Codex @codex  0

This modular-invariant function, initially defined for $\Re s>1$, has Fourier expansionwhere $\Xi (u)={\pi }^{-u/2}\Gamma (u/2)\zeta (u)$ is the completed Riemann zeta function, without the entire-xi polynomial factor. Poisson summation and the integral representation of the modified Bessel function of the second kind prove the expansion. It gives meromorphic continuation, $\mathcal{E}(z,s)=\mathcal{E}(z,1-s)$, and simple poles at zero and one with residues $-1/2,1/2$. Apparent poles at $s=1/2$ cancel between the two constant terms.

For the series over all nonzero integer pairs, the constant Fourier coefficient is $2\zeta (2s)y^s+2\sqrt{\pi }\Gamma (s-1/2)\zeta (2s-1)y^{1-s}/\Gamma (s)$. Integrating over one real period unfolds the translated intervals and then uses the Gaussian integral. Multiplication by ${\pi }^{-s}\Gamma (s)$ expresses it as $2\Lambda (2s)y^s+2\Lambda (2s-1)y^{1-s}$ in terms of the completed Riemann zeta function.

First alternative. The Euler product of the Riemann zeta function is absolutely convergent for $\sigma >1$. Its logarithm and the nonnegative trigonometric polynomialgiveConsequently $\zeta (\sigma {)}^3|\zeta (\sigma +it){|}^4|\zeta (\sigma +2it)|\ge 1$. Suppose $t\ne 0$ and the Riemann zeta function had a zero of order $m\ge 1$ at $1+it$. As $\sigma \downarrow 1$, its simple pole at one gives $\zeta (\sigma )=O((\sigma -1{)}^{-1})$, the alleged zero gives $\zeta (\sigma +it)=O((\sigma -1{)}^m)$, and $\zeta (\sigma +2it)$ remains bounded. The product would be $O((\sigma -1{)}^{4m-3})\to 0$, contradicting its lower bound. Thus there are no zeta zeros on $\Re s=1$; the point $s=1$ itself is a pole, not a zero.

For the first integral of the Chebyshev function, defineThe logarithmic derivative identity $-{\zeta }^{\prime }/\zeta (s)={\sum }_n\Lambda (n)n^{-s}$, valid for $\Re s>1$, leads toTo verify the kernel rather than assume an inversion formula, integrate $y^s/(s(s+1))$ on the vertical line $\Re s=c$. Closing to the left when $y>1$ gives the residues at zero and minus one, namely $1-y^{-1}$; closing to the right when $0<y<1$ gives zero. At $y=1$ the integral is zero as well. The horizontal integrals tend to zero because of the quadratic denominator, and the distant vertical integrals vanish in the indicated direction. Multiplying this kernel by $x$ with $y=x/n$ gives $(x-n{)}^{+}$. The absolute convergence of the Dirichlet series on $\Re s=c$, together with the integrable denominator, permits interchange with the integral and proves the formula.

Here is a quantitative contour route from this relation to the Prime number theorem. Question 3 establishes a zero-free region of the Riemann zeta function and the accompanying bounds for its logarithmic derivative, independently of the Prime number theorem. In a sufficiently narrow half-width region they implyon the left and horizontal sides of the rectangle with heights $\pm T$ and left edge ${\sigma }_L=1-b/(2\log (T+3))$, after making $b>0$ small enough also for bounded heights. The explanation of this bound is included at the end of Question 3. Take $c=1+1/\log x$ and $T=\exp (\sqrt{\log x})$. On the original line, the omitted tails aresince the Euler product bounds $|{\zeta }^{\prime }/\zeta (c+it)|$ by $-{\zeta }^{\prime }/\zeta (c)=O(\log x)$. Shift the truncated integral to the left edge. The only enclosed pole is at $s=1$, with residue $x^2/2$; the kernel poles at zero and minus one lie outside the rectangle. The left edge contributes $O(x^{2-b/(2\log (T+3))}{\log }^2(T+3))$, since $|s(s+1){|}^{-1}\ll (1+t^2{)}^{-1}$ there. The two horizontal edges contribute $O(x^2{\log }^2(T+3)/T^2)$. In particular, for some $a>0$,

It remains to remove the smoothing; differentiating an asymptotic formula would not justify this step. The monotonicity of the Second Chebyshev function gives, for $h=\epsilon x$ with $0<\epsilon <1$,First let $x\to \infty$ with $\epsilon$ fixed. After division by $x$, the lower and upper bounds tend to $1-\epsilon /2$ and $1+\epsilon /2$. Let $\epsilon \downarrow 0$ to obtain $\psi (x)\sim x$. Higher prime powers contribute $O(\sqrt{x}\log x)$ by the Chebyshev estimate, so $\theta (x)\sim x$. Finally partial summation givesThe integral is $O(x/{\log }^{2}x)$, by splitting at $\sqrt{x}$ and using $\theta (t)=O(t)$. This proves the Prime number theorem with $\pi (x)\sim x/\log x$.

