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Translating an integer interval by $xy$ changes a sum of unit-modulus terms by at most $2xy$. Averaging gives the displayed formula. For $n\asymp N$, $Z=N^{2/5}$, expand $-t\log (1+xy/n)$ to degree $r=\lfloor 5.01\log t/\log N\rfloor$. Its remainder is at most $t{N}^{-(r+1)/5}<t^{-1/500}$. This reduces the logarithmic exponential sum to bilinear polynomial sums while keeping explicit boundary and approximation errors.

Cancellation occurs when different complex numbers in an exponential sum have directions that reduce the absolute value of their sum. For example, the orthogonality of roots of unity makes ${\sum }_{a=0}^{p-1}e(a/p)=0$ for every integer $p>1$, despite all terms having absolute value one.

If ${\theta }_r$ have circular spacing at least $\delta$, thenMultiply the exponential sum by $e(-Mt)$, apply the Sobolev–Gallagher inequality on disjoint arcs of length $\delta$, and sum. The finite-interval Parseval identities and Cauchy-Schwarz inequality bound the derivative contribution by $2\pi N\sum |a_n{|}^2$.

For fixed $c,h>0$, $L\ge 1$ and $\epsilon \ge 0$, complete the square in the exponent. A translated Gaussian function summed on a fixed-spaced lattice has mass $O(\sqrt{L})$ uniformly in the translate, by comparison with its integral and its maximum. This keeps a square-root logarithm, rather than the full number of dyadic intervals, when summing exponentially damped exponential sum bounds.

For $s=\sigma +it$, $\sigma >0$, and $x\ge |t|/\pi$, the displayed approximation holds away from the pole. Apply the Van der Corput sum-integral lemma to $-t\log w/(2\pi )$ on $w\ge x$, where its derivative has modulus at most one half. Abel summation with $w^{-\sigma }$ converts the bounded unweighted discrepancy to $O(x^{-\sigma })$. Locally uniform convergence of this weighted discrepancy extends the identity from $\sigma >1$ to $\sigma >0$. It turns estimates for finite exponential sums into estimates for the Riemann zeta function.

The large sieve bounds how much an exponential sum can concentrate at separated points of the circle group. Its variance form of the large sieve also bounds simultaneous concentration in residue classes modulo many primes.

For an integer $m$, direct integration gives zero unless $m=0$, in which case it gives one. Products of this identity on the unit cube turn integrals of finite exponential sums into counts of integer solutions. This is the elementary orthogonality used in the Vinogradov mean value.

We use the Van der Corput sum-integral lemma. Put $e(u)=e^{2\pi iu}$ and $I_h={\int }_a^{b}e(f(x)-hx)dx$. The Fourier series of the periodization of $1_{[a,b]}(x)e(f(x))$ giveswhere integer endpoints have half weight. This is the Dirichlet-Jordan convergence theorem for a piecewise smooth, or more generally bounded-variation, periodic function. Here $f$ is $C^1$, so the periodized function has bounded variation. Changing to the requested endpoint convention costs at most one.

Write $u=f^{\prime }$. For $h\ne 0$, $|u-h|\ge |h|-\delta >0$. Since $u$ is continuous and monotone, the reciprocal has bounded variation, and integration by parts in the Riemann-Stieltjes sense yieldsThe variation of the reciprocal is at most $2\delta /(h^2-{\delta }^2)$. Summing over $h\ne 0$ gives $O((1-\delta {)}^{-1})$, separating $|h|=1$ and using convergence of ${\sum }_{h\ge 2}{h}^{-2}$. For each endpoint, useThe symmetric partial sums of the first term are a constant multiple of ${\sum }_{h=1}^H\sin (2\pi hx)/h$, uniformly bounded in $H$ and $x$; this standard Fourier series bound follows by splitting at $h\asymp 1/\Vert x\Vert$ and applying Abel summation to the remaining sine sum. The second term is absolutely summable with bound $O((1-\delta {)}^{-1})$. The same bound therefore holds for the whole sum of the $h\ne 0$ integrals. Since $I_0$ is the ordinary integral,No second derivative is required; monotonicity supplies the needed variation estimate.

For the Hardy-Littlewood approximation to the Riemann zeta function, take $f(w)=-t\log w/(2\pi )$. On $w\ge x\ge |t|/\pi$, $|f^{\prime }(w)|\le 1/2$ and $f^{\prime }$ is monotone. The proved lemma says that the difference between the partial sum of $w^{-it}$ and its integral over $(x,Y]$ is $O(1)$ uniformly in $Y$. Weighted Abel summation with the decreasing weight $w^{-\sigma }$ then makes the weighted difference $O(x^{-\sigma })$, since its total variation on $[x,\infty )$ is $x^{-\sigma }$. Initially for $\sigma >1$, the tail integral is $x^{1-s}/(s-1)$. The bounded primitive of the discrepancy gives a locally uniformly convergent weighted discrepancy integral for every $\sigma >0$, continuing the identity to that region. Thus, away from the pole,If $t=0$ the ordinary sum-integral comparison supplies the same estimate. At $s=1$ the formula is understood meromorphically. It approximates the Riemann zeta function by a finite Dirichlet polynomial, transfers exponential sum estimates to bounds in the critical strip, yields elementary near-one bounds for $\zeta$ and its derivative, and supports estimates for the mean value of Dirichlet polynomials and numerical calculations.

For large positive $t$ and $0<\sigma \le 1$, a fixed sufficiently large $C$ gives the displayed upper bound; to the right of one use $|\zeta (\sigma +it)|\ll {\log }^{2/3}t$. Its nonlinear dependence on $1-\sigma$ permits a wider zero-free region of the Riemann zeta function through the Landau zero-free-region theorem. Its proof uses estimates for exponential sums; its use as an upper-bound input is separate from a proof of a zero-free region.