Fluctuation of the Primitive of Hardy's Function

Statement of the Subject

The paper by Matti Jutila, published in Hardy-Ramanujan Journal 41 (2018), 32–39, studies the oscillatory and local behavior of the primitive

$$ F(T) = \int_0^T Z(t),\mathrm{d}t $$

of Hardy's function $Z(t) = \chi(\tfrac{1}{2}+it)^{-1/2},\zeta(\tfrac{1}{2}+it)$, where $\chi(s) = 2^s \pi^{s-1}\sin(\tfrac{1}{2}\pi s),\Gamma(1-s)$ is as in the functional equation $\zeta(s) = \chi(s)\zeta(1-s)$. The central objects of study are:

1. The local integrals $I(T_1,T_2) = \int_{T_1}^{T_2} Z(t),\mathrm{d}t = F(T_2) - F(T_1)$, which measure the net mass of $Z$ on an interval.
2. The occurrence of a Gibbs-type phenomenon for $F(t)$ at certain critical points, analogous to the classical Gibbs phenomenon for partial sums of Fourier series, first observed by M. A. Korolev.
3. An asymmetry in the behavior of $F(t)$ near the critical points $t(1/4)$ and $t(3/4)$, traced to the differing large-argument asymptotics of $K$-Bessel and $J$-Bessel functions.

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Prerequisites

Hardy's function and the χ-factor

The functional equation for the Riemann zeta-function is

$$ \zeta(s) = \chi(s),\zeta(1-s), \qquad \chi(s) = 2^s\pi^{s-1}\sin\left(\tfrac{1}{2}\pi s\right),\Gamma(1-s). $$

On the critical line $s = \tfrac{1}{2}+it$, the factor $\chi(\tfrac{1}{2}+it)$ has modulus 1, so $|\zeta(\tfrac{1}{2}+it)| = |Z(t)|$. Hardy's function $Z(t)$ is real-valued for real $t$, and its sign changes are the zeros of $\zeta$ on the critical line.

Korolev's bound

The sharpest global estimate for $F(T)$ is $F(T) = O(T^{1/4})$, due to M. A. Korolev. This is the best possible order. It says that the positive and negative values of $Z(t)$ nearly cancel on $[0,T]$, with remainder of order $T^{1/4}$.

Atkinson-type formula

Jutila proved in [Jut09] the decomposition

$$ F(T) = S_1(T,L) + S_2(T,L') + O\left((\log T)^{5/4}\right), $$

where $S_1$ and $S_2$ are certain finite sums of lengths $L, L' \simeq \sqrt{T}$. This is an analogue for $F(T)$ of Atkinson's formula for the error term in the mean-square of $\zeta$.

Step function K(x)

$$ K(x) = \begin{cases} 0 & \text{for } 0 \le x < 1/4 \text{ and } 3/4 < x \le 1, \ 2\pi & \text{for } 1/4 < x < 3/4. \end{cases} $$

This piecewise-constant function is the leading-order profile of $F(t)$ as $t$ varies through one "period" of the lattice structure determined by $\sqrt{T/2\pi}$.

Parametrization by ϑ

Write $\sqrt{T/2\pi} = P + \vartheta$ with $P \in \mathbb{N}$ and $0 \le \vartheta < 1$. Define $t(\vartheta) = 2\pi(P + \vartheta)^2$, so $\sqrt{t(\vartheta)/2\pi} = P + \vartheta$. Fix $T = t(0) = 2\pi P^2$. The paper studies $F(t(\vartheta))$ as a function of $\vartheta \in [0,1)$.

Van der Corput's lemma (Titchmarsh, Lemma 4.8)

If $f$ is a real-valued $C^1$-function on $[a,b]$ with $|f'(x)| \le \theta < 1$ for all $x \in [a,b]$, then

$$ \sum_{a \le n \le b} e^{2\pi i f(n)} = \int_a^b e^{2\pi i f(x)},\mathrm{d}x + O(1). $$

This converts sums over integers into integrals, with bounded error, when the phase has small derivative.

Airy function

$$ \text{Ai}(z) = \int_0^\infty \cos\left(\frac{x^3}{3} + xz\right)\mathrm{d}x. $$

This is one of the standard representations of the Airy function. It satisfies $\text{Ai}''(z) = z,\text{Ai}(z)$. For large positive $z$, $\text{Ai}(z)$ decays exponentially; for large negative $z$, it oscillates with slowly decaying amplitude.

