Is this Riemann sum formula for definite integral using of prime numbers true?

Solution 1:

Too long for a comment

Divide $[0,1]$ into $$[0,p_1/p_n],[p_1/p_n,p_2/p_n],\cdots,[p_{n-1}/p_n,1]$$

Then, using Riemann sum, we have $$I:=\int^1_0f(x)dx=\lim_{n\to\infty}\sum^n_{k=1}f\left(\frac{p_k}{p_n}\right)\frac{p_{k+1}-p_k}{p_n}$$

If we assume that $p_j=j\ln j$, $$I=\lim_{n\to\infty}\sum^n_{k=1}f\left(\frac{p_k}{p_n}\right)h(k,n)+\lim_{n\to\infty}\frac1n \sum^n_{k=1}f\left(\frac{p_k}{p_n}\right) \qquad{(1)}$$ where $$h(k,n)=\frac{(k+1)\ln(k+1)-k\ln k}{n\ln n}-\frac1n$$

It can be shown that $$h(k,n)\le h(n,n)=O(\frac1{n\ln n})$$

Therefore, the absolute value of the first term in $(1)$ is upper bounded by $$h(n,n)\cdot nM\to 0$$ where $M$ is a positive constant. This leads us to our desired result.

I am not sure if this argument can be made rigorous. I will review it when I have leisure time.

Solution 2:

Posting this as an answer rather than a comment because it contains the actual answer. Since I did not get a conclusive answer in MSE, I posted the question in MO where a rigorous proof was provided.

https://mathoverflow.net/questions/311085/calculating-limits-using-integration-for-sequence-of-prime-numbers