3 Second approach
3.1 Residue calculus on rectangles
This files gathers definitions and basic properties about rectangles.
The border of a rectangle is the union of its four sides.
A Rectangle’s border, given corners \(z\) and \(w\) is the union of the four sides.
A RectangleIntegral of a function \(f\) is one over a rectangle determined by \(z\) and \(w\) in \(\mathbb {C}\). We will sometimes denote it by \(\int _{z}^{w} f\). (There is also a primed version, which is \(1/(2\pi i)\) times the original.)
It is very convenient to define integrals along vertical lines in the complex plane, as follows.
Let \(f\) be a function from \(\mathbb {C}\) to \(\mathbb {C}\), and let \(\sigma \) be a real number. Then we define
We also have a version with a factor of \(1/(2\pi i)\).
If \(f\) is differentiable on a set \(s\) except at \(c\in s\), and \(f\) is bounded above on \(s\setminus \{ c\} \), then there exists a differentiable function \(g\) on \(s\) such that \(f\) and \(g\) agree on \(s\setminus \{ c\} \).
This is the Riemann Removable Singularity Theorem, slightly rephrased from what’s in Mathlib. (We don’t care what the function \(g\) is, just that it’s holomorphic.)
If \(f\) is holomorphic on a rectangle \(z\) and \(w\), then the integral of \(f\) over the rectangle with corners \(z\) and \(w\) is \(0\).
This is in a Mathlib PR.
The next lemma allows to zoom a big rectangle down to a small square, centered at a pole.
If \(f\) is holomorphic on a rectangle \(z\) and \(w\) except at a point \(p\), then the integral of \(f\) over the rectangle with corners \(z\) and \(w\) is the same as the integral of \(f\) over a small square centered at \(p\).
Chop the big rectangle with two vertical cuts and two horizontal cuts into smaller rectangles, the middle one being the desired square. The integral over each of the outer rectangles vanishes, since \(f\) is holomorphic there. (The constant \(c\) being “small enough” here just means that the inner square is strictly contained in the big rectangle.)
The rectangle (square) integral of \(f(s) = 1/s\) with corners \(-1-i\) and \(1+i\) is equal to \(2\pi i\).
This is a special case of the more general result above.
Suppose that \(f\) is a holomorphic function on a rectangle, except for a simple pole at \(p\). By the latter, we mean that there is a function \(g\) holomorphic on the rectangle such that, \(f = g + A/(s-p)\) for some \(A\in \mathbb {C}\). Then the integral of \(f\) over the rectangle is \(A\).
3.2 Perron Formula
In this section, we prove the Perron formula, which plays a key role in our proof of Mellin inversion.
The following is preparatory material used in the proof of the Perron formula, see Lemma 32.
TODO : Move to general section
Let \(a:\mathbb {R}\to \mathbb {C}\) be a function, and let \(\sigma {\gt}0\) be a real number. Suppose that, for all \(\sigma , \sigma '{\gt}0\), we have \(a(\sigma ')=a(\sigma )\), and that \(\lim _{\sigma \to \infty }a(\sigma )=0\). Then \(a(\sigma )=0\).
Let \(a:\mathbb {R}\to \mathbb {C}\) be a function, and let \(\sigma {\lt}-3/2\) be a real number. Suppose that, for all \(\sigma , \sigma '{\gt}0\), we have \(a(\sigma ')=a(\sigma )\), and that \(\lim _{\sigma \to -\infty }a(\sigma )=0\). Then \(a(\sigma )=0\).
Let \(x{\gt}0\) and \(x{\lt}1\). Then
Standard.
Let \(x{\gt}1\). Then
Standard.
Let \(x{\gt}0\). Then the function \(f(s) = x^s/(s(s+1))\) is holomorphic on the half-plane \(\{ s\in \mathbb {C}:\Re (s){\gt}0\} \).
Composition of differentiabilities.
The integral
is positive (and hence convergent - since a divergent integral is zero in Lean, by definition).