Bessel functions $J_{\pm 1/3}$ and $K_{1/3}$

For $0 < x \ll 1$:

$$ J_{\pm 1/3}(x) = \frac{x^{\pm 1/3}}{2^{\pm 1/3},\Gamma(1 \pm 1/3)}\left(1 + O(x)\right), \qquad K_{1/3}(x) = 2^{-2/3},\Gamma(1/3),x^{-1/3}\left(1 + O(x)\right). $$

For $x \gg 1$:

$$ K_{1/3}(x) = \sqrt{\frac{\pi}{2x}},e^{-x}\left(1 + O(x^{-1})\right). $$

The $J$-Bessel functions oscillate for large argument; the $K$-Bessel function decays exponentially. This dichotomy is the source of the asymmetry in sec. 6 of the paper.

The Gibbs phenomenon (classical)

For a function with a jump discontinuity, partial sums of its Fourier series overshoot the discontinuity by a fixed proportion (approximately 8.95%) that does not vanish as the number of terms increases. Jutila's paper demonstrates an analogous overshoot for $F(t)$ at the points where the step function $K(\vartheta)$ jumps.

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Complete Exegesis

1. The global formula for F(T): equation (1.2)

Let $\vartheta_0 = \min(|\vartheta - 1/4|, |\vartheta - 3/4|)$. For $\vartheta_0 \neq 0$:

$$ F(T) = (T/2\pi)^{1/4}(-1)^P,K(\vartheta) + O\left(T^{1/6}\log T\right) + O\left(\min\left(T^{1/4},, T^{1/8}\vartheta_0^{-3/4}\right)\right). $$

This says: away from the critical fractional parts $\vartheta = 1/4$ and $\vartheta = 3/4$, the primitive is well-approximated by the step function $K$ scaled by $(T/2\pi)^{1/4}(-1)^P$. The second error term blows up as $\vartheta_0 \to 0$, reflecting the discontinuity of $K$ versus the continuity of $F$.

At the exceptional values $\vartheta = j/4$ with $j = 1$ or $3$:

$$ F(T) = (T/2\pi)^{1/4}(-1)^P\left(\pi + (-1)^{(j-1)/2}(\pi/3)\right) + O(T^{1/6}\log T). $$

For $j = 1$: the value is $\pi + \pi/3 = 4\pi/3$, multiplied by the prefactor. For $j = 3$: the value is $\pi - \pi/3 = 2\pi/3$, multiplied by the prefactor. These are the midpoint values of the step function's jump, analogous to the Cesàro value of a Fourier series at a discontinuity, but with a correction from the cubic phase.

2. The approximate functional equation: equation (1.5)

For $T \le t \le T + 2\sqrt{2\pi T} + 2\pi$ (i.e., one "period" of the lattice):

$$ F(t) = S_1(t, L_0) + O(T^c), $$

where $L_0 \simeq T^{1/4}$ and $c = 27/164 < 1/6$. This is a shorter sum than in (1.1): only $T^{1/4}$ terms instead of $T^{1/2}$. The second sum $S_2$ is absorbed into the error. This comes from a generalized approximate functional equation proved in [Jut17], where the splitting parameter is allowed to vary.

3. Simplification to S(ϑ): equations (2.6)–(2.8)

The simplified form of $S_1(t(\vartheta), L_0)$ retains only the leading-order terms of the coefficients:

$$ S_1(t(\vartheta), L_0) = 2\sqrt{2},(T/2\pi)^{1/4}(-1)^P,S(\vartheta) + O(1), $$

where

$$ S(\vartheta) = \sum_{0 \le n \le L_0} (n+\tfrac{1}{2})^{-1},(-1)^{n(n+1)/2},\sin\left(2\pi(n+\tfrac{1}{2})\vartheta + A(n+\tfrac{1}{2})^3\right), $$

and $A = A(T) = \frac{1}{12}\sqrt{2\pi^3},T^{-1/2}$.

The simplification steps are: (i) replace error terms $O((n+\tfrac{1}{2})^2/T)$ by 1; (ii) replace $t$ by $T$ in $A(t)$ and in the $(t/2\pi)^{1/4}$ prefactor; (iii) drop the fifth-order phase correction $O((n+\tfrac{1}{2})^5 T^{-3/2})$. These are all legitimate because $n \le L_0 \simeq T^{1/4}$, so the dropped terms contribute $O(1)$ in total.