Second alternative. We prove the functional equation of the Riemann zeta function from a theta function. Set $\Theta (u)={\sum }_{n\in \mathbb{Z}}{e}^{-\pi n^{2}u}$ for $u>0$. With Fourier transform convention $\hat{f}(\xi )={\int }_{\mathbb{R}}f(v)e^{-2\pi iv\xi }dv$, the Gaussian function $f(v)=e^{-\pi u{v}^2}$ hasIndeed, differentiation under the integral and integration by parts give ${\hat{f}}^{\prime }(\xi )=-2\pi \xi \hat{f}(\xi )/u$, while the Gaussian integral gives $\hat{f}(0)=u^{-1/2}$. Periodize $f$: the smooth period-one function ${\sum }_{n}f(v+n)$ has Fourier coefficients $\hat{f}(k)$, by integrating on $[0,1]$ and interchanging its absolutely convergent sum with the integral. Its absolutely convergent Fourier series, evaluated at zero, proves the Poisson summation formula in this case. HenceFor $\Re s>1$, termwise integration and the defining integral of the Gamma function give the Mellin representation of the completed Riemann zeta function:Split at one and substitute $u=1/v$ in the lower integral. The elementary part isand the remaining part becomes an integral over $[1,\infty )$. ThusThe integral is an entire function of $s$, because $\Theta (u)-1$ decays exponentially and uniform convergence holds on every compact subset of the complex plane. This proves meromorphic continuation of the completed Riemann zeta function, with only the displayed simple poles, and the right side is invariant under $s\mapsto 1-s$. ThereforeThe Gamma reflection formula and Gamma duplication formula give the equivalent form $\zeta (s)=2^s{\pi }^{s-1}\sin (\pi s/2)\Gamma (1-s)\zeta (1-s)$, understood by meromorphic continuation. Multiplication by $s(s-1)/2$ gives the entire Riemann xi function $\xi (s)=\xi (1-s)$.

Let $N(T)$ count positive ordinates of Nontrivial zeros of the Riemann zeta function up to height $T$, including multiplicity. The Riemann–von Mangoldt formula isMore precisely, away from ordinates of zeros, the constant term is $7/8$, the additional term is $S(T)={\pi }^{-1}\arg \zeta (1/2+iT)$, and the remaining error is $O(T^{-1})$. Here the argument is continued along $2\to 2+iT\to 1/2+iT$, and $S(T)=O(\log T)$. The coarse formula displayed above holds at all large heights with the usual right-continuous counting convention. Subtract it at $T+1$ and $T$: the main term changes by $O(\log T)$ and each error is $O(\log T)$. Hence

For infinitely many reciprocal-logarithmic gaps between zeta zeros, fix any $0<c<2\pi$, for example $c=\pi$. Suppose all sufficiently late gaps satisfied ${\gamma }_{n+1}-{\gamma }_n\le c/\log n$. Summing and usinggives ${\gamma }_n\le (c+o(1))n/\log n$. The last elementary estimate follows by integral comparison with $1/\log t$ and one integration by parts. The Riemann–von Mangoldt formula would implya contradiction. This reasoning is valid whether repeated ordinates are listed according to multiplicity or only distinct ordinates are listed, since in either case $n\le N({\gamma }_n)$. Consequently ${\gamma }_{n+1}-{\gamma }_n>c/\log n$ infinitely often, for any fixed $0<c<2\pi$.

Use $\Xi (u)={\pi }^{-u/2}\Gamma (u/2)\zeta (u)$ for the completed Riemann zeta function, without the polynomial factor used to define the entire Riemann xi function. For complex $\nu$ putThis is the Modified Bessel function of the second kind. The substitution $t=e^v$ writes it as $\frac{1}{2}{\int }_{\mathbb{R}}{e}^{-Y\cosh v+\nu v}dv$, proving that it is entire in $\nu$, that $K_{\nu }=K_{-\nu }$, and that it decays exponentially as $Y\to \infty$, uniformly for $\nu$ in a compact set.

