Is it possible to show (the trivial statement)

$\sum _{n\leq x}1=x+\mathcal O\left (1\right )$

using Perron's formula?

For $c$ a little bigger than $1$ and $1>c'>0$, a quantitative form of Perron's formula and then the Residue Theorem implies (with some parameter $T>0$)

$\begin {eqnarray*} \sum _{n\leq x}1&=&\int _{c-iT}^{c+iT}\frac {\zeta (s)x^sds}{s}+E_1 \\ &=&x+\int _{c'-iT}^{c'+iT}\frac {\zeta (s)x^sds}{s}+E_2+E_1, \end {eqnarray*}$

where $E_1$ is around $x/T$ and $E_2$ is the two horizontal integrals between the above two.

Since $\zeta (s)$ is around $t^{1/2-\sigma }$ in size in $(0,1/2)$ we obviously can't take the modulus signs directly inside. Are there any results that take account this cancellation over the integral? (And just as a safety check: the horizontal contributions can't give extra cancellation, right?)

Perhaps applying the functional equation we may be able to compute the integral explicitly?

Cheers.

Is it possible to show (the trivial statement)

$\sum _{n\leq x}1=x+\mathcal O\left (1\right )$

using Perron's formula?

For $c$ a little bigger than $1$ and $1>c'>0$, a quantitative form of Perron's formula and then the Residue Theorem implies (with some parameter $T>0$)

$\begin {eqnarray*} \sum _{n\leq x}1&=&\frac {1}{2\pi i}\int _{c-iT}^{c+iT}\frac {\zeta (s)x^sds}{s}+E_1 \\ &=&x+\frac {1}{2\pi i}\int _{c'-iT}^{c'+iT}\frac {\zeta (s)x^sds}{s}+E_2+E_1, \end {eqnarray*}$

where $E_1$ is around $x/T$ and $E_2$ is the two horizontal integrals between the above two.

Since $\zeta (s)$ is around $t^{1/2-\sigma }$ in size in $(0,1/2)$ we obviously can't take the modulus signs directly inside. Are there any results that take account this cancellation over the integral? (And just as a safety check: the horizontal contributions can't give extra cancellation, right?)

Perhaps applying the functional equation we may be able to compute the integral explicitly?

Cheers.