Perron's formula

We know that:

$ψ (x) = \sum_{n \leq x} Λ (n)$ , where

Λ (n) = {\begin{cases} \log p & if n = p^{k} for some prime p and k \geq 1, \\ 0 & otherwise . \end{cases}

I.e., Chebyshev function is a kind of accumulation of primes up to $x$
2. $- \frac{ζ^{'} (s)}{ζ (s)} = \sum_{n} \frac{Λ (n)}{n^{s}}$ , i.e., it has information of the "total accumulation"
3. $\frac{1}{2 π i} \int_{c - i \infty}^{c + i \infty} y^{s} / s d s$ is 1 if $y > 1$ , and 0 otherwise . For $c > 1$ , to have convergence.

How can we obtain from here that

ψ (x) = \frac{1}{2 π i} \int_{c - i \infty}^{c + i \infty} (- \frac{ζ^{'} (s)}{ζ (s)}) \frac{x^{s}}{s} d s

We start with the right-hand side

I = \frac{1}{2 π i} \int_{c - i \infty}^{c + i \infty} (- \frac{ζ^{'} (s)}{ζ (s)}) \frac{x^{s}}{s} d s

Replace $- \frac{ζ^{'} (s)}{ζ (s)}$

I = \frac{1}{2 π i} \int_{c - i \infty}^{c + i \infty} (\sum_{n = 1}^{\infty} \frac{Λ (n)}{n^{s}}) \frac{x^{s}}{s} d s

but then

I = \sum_{n = 1}^{\infty} Λ (n) [\frac{1}{2 π i} \int_{c - i \infty}^{c + i \infty} \frac{{(\frac{x}{n})}^{s}}{s} d s]

The integral equals 1 if $y > 1$ , which means $\frac{x}{n} > 1$ , or $n < x$ .
The integral equals 0 if $0 < y < 1$ , which means $\frac{x}{n} < 1$ , or $n > x$ .

Because the integral acts as a "filter" that only returns 1 when $n < x$ and 0 otherwise, our infinite sum truncates beautifully into a finite sum:

I = \sum_{n \leq x} Λ (n)