This integral is between \(\frac{1}{2}\) and \(1\) of the integral of \(\frac{1}{1+t^2}\), which is \(\pi \).
Let \(x{\gt}0\) and \(\sigma {\gt}1\). Then
Triangle inequality and pointwise estimate.
Let \(x{\gt}1\) and \(\sigma {\lt}-3/2\). Then
Triangle inequality and pointwise estimate.
Let \(x{\gt}0\) and \(\sigma \in \mathbb {R}\). Then
is integrable.
By 25, \(f\) is continuous, so it is integrable on any interval.
Also, \(|f(x)| = \Theta (x^{-2})\) as \(x\to \infty \),
and \(|f(-x)| = \Theta (x^{-2})\) as \(x\to \infty \).
Since \(g(x) = x^{-2}\) is integrable on \([a,\infty )\) for any \(a{\gt}0\), we conclude.
Let \(x{\gt}0\) and \(\sigma ',\sigma ''\in \mathbb {R}\). Then
goes to \(0\) as \(t\to -\infty \).
The numerator is bounded and the denominator tends to infinity.
Let \(x{\gt}0\) and \(\sigma ',\sigma ''\in \mathbb {R}\). Then
goes to \(0\) as \(t\to \infty \).
The numerator is bounded and the denominator tends to infinity.
We are ready for the first case of the Perron formula, namely when \(x{\lt}1\):
For \(x{\gt}0\), \(\sigma {\gt}0\), and \(x{\lt}1\), we have
Let \(f(s) = x^s/(s(s+1))\). Then \(f\) is holomorphic on the half-plane \(\{ s\in \mathbb {C}:\Re (s){\gt}0\} \). The rectangle integral of \(f\) with corners \(\sigma -iT\) and \(\sigma +iT\) is zero. The limit of this rectangle integral as \(T\to \infty \) is \(\int _{(\sigma ')}-\int _{(\sigma )}\). Therefore, \(\int _{(\sigma ')}=\int _{(\sigma )}\).
But we also have the bound \(\int _{(\sigma ')} \leq x^{\sigma '} * C\), where
\(C=\int _\mathbb {R}\frac{1}{|(1+t)(1+t+1)|}dt\).
Therefore \(\int _{(\sigma ')}\to 0\) as \(\sigma '\to \infty \).
The second case is when \(x{\gt}1\). Here are some auxiliary lemmata for the second case. TODO: Move to more general section
Let \(x\in \mathbb {R}\) and \(s \ne 0, -1\). Then
By ring.
Let \(x{\gt}0\). Then for \(0 {\lt} c {\lt} 1 /2\), we have that the function
is bounded above on the rectangle with corners at \(-c-i*c\) and \(c+i*c\) (except at \(s=0\)).
Applying Lemma 33, the function \(s ↦ x^s/s(s+1) - 1/s = x^s/s - x^0/s - x^s/(1+s)\). The last term is bounded for \(s\) away from \(-1\). The first two terms are the difference quotient of the function \(s ↦ x^s\) at \(0\); since it’s differentiable, the difference remains bounded as \(s\to 0\).
Let \(x{\gt}0\). Then for \(0 {\lt} c {\lt} 1 /2\), we have that the function
is bounded above on the rectangle with corners at \(-1-c-i*c\) and \(-1+c+i*c\) (except at \(s=-1\)).
Applying Lemma 33, the function \(s ↦ x^s/s(s+1) - x^{-1}/(s+1) = x^s/s - x^s/(s+1) - (-x^{-1})/(s+1)\). The first term is bounded for \(s\) away from \(0\). The last two terms are the difference quotient of the function \(s ↦ x^s\) at \(-1\); since it’s differentiable, the difference remains bounded as \(s\to -1\).
Let \(x{\gt}0\). Then for all sufficiently small \(c{\gt}0\), we have that
Let \(x{\gt}0\). Then for all sufficiently small \(c{\gt}0\), we have that
Compute the integral.