For $\Re s>1$ all the lattice sums and the integrals below converge absolutely. The $m=0$ terms in the completed nonholomorphic Eisenstein series contribute $\Xi (2s)y^s$. For $m\ne 0$, use the Gamma integral and then Poisson summation in $n$:Here the Poisson identity is ${\sum }_n{e}^{-\pi u(n+mx{)}^2}=u^{-1/2}{\sum }_h{e}^{-\pi h^2/u}{e}^{2\pi ihmx}$. At $h=0$, evaluating the Gamma integral and summing over $m$ gives $\Xi (2s-1)y^{1-s}$. At $h\ne 0$, the substitution $u=|h|t/(|m|y)$ and the integral representation of the modified Bessel function of the second kind giveCollect the frequency $r=mh\ne 0$. There are two signed choices of $m$ for each positive divisor of $|r|$. Thus, with ${\sigma }_a(n)={\sum }_{d\mid n}{d}^a$, the full Fourier expansion isThe factor $2$ is important: the lattice sum includes both signs, while the defining factor $1/2$ was already used in its Mellin integral. This expansion uses exactly the completion in the question.

We can deduce both of the offered conclusions. First recall, with a short proof, the needed functional equation of the Riemann zeta function. Put $\Theta (t)={\sum }_{n\in \mathbb{Z}}{e}^{-\pi n^{2}t}$. Poisson summation gives $\Theta (t)=t^{-1/2}\Theta (1/t)$. Splitting the Mellin integral of $\Theta -1$ at one yieldsThe integral is entire and invariant under $u\mapsto 1-u$. Hence $\Xi (u)=\Xi (1-u)$, with only simple poles at $u=0,1$, of residues $-1,1$. This is the pole-bearing completion, not the entire xi function.

For fixed $y>0$, the nonconstant part of the Eisenstein expansion is entire in $s$ and locally normally convergent: the divisor sums grow at most polynomially on a compact set of $s$, whereas $K_{s-1/2}(2\pi |r|y)$ decays exponentially in $|r|$. The same argument gives local uniformity for $z$ in compact subsets of the complex upper half-plane. Thus only the constant terms can have poles. Their apparent poles at $s=1/2$ cancel: $\Xi (2s)$ has principal part $1/(2(s-1/2))$, while $\Xi (2s-1)$ has its negative, and both multiply $y^{1/2}$ there. The remaining poles are simple, at $s=0,1$, with residues $-1/2,1/2$.

Under $s\mapsto 1-s$, the two constant terms interchange by the zeta functional equation. The nonconstant terms are unchanged because $K_{\nu }=K_{-\nu }$ andwhich follows by replacing a divisor $d$ by $|r|/d$. Therefore

For the Kronecker limit formula, evaluate the finite Laurent coefficient at $s=1$. Write ${\gamma }_E$ for the Euler--Mascheroni constant. From $\zeta (u)=1/(u-1)+{\gamma }_E+O(u-1)$ and ${\Gamma }^{\prime }(1/2)/\Gamma (1/2)=-{\gamma }_E-2\log 2$ we obtainThe gamma derivative follows by differentiating the Gamma duplication formula and using ${\Gamma }^{\prime }(1)=-{\gamma }_E$. Multiplication by $y^{1-s}$ adds $-\frac{1}{2}\log y$ to the constant term; the other constant term contributes $\Xi (2)y=\pi y/6$.

The half-order Modified Bessel function of the second kind is $K_{1/2}(Y)=\sqrt{\pi /(2Y)}{e}^{-Y}$. For example, in its even integral representation set $v/2$ as variable and then $u=\sinh (v/2)$: $\cosh v=1+2u^2$ and the $\cosh (v/2)$ factor converts the integral to a Gaussian. At $s=1$ the nonconstant Eisenstein coefficients are consequently ${\sigma }_{-1}(|r|)e^{-2\pi |r|y}$. On the other hand, the absolutely convergent logarithmic series givesAdding this identity to its conjugate identifies the nonconstant sum as $-2\log |{\prod }_{n\ge 1}(1-q^n)|$. With the analytic Dedekind eta function product $\eta (z)=q^{1/24}{\prod }_{n\ge 1}(1-q^n)$, the resulting answer isThis derivation of the Kronecker limit formula uses the analytic eta product only; it does not assume its modular transformation. The lattice series is invariant for $\Re s>1$ by an integral change of the pair $(m,n)$ and $\Im (\gamma z)=y/|cz+d{|}^2$. Analytic continuation preserves this invariance, so its finite Laurent coefficient is invariant too. This supplies the independent weight-two transformation argument used in Question 2.