Combining (1.5) and (2.6):

$$ F(t(\vartheta)) = 2\sqrt{2},(T/2\pi)^{1/4}(-1)^P,S(\vartheta) + O(T^c). $$

The change in $F$ is then

$$ F(t_2) - F(t_1) = 2\sqrt{2},(T/2\pi)^{1/4}(-1)^P \int_{\vartheta_1}^{\vartheta_2} S'(\vartheta),\mathrm{d}\vartheta + O(T^c). $$

4. The derivative S′(ϑ): equation (2.10)

Differentiating (2.7) term-by-term with respect to $\vartheta$:

$$ S'(\vartheta) = 2\pi \sum_{0 \le n \le L_0} (-1)^{n(n+1)/2},\cos\left(2\pi(n+\tfrac{1}{2})\vartheta + A(n+\tfrac{1}{2})^3\right). $$

The factor $(n+\tfrac{1}{2})^{-1}$ in $S(\vartheta)$ is cancelled by the derivative of the sine, which brings down the factor $2\pi(n+\tfrac{1}{2})$. Term-by-term differentiation is justified because the sum is finite (length $L_0$).

5. Arithmetic of the sign factors: equations (3.11)–(3.14)

Setting $\vartheta = j/4 + \delta$ and expanding the cosine using the angle-addition identity:

$$ \cos\left(2\pi(n+\tfrac{1}{2})\tfrac{j}{4}\right)\cos\left(2\pi(n+\tfrac{1}{2})\delta + A(n+\tfrac{1}{2})^3\right) - \sin\left(2\pi(n+\tfrac{1}{2})\tfrac{j}{4}\right)\sin\left(2\pi(n+\tfrac{1}{2})\delta + A(n+\tfrac{1}{2})^3\right). $$

The key computation: the products

$$ (-1)^{n(n+1)/2}\cos\left(2\pi(n+\tfrac{1}{2})\tfrac{j}{4}\right) \quad \text{and} \quad (-1)^{n(n+1)/2}\sin\left(2\pi(n+\tfrac{1}{2})\tfrac{j}{4}\right) $$

depend only on $n \bmod 4$. Computing for $n = 0,1,2,3$:

• The sequence $(-1)^{n(n+1)/2}$ takes values $1, -1, -1, 1$ (for $n = 0,1,2,3$).
• For $j = 1$: $\cos(2\pi(n+\tfrac{1}{2})/4)$ at half-integers gives $\cos(\pi/4), \cos(3\pi/4), \cos(5\pi/4), \cos(7\pi/4) = \tfrac{1}{\sqrt{2}}(-1, -1, 1, 1)$ after the shift.
• Multiplying entry-by-entry: the products for the cosine factor yield $\tfrac{1}{\sqrt{2}}(1,1,1,1)$ up to the sign $(-1)^{(j-1)/2}$.
• For the sine factor, the products yield $\tfrac{1}{\sqrt{2}}(1,-1,1,-1)$ for both $j = 1$ and $j = 3$.

Therefore:

$$ S'\left(\tfrac{j}{4}+\delta\right) = \sqrt{2},\pi,(-1)^{(j-1)/2}\sum_{0 \le n \le L_0}\cos\left(2\pi(n+\tfrac{1}{2})\delta + A(n+\tfrac{1}{2})^3\right) $$

$$ \quad - \sqrt{2},\pi\sum_{0 \le n \le L_0}(-1)^n\sin\left(2\pi(n+\tfrac{1}{2})\delta + A(n+\tfrac{1}{2})^3\right). $$

The first sum has all signs aligned (the arithmetic of residues modulo 4 produced a constant vector). The second sum has alternating signs via the $(-1)^n$ factor.