For \(x{\gt}1\) (of course \(x{\gt}0\) would suffice) and \(\sigma {\gt}0\), we have
We pull to a square with corners at \(-c-i*c\) and \(c+i*c\) for \(c{\gt}0\) sufficiently small. By Lemma 36, the integral over this square is equal to \(1\).
For \(x{\gt}1\), we have
Pull contour from \((-1/2)\) to \((-3/2)\).
For \(x{\gt}1\) and \(\sigma {\lt}-3/2\), we have
Pull contour from \((-3/2)\) to \((\sigma )\).
For \(x{\gt}1\) and \(\sigma {\gt}0\), we have
Let \(f(s) = x^s/(s(s+1))\). Then \(f\) is holomorphic on \(\mathbb {C}\setminus {0,-1}\).
First pull the contour from \((\sigma )\) to \((-1/2)\), picking up a residue \(1\) at \(s=0\).
Next pull the contour from \((-1/2)\) to \((-3/2)\), picking up a residue \(-1/x\) at \(s=-1\).
Then pull the contour all the way to \((\sigma ')\) with \(\sigma '{\lt}-3/2\).
For \(\sigma ' {\lt} -3/2\), the integral is bounded by \(x^{\sigma '}\int _\mathbb {R}\frac{1}{|(1+t ^2)(2+t ^2)|^{1/2}}dt\).
Therefore \(\int _{(\sigma ')}\to 0\) as \(\sigma '\to \infty \).
The two together give the Perron formula. (Which doesn’t need to be a separate lemma.)
For \(x{\gt}0\) and \(\sigma {\gt}0\), we have
3.3 Mellin transforms
Let \(f, g\) be once differentiable functions from \(\mathbb {R}_{{\gt}0}\) to \(\mathbb {C}\) so that \(fg'\) and \(f'g\) are both integrable, and \(f\cdot g (x)\to 0\) as \(x\to 0^+,\infty \). Then
Partial integration.
In this section, we define the Mellin transform (already in Mathlib, thanks to David Loeffler), prove its inversion formula, and derive a number of important properties of some special functions and bumpfunctions.
Def: (Already in Mathlib) Let \(f\) be a function from \(\mathbb {R}_{{\gt}0}\) to \(\mathbb {C}\). We define the Mellin transform of \(f\) to be the function \(\mathcal{M}(f)\) from \(\mathbb {C}\) to \(\mathbb {C}\) defined by
[Note: My preferred way to think about this is that we are integrating over the multiplicative group \(\mathbb {R}_{{\gt}0}\), multiplying by a (not necessarily unitary!) character \(|\cdot |^s\), and integrating with respect to the invariant Haar measure \(dx/x\). This is very useful in the kinds of calculations carried out below. But may be more difficult to formalize as things now stand. So we might have clunkier calculations, which “magically” turn out just right - of course they’re explained by the aforementioned structure...]
Let \(f\) be a function from \(\mathbb {R}_{{\gt}0}\) to \(\mathbb {C}\). We define the Mellin transform of \(f\) to be the function \(\mathcal{M}(f)\) from \(\mathbb {C}\) to \(\mathbb {C}\) defined by
[Note: already exists in Mathlib, with some good API.]
Let \(F\) be a function from \(\mathbb {C}\) to \(\mathbb {C}\). We define the Mellin inverse transform of \(F\) to be the function \(\mathcal{M}^{-1}(F)\) from \(\mathbb {R}_{{\gt}0}\) to \(\mathbb {C}\) defined by
where \(\sigma \) is sufficiently large (say \(\sigma {\gt}2\)).
Let \(0 {\lt} t {\lt} x\) and \(\sigma {\gt}0\). Then the inverse Mellin transform of the Perron function
is equal to
This is a straightforward calculation.
Let \(0 {\lt} x {\lt} t\) and \(\sigma {\gt}0\). Then the inverse Mellin transform of the Perron function is equal to
This is a straightforward calculation.
