ilk olarak $\psi$ fonksiyonunu tanimlayalim ($p$ asal, $m$ pozitif tamsayi): $$\psi(x)=\sum\limits_{p^m\leq x}\log p.$$ Aslinda bu fonksiyon ($[x]$ tak deger fonksiyonu "asagiya yuvarlayan" olmak uzere) $$\psi(x)=\sum\limits_{p \leq x}\bigg[\frac{\log x}{\log p}\bigg]\log p.$$ $\pi(x)$ fonksiyonu da $x$'den kucuk asal sayilarin sayisi olsun.
Theorem: Eger $x \rightarrow \infty$ iken $\psi(x) \sim x$ ise $\pi(x)\sim \frac {\log x}x$
ispat: ilk olarak $$\psi(x)=\sum\limits_{p \leq x}\bigg[\frac{\log x}{\log p}\bigg]\log p \leq \sum\limits_{p \leq x}\frac{\log x}{\log p}\log p \leq \pi(x) \log x$$ oldugundan $$\frac {\psi(x)}{x} \leq \pi(x) \frac{\log x}{x}$$ gelir. Burdan da $\lim\limits_{x \rightarrow\infty}\frac{\psi(x)}{x}=1$ oldugundan ilk gozlemimiz $$1 \leq \lim\limits_{x\rightarrow \infty}\inf \pi(x) \frac{\log x}{x}.$$ Simdi $0 < \alpha < 1$ olacak sekilde bir $\alpha$ secelim (sabitleyelim). O zaman $$\psi(x) \geq \sum\limits_{p \leq x}\log p \geq \sum\limits_{x^\alpha \leq p \leq x}\log p \geq (\pi(x)-\pi(x^\alpha))\log x^\alpha$$. Yani $$\frac{\psi(x)}{x} \geq \alpha \frac{ \log x(\pi(x) -\pi(x^\alpha))}{x}=\alpha \frac{ \log x(\pi(x) -x^\alpha)}{x}=\alpha\pi(x) \frac{ \log x}{x} -\alpha\frac{\log x}{x^{1-\alpha}} $$ yani $$1 \geq \alpha\lim\limits_{x\rightarrow \infty}\sup \pi(x) \frac{\log x}{x}.$$ O halde $$\pi(x)\sim \frac {\log x}x.$$