $s$ the zero is trivial zero iff $s$ is a pole from the $\Gamma\left(\frac{s}{2}\right)$, that is $s=0,-2,-4,\cdots$. The other zeros is known to be located on $0\leq \Re(z) \leq 1$, are called nontrivial zeros.  Therefore we have got the idea now, that the Riemann Hypothesis asks wether these nontrivial zeros always located on the $\Re(s)=\frac{1}{2}$.

It is known that more than billion of the first nontrivial zeros are all agree with the hypothesis, which seems likely the hypothesis to be true. But, as the problem is complicated enough to describe, the proof challenges many greatest minds in the world(s), as the most difficult problem in Number Theory.

Writing this post remind me with the quote from Albert Einstein, which is put on the wall of the seminar room in our university :D  , "If you can't explain it simply, you don't understand it well enough" ,  well maybe I don't really understand this very well,  because even to describe it, the miscellaneous properties of the other special functions are being used.

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