Let \(f\) be a twice differentiable function from \(\mathbb {R}_{{\gt}0}\) to \(\mathbb {C}\), and let \(\sigma \) be sufficiently large. Then
The proof is from [Goldfeld-Kontorovich 2012]. Integrate by parts twice (assuming \(f\) is twice differentiable, and all occurring integrals converge absolutely, and boundary terms vanish).
We now have at least quadratic decay in \(s\) of the Mellin transform. Inserting this formula into the inversion formula and Fubini-Tonelli (we now have absolute convergence!) gives:
Apply the Perron formula to the inside:
where we integrated by parts (undoing the first partial integration), and finally applied the fundamental theorem of calculus (undoing the second).
Finally, we need Mellin Convolutions and properties thereof.
Let \(f\) and \(g\) be functions from \(\mathbb {R}_{{\gt}0}\) to \(\mathbb {C}\). Then we define the Mellin convolution of \(f\) and \(g\) to be the function \(f\ast g\) from \(\mathbb {R}_{{\gt}0}\) to \(\mathbb {C}\) defined by
Let us start with a simple property of the Mellin convolution.
Let \(f\) and \(g\) be functions from \(\mathbb {R}_{{\gt}0}\) to \(\mathbb {R}\) or \(\mathbb {C}\), for \(x\neq 0\),
By Definition 6,
in which we change variables to \(z=x/y\):
The Mellin transform of a convolution is the product of the Mellin transforms.
Let \(f\) and \(g\) be functions from \(\mathbb {R}_{{\gt}0}\) to \(\mathbb {C}\) such that
is absolutely integrable on \([0,\infty )^2\). Then
By (??) and Fubini’s theorem,
in which we change variables from \(x\) to \(z=x/y\):
which, by Definition 4, is
Let \(\nu \) be a bumpfunction.
There exists a smooth (once differentiable would be enough), nonnegative “bumpfunction” \(\nu \), supported in \([1/2,2]\) with total mass one:
Same idea as Urysohn-type argument.
The \(\nu \) function has Mellin transform \(\mathcal{M}(\nu )(s)\) which is entire and decays (at least) like \(1/|s|\).
The Mellin transform of \(\nu \) is
as \(|s|\to \infty \) with \(\sigma _1 \le \Re (s) \le 2\).
[Of course it decays faster than any power of \(|s|\), but it turns out that we will just need one power.]
Integrate by parts:
Since \(\Re (s)\) is bounded, the right-hand side is bounded by a constant times \(1/|s|\).
We can make a delta spike out of this bumpfunction, as follows.
Let \(\nu \) be a bumpfunction supported in \([1/2,2]\). Then for any \(\epsilon {\gt}0\), we define the delta spike \(\nu _\epsilon \) to be the function from \(\mathbb {R}_{{\gt}0}\) to \(\mathbb {C}\) defined by
This spike still has mass one:
For any \(\epsilon {\gt}0\), we have
Substitute \(y=x^{1/\epsilon }\), and use the fact that \(\nu \) has mass one, and that \(dx/x\) is Haar measure.
The Mellin transform of the delta spike is easy to compute.
For any \(\epsilon {\gt}0\), the Mellin transform of \(\nu _\epsilon \) is
Substitute \(y=x^{1/\epsilon }\), use Haar measure; direct calculation.
In particular, for \(s=1\), we have that the Mellin transform of \(\nu _\epsilon \) is \(1+O(\epsilon )\).
For any \(\epsilon {\gt}0\), we have
This is immediate from the above theorem.
As \(\epsilon \to 0\), we have
By Lemma 7,
which by Definition 4 is
Since \(\nu (x) x^{\epsilon -1}\) is integrable (because \(\nu \) is continuous and compactly supported),
By Taylor’s theorem,
so, since \(\nu \) is absolutely integrable,
We conclude the proof using Theorem 7.
