Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 28 / 2

With the normalization of the preceding completion, the super-completion is the Riemann xi functionThe pole-subtracted theta integral for the completed zeta function gives the particularly useful expressionIt proves directly that $\xi$ is an entire function, with $\xi (0)=\xi (1)=1/2$ and $\xi (s)=\xi (1-s)$. Here “integral function” means entire, not integer-valued.

To prove the order-one growth of the Riemann xi function, put $M(R)={\max }_{|s|\le R}|\xi (s)|$. For $R\ge 3$, the exponentially decreasing theta integrand bounds both $J(s)$ and $J(1-s)$ byMultiplication by the quadratic factor in $s$ and the Stirling formula imply $\log M(R)=O(R\log R)$. Hence the order of an entire function is at most one. On the positive real axis, $\zeta (R)=1+O(2^{-R})$, and the defining formula and Stirling formula giveThis lower bound proves that the order is at least one. Consequently $\xi$ is entire of order exactly one, generally not of finite exponential type.

Its zeros are precisely the Nontrivial zeros of the Riemann zeta function, counted with multiplicity. The values at zero and one are nonzero; the Gamma function has no zeros; and the completed functional equation together with the Euler product shows that there are no further completed zeros outside the closed critical strip. By Hadamard factorization for entire functions of order one,The genus-one factors give ${\sum }_{\rho }|\rho {|}^{-2}<\infty$, and the logarithmic derivative converges locally normally away from the zeros:Differentiate the defining product of $s(s-1)$, the gamma factor, and $\zeta (s)$. Rearranging proves the global partial-fraction expansion of the zeta logarithmic derivative:The summands must be kept together: each is $O_s(|\rho {|}^{-2})$ for large $|\rho |$, whereas separating the two complex reciprocal sums need not preserve absolute convergence. The gamma term accounts for the trivial zeros as well as cancellation at zero.

A precise endpoint version of the Perron formula is as follows. If $F(s)={\sum }_{n\ge 1}{a}_n{n}^{-s}$ converges absolutely on $\Re s=c>0$, then, for $x>0$,When $x$ is not an integer, the left side is simply the sum over $n\le x$.

Prove first the scalar kernel formula, with symmetric vertical limits:For $y>1$, close the contour to the left and apply the residue theorem at $s=0$; the exponential $e^{s\log y}$ makes the closing integral vanish. For $y<1$, close to the right, where there is no enclosed pole and the exponential decreases. The usual semicircle estimate, or integration by parts on the closing arcs, justifies these limits. At $y=1$ the truncated integral is ${\pi }^{-1}\arctan (T/c)$, tending to $1/2$.

One also needs a uniform bound to pass through the Dirichlet series. Pairing the two halves of the vertical integral gives, with $u=\log y$,The cosine term is absolutely bounded. The sine term is uniformly bounded by integration by parts against the bounded sine integral ${\int }_0^v\sin (tu)dt/t$, since $t^2/(c^2+t^2)$ is increasing from zero to one. Thus $|K_T(y)|\le C_c{y}^c$, independently of $T,y$.

For finite $T$, absolute convergence permits termwise integration, giving ${\sum }_n{a}_n{K}_T(x/n)$. The uniform bound dominates this by $C_c{x}^c{\sum }_n|a_n|n^{-c}<\infty$. The dominated convergence theorem and the scalar kernel formula now prove the boxed Perron formula, including its half-weight endpoint convention.

For the link with primes, the logarithmic derivative of the Euler product givesApplying Perron formula therefore expresses the second Chebyshev function ${\psi }_0(x)$, with half weight at a prime-power endpoint, as the vertical integral of $-{\zeta }^{\prime }/\zeta (s)x^s/s$. Shift this contour left, choosing heights away from zeros and using the partial-fraction expansion to control the integrand. The pole at one contributes $x$, a nontrivial zero $\rho$ of multiplicity $m$ contributes $-m{x}^{\rho }/\rho$, the trivial zeros contribute ${\sum }_{k\ge 1}{x}^{-2k}/(2k)$, and the pole from $1/s$ contributes $-{\zeta }^{\prime }(0)/\zeta (0)=-\log (2\pi )$. Thus the resulting Riemann–von Mangoldt explicit formula iswith zeros summed symmetrically in height. A truncated contour gives the corresponding truncated formula and remainder. The pole at one gives the main term for primes; zeros farther left give smaller oscillating corrections. Zero-free regions therefore translate into error bounds for $\psi (x)$ and, by partial summation, for the prime-counting function.