# A proof for prime number theorem!

From here we get:

$\ln(N)-1$ ~ $\sum_{P≤N}\frac{\ln(P)}{P}$

Transform this into following using function $p(x)$ defined as the $x_{th}$ prime, and let $k$ be a variable

$∫_{0}^{k}\frac{\ln(p(x))}{p(x)}dx$ ~ $\ln(p(k)-1)$

Take derivative with respect to $k$ on both sides

$\frac{\ln(p(k))}{\cancel{p(k)}}$ ~ $\frac{dp(k)}{d(k)} \times \frac{1}{\cancel{p(k)}}$

$\ln(p(k))$ ~ $\frac{dp(k)}{dk}$

Now consider instead of taking derivative of $p(k)$ we take derivative of $p(\frac{k}{\ln(k)})$

$\frac{dp(\frac{k}{\ln(k)})}{dk}$ ~ $\ln(p(\frac{k}{\ln(k)})) \times (\frac{1}{\ln(k)}-\frac{1}{(\ln(k))^2})$ ~ $\frac{\ln(p(\frac{k}{\ln(k)}))}{\ln(k)}(1-\frac{1}{\ln(k)})$ ~ $\frac{\ln(p(\frac{k}{\ln(k)}))}{\ln(k)}$

Let $p(\frac{k}{\ln(k)})=f(k)$

$f'(k)$ ~ $\frac{\ln(f(k))}{\ln(k)}$

We can observe that if $f(k)=k$ then $f'(k)=\frac{\ln(f(k))}{\ln(k)}$

So $p(\frac{k}{\ln(k)})$ ~ $k$

$\pi(k)$ ~ $\frac{k}{\ln(k)}$

Note by Zakir Husain
4 months ago

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You can use the above result to find the Euler constant 0.577....

- 3 months, 3 weeks ago

@Aruna Yumlembam - Can you tell me how? I have tried a lot.

- 3 months, 2 weeks ago

The following is the argument,since ln(N)-1 is approximately ln(2)/2+ln(3)/3+ln(5)/5+....ln(p)/p ,this implies -ln(N)+1 is approx. -(ln(2)/2+ln(3)/3+ln(5)/5+.....ln(p)/p),then add 1/2+1/3+1/4+1/5+....+1/N to both sides and take the limit of N approaches infinity. Then limit as N approaches infinity (1/2+1/3+1/4+1/5+....1/N)-(ln(2)/2+ln(3)/3+ln(5)/5+....+ln(p)/p) must approximately be 0.577..,where p<N. I also did a test ,the result came out correctly up to the second decimal place.

- 3 months, 2 weeks ago

Very cool! Also, what is the Euler constant? @Zakir Husain @Aruna Yumlembam

- 3 months, 2 weeks ago

@Yajat Shamji See this and this

- 3 months, 2 weeks ago