Let \(1_{(0,1]}\) be the function from \(\mathbb {R}_{{\gt}0}\) to \(\mathbb {C}\) defined by
This has Mellin transform
The Mellin transform of \(1_{(0,1]}\) is
[Note: this already exists in mathlib]
This is a straightforward calculation.
What will be essential for us is properties of the smooth version of \(1_{(0,1]}\), obtained as the Mellin convolution of \(1_{(0,1]}\) with \(\nu _\epsilon \).
Let \(\epsilon {\gt}0\). Then we define the smooth function \(\widetilde{1_{\epsilon }}\) from \(\mathbb {R}_{{\gt}0}\) to \(\mathbb {C}\) by
Let \(c:=2^\epsilon {\gt} 1\), in terms of which we wish to prove
Letting \(f(x):=x\log x - x\), we can rewrite this as \(f(1) {\lt} f(c)\). Since
\(f\) is monotone increasing on [1, ∞), and we are done.
In particular, we have the following two properties.
Fix \(\epsilon {\gt}0\). There is an absolute constant \(c{\gt}0\) so that: If \(0 {\lt} x \leq (1-c\epsilon )\), then
Opening the definition, we have that the Mellin convolution of \(1_{(0,1]}\) with \(\nu _\epsilon \) is
The support of \(\nu _\epsilon \) is contained in \([1/2^\epsilon ,2^\epsilon ]\), so it suffices to consider \(y \in [1/2^\epsilon x,2^\epsilon x]\) for nonzero contributions. If \(x {\lt} 2^{-\epsilon }\), then the integral is the same as that over \((0,\infty )\):
in which we change variables to \(z=x/y\) (using \(x{\gt}0\)):
which is equal to one by Lemma 46. We then choose
which satisfies
by Lemma ??, so
Fix \(0{\lt}\epsilon {\lt}1\). There is an absolute constant \(c{\gt}0\) so that: if \(x\geq (1+c\epsilon )\), then
Again the Mellin convolution is
but now if \(x {\gt} 2^\epsilon \), then the support of \(\nu _\epsilon \) is disjoint from the region of integration, and hence the integral is zero. We choose
By Lemma ??,
so
If \(\nu \) is nonnegative, then \(\widetilde{1_{\epsilon }}(x)\) is nonnegative.
If \(\nu \) is nonnegative and has mass one, then \(\widetilde{1_{\epsilon }}(x)\le 1\), \(\forall x{\gt}0\).
and since \(1_{(0,1]}(y)\le 1\), and all the factors in the integrand are nonnegative,
(because in mathlib the integral of a non-integrable function is \(0\), for the inequality above to be true, we must prove that \(\nu ((x/y)^{\frac1\epsilon })/y\) is integrable; this follows from the computation below). We then change variables to \(z=(x/y)^{\frac1\epsilon }\):
which by Theorem 7 is 1.
Combining the above, we have the following three Main Lemmata of this section on the Mellin transform of \(\widetilde{1_{\epsilon }}\).
Fix \(\epsilon {\gt}0\). Then the Mellin transform of \(\widetilde{1_{\epsilon }}\) is
By Definition 8,
We wish to apply Theorem 6. To do so, we must prove that
is integrable on \([0,\infty )^2\). It is actually easier to do this for the convolution: \(\nu _\epsilon \ast 1_{(0,1]}\), so we use Lemma 45: for \(x\neq 0\),
Now, for \(x=0\), both sides of the equation are 0, so the equation also holds for \(x=0\). Therefore,
Now,
has compact support that is bounded away from \(y=0\) (specifically \(y\in [2^{-\epsilon },2^\epsilon ]\) and \(x\in (0,y]\)), so it is integrable. We can thus apply Theorem 6 and find
Given \(0{\lt}\sigma _1\le \sigma _2\), for any \(s\) such that \(\sigma _1\le \mathcal Re(s)\le \sigma _2\), we have
At \(s=1\), we have
Fix a nonnegative, continuously differentiable function \(F\) on \(\mathbb {R}\) with support in \([1/2,2]\). Then for any \(\epsilon {\gt}0\), the function \(x \mapsto \int _{(0,\infty )} x^{1+it} \widetilde{1_{\epsilon }}(x) dx\) is continuous at any \(y{\gt}0\).