Use the Fourier transform convention $\hat{f}(u)={\int }_{\mathbb{R}}f(x)e^{-2\pi iux}dx$. A version of the Poisson summation formula sufficient here isBoth sums converge absolutely. To prove it, form the periodization of a Schwartz function $P(x)={\sum }_{n\in \mathbb{Z}}f(x+n)$. The rapid decay of $f$ and all its derivatives makes this series and its differentiated series uniformly convergent on $[0,1]$, so $P$ is smooth and one-periodic. Its Fourier coefficients areRepeated integration by parts makes these coefficients rapidly decreasing. Thus ${\sum }_m\hat{f}(m)e^{2\pi imx}$ converges absolutely and uniformly and has the same coefficients as $P$. By uniqueness of Fourier series it equals $P$. Evaluation at zero proves the formula.

For $u>0$, the Gaussian $f(x)=e^{-\pi u{x}^2}$ has transform $u^{-1/2}{e}^{-\pi m^2/u}$. This follows from the Gaussian integral at frequency zero and integration by parts, which gives the differential equation ${\hat{f}}^{\prime }(v)=-(2\pi v/u)\hat{f}(v)$. Hence the Jacobi theta functionsatisfies $\Theta (u)=u^{-1/2}\Theta (1/u)$.

Use the normalization of the completed Riemann zeta functionInitially, for $\Re s>1$, the Gamma function integral and absolute convergence give its Mellin transform representationIndeed, the two equal terms for $n$ and $-n$ cancel the factor $1/2$, and the $n$th integral is ${\pi }^{-s/2}\Gamma (s/2)n^{-s}$.

Split at $u=1$ and use the theta transformation in the integral over $(0,1)$. Its elementary part isIn the remaining part substitute $v=1/u$. DefineThe pole-subtracted theta integral for the completed zeta function is thereforeSince $\Theta (u)-1=O(e^{-\pi u})$ for $u\ge 1$, both integrals and every derivative in $s$ converge uniformly on compact subsets of $\mathbb{C}$. Thus $J$ is an entire function. The boxed identity supplies the meromorphic continuation of $\Xi$ to the whole plane, with no other singularities. Its poles really are simple, because the residues of the displayed rational terms are nonzero:The identity also gives $\Xi (s)=\Xi (1-s)$, the completed functional equation of the Riemann zeta function.

The constant Fourier coefficient is $A_0(y,s)={\int }_0^{1}G(x+iy,s)dx$. The $m=0$ terms give $2\zeta (2s)y^s$. For fixed $m\ne 0$, unfolding the sum over $n$ givesIndeed, the substitution $u=mx+n$ introduces a factor $1/|m|$, while the translated intervals of length $|m|$ cover the real line exactly $|m|$ times, cancelling that factor. Now scale $u=|m|yt$. The beta function integral yieldswhich can be checked by inserting $(1+t^2{)}^{-s}=\Gamma (s{)}^{-1}{\int }_0^{\infty }{v}^{s-1}{e}^{-v(1+t^2)}dv$ and evaluating the inner Gaussian integral. Thus the fixed-$m$ contribution is $|m{|}^{1-2s}{y}^{1-s}\sqrt{\pi }\Gamma (s-1/2)/\Gamma (s)$. Summing over positive and negative $m$ proves the constant term of a nonholomorphic Eisenstein seriesWith the printed normalization $\xi (u)={\pi }^{-u/2}\Gamma (u/2)\zeta (u)$ of the completed Riemann zeta function, this is exactlyBoth the unfolding and the original series calculation are justified for $\mathrm{Re}s>1$. Beyond this region the identities are understood meromorphically: $G(\tau ,s)$ is the Epstein zeta function of the unit-covolume lattice $(\mathbb{Z}\tau +\mathbb{Z})/\sqrt{y}$, so question 4 supplies its continuation. The symbol $\xi$ here has the completed-zeta normalization displayed above, which has poles at zero and one; it does not include the extra polynomial factor sometimes used to define an entire xi function.