6. Euler–Maclaurin / van der Corput replacement: equations (3.16)–(3.17)

Define $f(x) = (x+\tfrac{1}{2})\delta + (A/2\pi)(x+\tfrac{1}{2})^3$. For the constraints $|\delta| \le a$ and $L_0 = bT^{1/4}$ with $a, b$ sufficiently small:

$$ f'(x) = \delta + \tfrac{3A}{2\pi}(x+\tfrac{1}{2})^2. $$

Since $A \simeq T^{-1/2}$ and $x \le L_0 = bT^{1/4}$, we get $f'(x) \le a + \tfrac{3b^2}{2\pi}\cdot\text{const}$, which can be made $\le \theta < 1$ by choosing $a$ and $b$ small. Van der Corput's lemma then gives:

$$ \sum_{0 \le n \le L_0}\cos\left(2\pi(n+\tfrac{1}{2})\delta + A(n+\tfrac{1}{2})^3\right) = \Re\int_0^{L_0} e(f(x)),\mathrm{d}x + O(1). $$

For the alternating sum (3.14): splitting into even and odd $n$ produces two integrals whose difference is $O(1)$ (by the same lemma, since the phase functions for even and odd subsequences are close). Hence:

$$ \sum_{0 \le n \le L_0}(-1)^n\sin\left(2\pi(n+\tfrac{1}{2})\delta + A(n+\tfrac{1}{2})^3\right) = O(1). $$

The alternating sign $(-1)^n$ introduces a half-integer shift in the effective frequency, pushing $|f'|$ away from zero and making the sum bounded.

7. Extension to infinity and the Airy integral B(u): equations (3.18)–(3.22)

Extending the integral from $[0, L_0]$ to $[0, \infty)$ is justified by integration by parts: for $x \ge L_0$, $|f'(x)| \gg 1$ (the cubic term dominates), provided $a$ is small relative to $b^2$. The tail integral is $O(L_0^{-2}T^{1/2}) = O(1)$.

Combining (3.13), (3.14), (3.16), (3.17):

$$ S'\left(\tfrac{j}{4}+\delta\right) = \sqrt{2},\pi,(-1)^{(j-1)/2},B(\delta) + O(1), $$

where

$$ B(u) = \int_0^\infty \cos(Ax^3 + 2\pi u x),\mathrm{d}x. $$

This integral is a scaled Airy function. The substitution $x = (3/A)^{1/3}y$ transforms the cubic into $y^3/3$, yielding:

$$ B(u) = \sqrt{2/\pi},\text{Ai}\left(2\sqrt{2\pi},T^{1/6}u\right),T^{1/6}. $$

Alternatively, via Watson's formula (p. 190 of [Wat66]):

For $u > 0$:

$$ B(u) = \frac{\sqrt{2\pi u}}{3\sqrt{A}},K_{1/3}\left(\frac{2(2\pi u)^{3/2}}{3\sqrt{3A}}\right). $$

For $u < 0$:

$$ B(u) = \frac{\pi\sqrt{2\pi|u|}}{3\sqrt{3A}}\left(J_{1/3}\left(\frac{2(2\pi|u|)^{3/2}}{3\sqrt{3A}}\right) + J_{-1/3}\left(\frac{2(2\pi|u|)^{3/2}}{3\sqrt{3A}}\right)\right). $$

At $u = 0$:

$$ B(0) = \sqrt{2\pi},3^{-2/3},\Gamma(2/3)^{-1},T^{1/6}. $$

This follows from the known value $\text{Ai}(0) = 3^{-2/3}/\Gamma(2/3)$ and the relation (3.22).

8. Theorem 1: the variation formula

Combining (2.9) and (3.18):

Theorem 1. Define $t_{i,j} = t(j/4 + \delta_i)$ for $i = 1,2$ and $j = 1,3$. Suppose $|\delta_i| \le a$ with $a$ sufficiently small. Then:

$$ F(t_{2,j}) - F(t_{1,j}) = 4\pi(T/2\pi)^{1/4}(-1)^P(-1)^{(j-1)/2}\int_{\delta_1}^{\delta_2} B(u),\mathrm{d}u + O(T^c) + O(|\delta_2-\delta_1|,T^{1/4}). $$

The coefficient $4\pi$ arises as $2\sqrt{2}\cdot\sqrt{2}\pi = 4\pi$. The last error term comes from the $O(1)$ in (3.18), integrated over $[\delta_1, \delta_2]$ and multiplied by the prefactor $2\sqrt{2}(T/2\pi)^{1/4}$.