Use Lemma ?? to write \(\widetilde{1_{\epsilon }}(x)\) as an integral over an integral near \(1\), in particular avoiding the singularity at \(0\). The integrand may be bounded by \(2^{\epsilon }\nu _\epsilon (t)\) which is independent of \(x\) and we can use dominated convergence to prove continuity.
3.4 Zeta Bounds
We record here some prelimiaries about the zeta function and general holomorphic functions.
If a function \(f\) is holomorphic in a neighborhood of \(p\) and \(\lim _{s\to p} (s-p)f(s) = A\), then \(f(s) = \frac{A}{s-p} + O(1)\) near \(p\).
The function \((s - p)\cdot f(s)\) bounded, so by Theorem 3, there is a holomorphic function, \(g\), say, so that \((s-p)f(s) = g(s)\) in a neighborhood of \(s=p\), and \(g(p)=A\). Now because \(g\) is holomorphic, near \(s=p\), we have \(g(s)=A+O(s-p)\). Then when you divide by \((s-p)\), you get \(f(s) = A/(s-p) + O(1)\).
The Riemann zeta function \(\zeta (s)\) has a simple pole at \(s=1\) with residue \(1\). In particular, the function
is bounded in a neighborhood of \(s=1\).
From ‘riemannZeta_residue_one‘ (in Mathlib), we know that \((s-1)\zeta (s)\) goes to \(1\) as \(s\to 1\). Now apply Theorem 11. (This can also be done using \(\zeta _0\) below, which is expressed as \(1/(s-1)\) plus things that are holomorphic for \(\Re (s){\gt}0\)...)
If a function \(f\) has a simple pole at a point \(p\) with residue \(A \neq 0\), then \(f\) is nonzero in a punctured neighborhood of \(p\).
We know that \(f(s) = \frac{A}{s-p} + O(1)\) near \(p\), so we can write
The first term is bounded, say by \(M\), and the second term goes to \(\infty \) as \(s \to p\). Therefore, there exists a neighborhood \(V\) of \(p\) such that for all \(s \in V \setminus \{ p\} \), we have \(f(s) \neq 0\).
If \(f\) is holomorphic in a neighborhood of \(p\), and there is a simple pole at \(p\), then \(f'/ f\) has a simple pole at \(p\) with residue \(-1\):
Using Theorem 3, there is a function \(g\) holomorphic near \(p\), for which \(f(s) = A/(s-p) + g(s) = h(s)/ (s-p)\). Here \(h(s):= A + g(s)(s-p)\) which is nonzero in a neighborhood of \(p\) (since \(h\) goes to \(A\) which is nonzero). Then \(f'(s) = (h'(s)(s-p) - h(s))/(s-p)^2\), and we can compute the quotient:
Since \(h\) is nonvanishing near \(p\), this remains bounded in a neighborhood of \(p\).
If \(f\) is bounded above in a punctured neighborhood of \(p\), then \(f\) is \(O(1)\) in that neighborhood.
Elementary...
Let’s also record that if a function \(f\) has a simple pole at \(p\) with residue \(A\), and \(g\) is holomorphic near \(p\), then the residue of \(f \cdot g\) is \(A \cdot g(p)\).
If \(f\) has a simple pole at \(p\) with residue \(A\), and \(g\) is holomorphic near \(p\), then the residue of \(f \cdot g\) at \(p\) is \(A \cdot g(p)\). That is, we assume that
near \(p\), and that \(g\) is holomorphic near \(p\). Then
Elementary calculation.
The first term is \(g(s)(f(s) - \frac{A}{s - p})\), which is bounded near \(p\) by the assumption on \(f\) and the fact that \(g\) is holomorphic near \(p\). The second term is \(A\) times the log derivative of \(g\) at \(p\), which is bounded by the assumption that \(g\) is holomorphic.
