Past exam of the analytic number theory and elliptic curves course of the University of Oxford / 2010 / A / 3

The vertical integral is absolutely convergent because the denominator is quadratic in the imaginary part. If $x<1$, move its line to the right: no poles are crossed, and the integral on $\mathrm{Re}s=R$ is $O(x^R/R)$, tending to zero. If $x>1$, move left across the simple poles at $0,-1$. Their residues are $1$ and $-x^{-1}$, and the far-left integral tends to zero. Horizontal segments vanish by their $O(|\mathrm{Im}s{|}^{-2})$ decay for a fixed pair of vertical lines. At $x=1$, dominated convergence on the original line gives the limiting value zero. Hence the kernel is $0$ below one and $1-x^{-1}$ at or above one.

On a line $c>2$, absolute convergence permits interchanging the totient Dirichlet series and this integral. The $n$th term is $x\varphi (n)$ times the kernel at $x/n$, giving exactly $\varphi (n)(x-n)$ for $n\le x$. ThusWe may take $c=5/2$ and shift to $\mathrm{Re}s=3/2$. The denominator has no zeros in this strip; the only pole crossed is $s=2$, where $\zeta (s-1)$ has residue one. Its residue in the integrand isOn the new line, the supplied estimates bound the integrand by $O(x^{5/2}|t{|}^{-3/2}{\log }^8|t|)$ for $|t|\ge 2$, an integrable function. The compact portion is bounded as well. The same estimates, uniformly across the strip, make the horizontal integrals tend to zero. ConsequentlyFor the Tauberian step let $A(x)={\sum }_{n\le x}\varphi (n)$. It is nondecreasing because the summands are nonnegative, and $S(x)={\int }_0^{x}A(t)dt$. For fixed $0<ϵ<1$,After division by $x^2$ and passage to lower and upper limits, the asymptotic for $S$ gives bounds $(3-3ϵ+{ϵ}^2)/{\pi }^2$ and $(3+3ϵ+{ϵ}^2)/{\pi }^2$. Letting $ϵ\downarrow 0$ squeezes them together. ThereforeThis uses monotonicity of the summatory function; differentiating an asymptotic formula without such control would not justify the conclusion.