9. Theorem 2: controlled bias of Z(t) near the critical points

Theorem 2. Let $t_1 < t_2$ with both $t_i$ in the interval $[t(j/4) - dT^{1/3},, t(j/4) + dT^{1/3}]$ for $j = 1$ or $3$ and $d$ sufficiently small. Suppose $U = t_2 - t_1 \gg c_4 T^{c+1/12}$. Then:

$$ c_5,U,T^{-1/12} \le (-1)^P(-1)^{(j-1)/2},I(t_1,t_2) \le c_6,U,T^{-1/12}. $$

Proof mechanism. The constraint $|t_i - t(j/4)| \le dT^{1/3}$ translates to $|\delta_i| \le d_1 T^{-1/6}$ (since $t(\vartheta) = 2\pi(P+\vartheta)^2$ and the derivative is $\approx 4\pi P \simeq T^{1/2}$). Then $\delta_2 - \delta_1 \simeq T^{-1/2}U$. The arguments of the Bessel functions in (3.20)–(3.21) are $O(1)$ (small), so the small-argument expansions apply, giving $B(u) \simeq T^{1/6}$ (positive). Hence the leading term in (4.24) is $\simeq T^{1/4}\cdot T^{1/6}\cdot T^{-1/2}U = T^{-1/12}U$. This dominates the error terms when $c_4$ is sufficiently large. The positivity of $B(u)$ in this range is essential: it ensures $I(t_1,t_2)$ has a definite sign.

The mean bias of $Z(t)$ over $[t_1,t_2]$ is therefore

$$ \frac{I(t_1,t_2)}{t_2 - t_1} \simeq T^{-1/12}, $$

confirming that Hardy's function has a persistent (though small) directional bias in these intervals.

10. The Gibbs phenomenon for F(t): Section 5

Let $u_0 < 0$ with $|u_0|$ minimal such that $B(u_0) = 0$. By (3.22), this corresponds to $v_0 = 2\sqrt{2\pi},T^{1/6}u_0$, where $v_0$ is the first negative zero of the Airy function. Since $v_0 \approx -2.338$ (a numerical constant), $|u_0| \simeq T^{-1/6}$.

The integral from $u_0$ to 0:

$$ \int_{u_0}^0 B(u),\mathrm{d}u = \frac{1}{3} + c_0, $$

where $c_0 > 0$. This is established as follows: by (3.19) and a known identity (eq. (2.13) in [Jut11]),

$$ \int_{-\infty}^0 B(u),\mathrm{d}u = \frac{1}{3}. $$

Therefore

$$ \int_{u_0}^0 B(u),\mathrm{d}u = \frac{1}{3} - \int_{-\infty}^{u_0} B(u),\mathrm{d}u. $$

The integral $\int_{-\infty}^{u_0}\text{Ai}(v),\mathrm{d}v$ is negative (the Airy function oscillates for negative argument, and the integral from $-\infty$ to the first zero undershoots zero). Hence subtracting a negative quantity from $1/3$ gives $1/3 + c_0$ with $c_0 > 0$.

The computation of the overshoot: write

$$ F(t(j/4+u_0)) = F(t(j/4)) - \left(F(t(j/4)) - F(t(j/4+u_0))\right). $$

The first term is given by (1.3). The second term is given by Theorem 1 with $\delta_1 = u_0$, $\delta_2 = 0$:

$$ F(t(j/4)) - F(t(j/4+u_0)) = 4\pi(T/2\pi)^{1/4}(-1)^P(-1)^{(j-1)/2}\int_{u_0}^0 B(u),\mathrm{d}u + O(T^c). $$

Using (3.23) and combining:

$$ F(t(j/4+u_0)) = \pi(T/2\pi)^{1/4}(-1)^P\left(1 - (-1)^{(j-1)/2}(1+4c_0)\right) + O(T^{1/6}\log T). $$

For $P$ even and $j = 3$:

$$ F(t(3/4+u_0)) = 2\pi(T/2\pi)^{1/4}(1+2c_0) + O(T^{1/6}\log T). $$

Comparing with (1.2): away from the jump, $F$ would be $2\pi(T/2\pi)^{1/4}$. The factor $(1+2c_0)$ is the overshoot — exactly analogous to the classical Gibbs overshoot for Fourier series. Korolev showed numerically that there exist large $T$ with

$$ F(T) = 2\pi(T/2\pi)^{1/4}(1 + 0.277435\ldots + o(1)), $$

identifying $2c_0 \approx 0.277435$.

