Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 8 / 4

Use the Chebyshev estimate from central binomial coefficients. DefineThe first is the Chebyshev theta function; the second counts prime powers with the same logarithmic prime weight. For a positive integer $m$, every prime $m<p\le 2m$ divides $\left(\genfrac{}{}{0ex}{}{2m}{m}\right)$, henceSumming over dyadic intervals yields $\vartheta (2^j)\le 2^{j+1}\log 2$. By monotonicity and rounding $x$ upward to a power of two,

For the lower bound, the central binomial coefficient is the largest of the $2m+1$ coefficients whose sum is $4^m$, soThe exponent of a prime in the central coefficient isEach summand is zero or one. Therefore $\log \left(\genfrac{}{}{0ex}{}{2m}{m}\right)\le \psi (2m)$. Higher prime powers contribute onlyCombining the lower bound with $2m$ just below $x$ shows $\vartheta (x)\ge cx$ for sufficiently large $x$, with, for example, $c=(\log 2)/2$.

Let $\pi (x)$ denote the number of primes at most $x$. Since every prime weight is at most $\log x$,For the upper bound, separate the primes at $\sqrt{x}$. The small ones number at most $\sqrt{x}$, and every larger one has weight at least $(\log x)/2$, givingSince $\sqrt{x}=o(x/\log x)$, this provesfor all sufficiently large $n$, for instance with $A=(\log 2)/2$ and $B=8\log 2+1$. This elementary Chebyshev estimate does not assume the Prime number theorem.

The printed differential in the transform integral is $dz$; it must be $dt$, since $z$ is the transform parameter. We prove the Newman Tauberian theorem in this corrected interpretation. Setalmost everywhere. The finite-interval transform $F_T$ is entire.

Fix $R>0$. Because the analytic domain contains the whole imaginary axis, compactness supplies a $\delta >0$ such that the thin rectangle $-\delta \le \mathrm{Re}z\le 0$, $|\mathrm{Im}z|\le R$ lies in the domain. Let $C_{+}$ be the right semicircle of radius $R$, oriented from $-iR$ to $iR$. Join its endpoints by a leftward path $L$ along the other three sides of that rectangle. This is a closed positively oriented contour.

Use contour damping for bounded Laplace transforms withThe residue theorem givesOn $C_{+}$, with $x=\mathrm{Re}z>0$,Thus the integrand has modulus at most $2M/R^2$, and this half-circle contributes at most $M/R$ after division by $2\pi$.

On $L$, the $F$ term tends to zero as $T\to \infty$: every interior point of the path has negative real part, $F$ is bounded on this fixed compact path, and the remaining kernel factor is bounded because the path avoids zero. Dominated convergence applies, with the two endpoints irrelevant to the path integral.

For the $F_T$ term, deform $L$ to the left semicircle $C_{-}$ of radius $R$. This deformation uses only the entire function $F_T$; it does not demand that $F$ extend across a large left half-disk. Both paths lie to the left of zero and their enclosed deformation region avoids the kernel pole. On $C_{-}$, for $x<0$,The same circle factor gives another bound $M/R$. ConsequentlyThe radius $R$ is arbitrary, soThis proves convergence of the ordinary improper integral, not merely a damped limit.

For the weakened domain hypothesis, take $f(t)=1$. Its Laplace transform is $F(z)=1/z$, analytic on the open right half-plane, but ${\int }_0^{T}f(t)dt=T$ diverges. Analyticity only in the open right half-plane is insufficient.