I am studying the proof of the Prime Number Theorem from Introduction to Analytic Number Theory by Tom M. Apostol and I came across this result:
If $c>1$ and $x \geq 1$, we have:
$$\frac{\psi_1(x)}{x^2} = \frac{1}{2\pi i}\int_{c-\infty i}^{c+\infty i} \frac{x^{s-1}}{s(s+1)}\left(-\frac{\zeta'(s)}{\zeta(s)}\right)ds$$
Here, $\psi_1(x) = \int_1^x \psi(t)dt$ and $\psi(t)$ is the Chebyshev $\psi$ function. I understood its proof but I am having trouble understanding the motivation behind the next result. The book says :
The quotient $-\frac{\zeta'(s)}{\zeta(s)}$ has a first order pole at $s=1$ with residue 1. So, if we subtract this pole we get the formula $$\frac{\psi_1(x)}{x^2} – \frac{1}{2}\left(1-\frac{1}{x}\right)^2 = \frac{1}{2\pi i}\int_{c-\infty i}^{c+\infty i} \frac{x^{s-1}}{s(s+1)}\left(-\frac{\zeta'(s)}{\zeta(s)}-\frac{1}{s-1}\right)ds$$
Specifically, I am not able to see why we need to subtract the pole. Clearly we have $c>1$, so the limits of the integral have $Re(s)>1$. Hence the term $-\frac{\zeta'(s)}{\zeta(s)}$ in the integrand would never take the input $s=1$. So, why do we bother subtracting the pole ?
This might seem like a silly question but I am completely confused. Any help shall be highly appreciated.
Best Answer
This is a preparation step to apply Riemann-Lebesgue lemma. To elaborate, let's just consider a general situation where the integrand other than $x^s$ is replaced by some function $g(s)$:
$$ {1\over2\pi i}\int_{c-i\infty}^{c+i\infty}x^sg(s)\mathrm ds={1\over2\pi}\int_{-\infty}^\infty x^{c-1}g(c+it)e^{it\log x}\mathrm dt\tag1 $$
By Riemann-Lebesgue lemma, we know that if $g(c+it)$ is integrable then
$$ \lim_{x\to\infty}\int_{-\infty}^\infty g(c+it)e^{it\log x}\mathrm dt=0\tag2 $$
In order to apply (2) to (1), we need to ensure that the integrand $g(s)$ must also be integrable on $s=1+it$. This is why we wish to remove the pole from $\zeta'/\zeta(s)$ before taking limits.
Hope this can address your concern!