11. Asymmetry near t(1/4) vs. t(3/4): Section 6

Decompose $F(t) = (T/2\pi)^{1/4}(-1)^P K(\vartheta) + G(t)$, where

$$ G(t) \ll T^{1/6}\log T + \min(T^{1/4},, T^{1/8}\vartheta_0^{-3/4}). $$

The behavior of $G(t)$ depends on the sign of $\delta$ (where $\vartheta = j/4 + \delta$), because $B(u)$ is expressed via $K_{1/3}$ for $u > 0$ and via $J_{\pm 1/3}$ for $u < 0$.

Case δ > 0: Using the exponential decay of $K_{1/3}$ (eq. 6.28):

$$ B(u) = 2^{-5/4}3^{-1/4}\pi^{1/4}A^{-1/4}u^{-1/4}\exp\left(-\frac{2(2\pi u)^{3/2}}{3\sqrt{3A}}\right)\left(1 + O(T^{-1/4}u^{-3/2})\right). $$

The error term in (6.25) is sharpened to

$$ E \ll T^{5/28} + T^{1/8}\vartheta_0^{-3/4}\exp\left(-\frac{2(2\pi\vartheta_0)^{3/2}}{3\sqrt{3A}}\right), $$

in the range $c_7 T^{-1/6} \le \delta \ll T^{-1/14}$. The exponential suppression means the error decays much faster than the estimate (6.26) suggests. The balancing parameter $\delta' = T^{-1/14}$ is chosen to equalize $T^{1/4}\delta' = T^{1/8}(\delta')^{-3/4} = T^{5/28}$.

Case δ < 0: The $J$-Bessel functions oscillate, and so does $B(u)$. Using the Bombieri–Iwaniec asymptotic for the Airy integral:

$$ B(u) = 2^{-1/4}3^{-1/4}\pi^{1/4}A^{-1/4}|u|^{-1/4}\cos\left(-\frac{2(2\pi|u|)^{3/2}}{3\sqrt{3A}} + \pi/4\right) + O(D^{-1}) $$

for $|u| \simeq D$. By choosing $\delta_i$ to lie on a single arch of this cosine, with $|\delta_2 - \delta_1| \simeq D^{-1/2}T^{-1/4}$ and $|B(u)| \simeq D^{-1/4}T^{1/8}$ throughout, Theorem 1 gives

$$ |F(t_{2,j}) - F(t_{1,j})| \simeq T^{1/8}D^{-3/4}. $$

Since the main terms in (6.25) agree at $t_{1,j}$ and $t_{2,j}$ (both have the same value of $K(\vartheta)$), the full difference must come from $G$, proving $G(t(j/4-\vartheta_0)) \simeq T^{1/8}D^{-3/4}$ for some $\vartheta_0 \simeq D$. This shows (6.26) is best possible for $\delta < 0$.

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Assumptions and Their Roles

Assumption	Where Used	Consequence of Removal
$|\delta| \le a$, $a$ small	Van der Corput application (3.16)	$|f'|$ may exceed 1; sum-to-integral conversion fails
$L_0 = bT^{1/4}$, $b$ small	Same	Same
$a$ small relative to $b^2$	Extending integral to $\infty$	$|f'(x)|$ may not be $\gg 1$ for $x \ge L_0$; tail not negligible
$U \gg c_4 T^{c+1/12}$ in Thm. 2	Ensuring main term dominates error	For shorter intervals, the $O(T^c)$ error may swamp the integral
$c_7 T^{-1/6} \le \vartheta_0 \ll T^{-1/18-\varepsilon}$ in sec. 6	Bessel asymptotics for large argument	Outside this range, either the trivial bound suffices or the $\vartheta_0$-dependence vanishes

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Consequences

Corollary (Persistent bias)

In intervals of length $U \gg T^{c+1/12}$ centered at $t(j/4)$, Hardy's function $Z(t)$ has a net integral of definite sign, with magnitude $\simeq UT^{-1/12}$. This exceeds the "expected" order $U^{1/2}$ by a factor $\simeq T^{1/12}$ when $U \simeq T^{1/3}$ (recovering estimate (1.4)).

Corollary (Gibbs overshoot is quantified)

The overshoot constant $2c_0 \approx 0.277435$ is determined by the integral of the Airy function from its first negative zero to zero. This is a universal constant, independent of $T$.