As a corollary, the log derivative of the Riemann zeta function has a simple pole at \(s=1\):
The log derivative of the Riemann zeta function \(\zeta (s)\) has a simple pole at \(s=1\) with residue \(-1\):
For any natural \(N\ge 1\), we define
Let \(k \le a {\lt} b\le k+1\), with \(k\) an integer, and let \(\phi \) be continuously differentiable on \([a, b]\). Then
Partial integration.
Let \(a {\lt} b\), and let \(\phi \) be continuously differentiable on \([a, b]\). Then
Apply Lemma 56 in blocks of length \(\le 1\).
Let \(0 {\lt} a {\lt} b\) be natural numbers and \(s\in \mathbb {C}\) with \(s \ne 1\) and \(s \ne 0\). Then
Apply Lemma 57 to the function \(x \mapsto x^{-s}\).
For any \(0 {\lt} a {\lt} b\) and \(s \in \mathbb {C}\) with \(\sigma =\Re (s){\gt}0\),
Apply the triangle inequality
and evaluate the integral.
Let \(N\) be a natural number and \(s\in \mathbb {C}\), \(\Re (s){\gt}1\). Then
Apply Lemma 58 with \(a=N\) and \(b\to \infty \).
For any \(N\ge 1\) and \(s = \sigma + tI \in \mathbb {C}\), \(\sigma {\gt} 0\),
Apply Lemma 59 with \(a=N\) and \(b\to \infty \).
For any \(N\ge 1\) and \(s = \sigma + tI \in \mathbb {C}\), \(\sigma =\in (0,2], 2 {\lt} |t|\),
Apply Lemma 61 and estimate \(|s|\ll |t|\).
Big-Oh version of Lemma 62.
For any \(N\ge 1\) and \(s = \sigma + tI \in \mathbb {C}\), \(\sigma =\in (0,2], 2 {\lt} |t|\),
Apply Lemma 61 and estimate \(|s|\ll |t|\).
For any \(N\ge 1\), the function \(\zeta _0(N,s)\) is holomorphic on \(\{ s\in \mathbb {C}\mid \Re (s){\gt}0 ∧ s \ne 1\} \).
The function \(\zeta _0(N,s)\) is a finite sum of entire functions, plus an integral that’s absolutely convergent on \(\{ s\in \mathbb {C}\mid \Re (s){\gt}0 ∧ s \ne 1\} \) by Lemma 61.
The set \(\{ s\in \mathbb {C}\mid \Re (s){\gt}0 ∧ s \ne 1\} \) is path-connected.
Construct explicit paths from \(2\) to any point, either a line segment or two joined ones.
For \(\Re (s){\gt}0\), \(s\ne 1\), and for any \(N\),
Given \(n ≤ t\) and \(\sigma \) with \(1-A/\log t \le \sigma \), we have that
Use \(|n^{-s}| = n^{-\sigma } = e^{-\sigma \log n} \le \exp (-\left(1-\frac{A}{\log t}\right)\log n) \le n^{-1} e^A\), since \(n\le t\).
For any \(s = \sigma + tI \in \mathbb {C}\), \(1/2 \le \sigma \le 2, 3 {\lt} |t|\) and any \(0 {\lt} A {\lt} 1\) sufficiently small, and \(1-A/\log |t| \le \sigma \), we have
First replace \(\zeta (s)\) by \(\zeta _0(N,s)\) for \(N = \lfloor |t| \rfloor \). We estimate:
, where we used Lemma 67 and Lemma 62. The first term is \(\ll \log |t|\). For the second term, estimate
For any \(s = \sigma + tI \in \mathbb {C}\), \(1/2 \le \sigma \le 2, 3 {\lt} |t|\), and any \(0 {\lt} A {\lt} 1\) sufficiently small, and \(1-A/\log |t| \le \sigma \), we have
Estimate \(|s|= |\sigma + tI|\) by \(|s|\le 2 +|t| \le 2|t|\) (since \(|t|{\gt}3\)). Estimating \(|\left\lfloor x \right\rfloor +1/2-x|\) by \(1\), and using \(|x^{-s-1}| = x^{-\sigma -1}\), we have
For the last integral, integrate by parts, getting:
Now use \(\log N \le \log |t|\) to get the result.
