# Equivalent form of prime number theorem

In the book The elementary proof of Prime Number Theorem it says that

The prime number theorem $\psi (x)\sim x$ is equivalent to $\int_1^\infty \frac{\psi (t)-t}{t^2} =-\gamma -1$, where $\psi (x)=\sum_{n\le x} \Lambda (n)$, $\gamma$ is the Euler constant.

The book also gives a hint to prove $\sum_{n\le x}\frac{\Lambda (n)}{n}=\int_1^x \frac{\psi(t)}{t^2} dt+\frac{\psi (x)}{x}$ first, but I have no idea on how to prove this hint or prove the original equivalence.

All I know now is the formula in this question $\sum_{n\le x}\frac{\Lambda (n)}{n}=\ln x-\gamma +o(1)$ which is exactly another form of prime number theorem and the relation $\psi (x)=x-1+o(x)-\gamma -\int_1^x o(t)dt$ proved before in the book.

Can the proof obtain from these two identites or I need to find a new approach? Thanks in advance.

• From $\sum_{n \le x} \frac{\Lambda(n)}{n} = \ln x+C+o(1)$ you know that $$\sum_{n \le x} \frac{\Lambda(n)-1}{n} = -H_n+\ln x+C+o(1) = C-\gamma+o(1)$$ With the Abel summation formula you also have for $x \in \mathbb{N}$ : $$\sum_{n \le x} \frac{\Lambda(n)-1}{n} = \int_1^x \frac{\psi(t)-\lfloor t\rfloor}{t^2} dt+ \frac{\psi(x)}{x}-1$$
• And with $\int_1^x \frac{\psi(t)-t}{t^2}dt = A+o(1)$ then $$\int_1^x \frac{\psi(t)-\lfloor t\rfloor}{t^2}dt = A+o(1)+\int_1^x \frac{t-\lfloor t\rfloor}{t^2}dt = A+o(1)+1-\gamma$$ since $\displaystyle\int_1^x \frac{t-\lfloor t\rfloor}{t^2}dt = \int_1^\infty \frac{t-\lfloor t\rfloor}{t^2}dt+o(1) = 1-\gamma+o(1)$
($\zeta(s) = \frac{s}{s-1}-s\int_1^\infty (t-\lfloor t\rfloor)t^{-s-1}dt=\frac{1}{s-1}+1-s\int_1^\infty (t-\lfloor t\rfloor)t^{-s-1}dt$ and $\lim_{s \to 1}\zeta(s) - \frac{1}{s-1} = \gamma$)
Put altogether $$C-\gamma+o(1) =\frac{\psi(x)}{x}-1 +A+1-\gamma+o(1)$$ which means $\frac{\psi(x)}{x} = C-A+o(1)$ and with $C = -\gamma,A = -1-\gamma$ you get the prime number theorem $$\frac{\psi(x)}{x} = 1+o(1)$$