Connection to mean-value results

The mean-value result $\frac{1}{T}\int_T^{2T}I(t,t+U)^2,\mathrm{d}t \simeq U$ from [Jut15] would suggest $|I|$ is typically $\simeq U^{1/2}$. The present paper shows this heuristic fails because the jumps of $F$ near $t(j/4)$ provide atypically large contributions to the integral $I$. These rare, large jumps inflate the mean square.

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Example: A Concrete Instance

Take $P = 1000$, so $T = 2\pi \cdot 10^6$. Then:

• $A = \frac{1}{12}\sqrt{2\pi^3},T^{-1/2} \approx 1.04 \times 10^{-4}$.
• $L_0 \simeq T^{1/4} \approx 63$.
• The critical points are $t(1/4) = 2\pi(1000.25)^2$ and $t(3/4) = 2\pi(1000.75)^2$.
• The Gibbs overshoot occurs at $t(3/4 + u_0)$ where $|u_0| \simeq T^{-1/6} \approx 0.023$, i.e., about 0.023 in the $\vartheta$-parameter past the jump.
• $F$ overshoots its "plateau" value $2\pi(T/2\pi)^{1/4}$ by a factor $1 + 0.2774\ldots$, reaching approximately $1.2774 \times 2\pi(T/2\pi)^{1/4}$ before oscillating back.
• For $\delta > 0$ (after the jump), the exponential decay in (6.29) causes the error $G(t)$ to be vastly smaller than the worst-case bound.
• For $\delta < 0$ (before the jump), the oscillatory $J$-Bessel behavior means the bound (6.26) is tight, and fluctuations of order $T^{1/8}D^{-3/4}$ genuinely occur.

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Demythologization

The paper's central mechanism is straightforward once unpacked. The primitive $F(t)$ is well-approximated by a step function in the parameter $\vartheta$. Step functions have jumps. The natural question is: how does the continuous function $F$ negotiate these jumps?

The answer is given entirely by the Airy function. The sum $S'(\vartheta)$, after arithmetic simplification modulo 4 and sum-to-integral conversion, collapses to a single Airy integral $B(\delta)$. The Airy function is the universal transition profile for integrals with a cubic-phase stationary point — it governs how the sum transitions from constructive to destructive interference as $\delta$ passes through zero.

The asymmetry ($\delta > 0$ vs. $\delta < 0$) is not mysterious either: it reflects the elementary fact that the $K$-Bessel function decays exponentially while the $J$-Bessel functions oscillate. On one side of the jump, interference dies out exponentially fast; on the other, it persists as oscillation. This dichotomy is built into the Airy function itself (exponential decay for positive argument, oscillation for negative argument).

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Summary

Jutila's paper establishes the following about the primitive $F(T) = \int_0^T Z(t),\mathrm{d}t$ of Hardy's function:

1. $F(t)$ is approximated to order $O(T^c)$ (with $c < 1/6$) by a sum $S(\vartheta)$ of length $T^{1/4}$, via the approximate functional equation (1.5).
2. The derivative $S'(\vartheta)$ near the critical fractional parts $\vartheta = 1/4, 3/4$ reduces to a scaled Airy integral $B(\delta)$, after: (a) angle-addition and mod-4 arithmetic to disentangle the sign factor $(-1)^{n(n+1)/2}$; (b) van der Corput's lemma to convert the sum to an integral; (c) extension to infinity.
3. The positivity of $B(\delta)$ for small $|\delta|$ implies that $F(t)$ is monotonic through the critical points, yielding Theorem 2: Hardy's function has a definite-sign bias of magnitude $\simeq T^{-1/12}$ in intervals of appropriate length near $t(j/4)$.
4. The integral $\int_{u_0}^0 B(u),\mathrm{d}u = 1/3 + c_0$ with $c_0 > 0$ produces an overshoot at the jump — a Gibbs phenomenon for $F(t)$ — quantified by Korolev as a multiplicative factor $1 + 0.277435\ldots$.
5. The error term in the step-function approximation (1.2) is best possible for $\delta < 0$ (the oscillatory Bessel side) but exponentially improved for $\delta > 0$ (the decaying Bessel side), revealing a fundamental asymmetry in the local structure of $F(t)$.