For any \(s = \sigma + tI \in \mathbb {C}\), \(1/2 \le \sigma \le 2, 3 {\lt} |t|\), there is an \(A{\gt}0\) so that for \(1-A/\log t \le \sigma \), we have
First replace \(\zeta (s)\) by \(\zeta _0(N,s)\) for \(N = \lfloor |t| \rfloor \). Differentiating term by term, we get:
Estimate as before, with an extra factor of \(\log |t|\).
As \(\sigma \to 1^+\),
Zeta has a simple pole at \(s=1\). Equivalently, \(\zeta (s)(s-1)\) remains bounded near \(1\). Lots of ways to prove this. Probably the easiest one: use the expression for \(\zeta _0 (N,s)\) with \(N=1\) (the term \(N^{1-s}/(1-s)\) being the only unbounded one).
There exists a \(c{\gt}0\) such that for all \(1 {\lt} \sigma ≤ 2\),
Split into two cases, use Lemma 71 for \(\sigma \) sufficiently small and continuity on a compact interval otherwise.
For all \(\sigma {\gt}1\),
The identity
for \(\sigma {\gt}1\) is already proved by Michael Stoll in the EulerProducts PNT file.
For \(\sigma {\gt}1\) (and \(\sigma \le 2\)),
as \(|t|\to \infty \).
For any \(t\ne 0\) (so we don’t pass through the pole), and \(\sigma _1 {\lt} \sigma _2\),
This is the fundamental theorem of calculus.
For any \(A{\gt}0\) sufficiently small, there is a constant \(C{\gt}0\) so that whenever \(1- A / \log t \le \sigma _1 {\lt} \sigma _2\le 2\) and \(3 {\lt} |t|\), we have that:
For any \(A{\gt}0\) sufficiently small, there is a constant \(C{\gt}0\) so that whenever \(1- A / \log ^9 |t| \le \sigma {\lt} 1\) and \(3 {\lt} |t|\), we have that:
Let \(\sigma \) be given in the prescribed range, and set \(\sigma ' := 1+ A / \log ^9 |t|\). Then
where we used Lemma 74 and Lemma 76. Now by making \(A\) sufficiently small (in particular, something like \(A = 1/16\) should work), we can guarantee that
as desired.
Annoyingly, it is not immediate from this that \(\zeta \) doesn’t vanish there! That’s because \(1/0 = 0\) in Lean. So we give a second proof of the same fact (refactor this later), with a lower bound on \(\zeta \) instead of upper bound on \(1 / \zeta \).
For any \(A{\gt}0\) sufficiently small, there is a constant \(C{\gt}0\) so that whenever \(1- A / \log ^9 |t| \le \sigma {\lt} 1\) and \(3 {\lt} |t|\), we have that:
Follow same argument.
Now we get a zero free region.
There is an \(A{\gt}0\) so that for \(1-A/\log ^9 |t| \le \sigma {\lt} 1\) and \(3 {\lt} |t|\),
Apply Lemma 78.
There is an \(A{\gt}0\) so that for \(1-A/\log ^9 |t| \le \sigma {\lt} 1\) and \(3 {\lt} |t|\),
There is an \(A{\gt}0\) so that for \(1-A/\log ^9 T \le \sigma {\lt} 1\) and \(3 {\lt} |t| ≤ T\),
This Lemma 80, but uniform in \(t\). The point is that the upper bound on \(\zeta ' / \zeta \) and the lower bound on \(\sigma \) only improve as \(|t|\) increases.
The zeta function does not vanish on the 1-line.
This fact is already proved in Stoll’s work.
Then, since \(\zeta \) doesn’t vanish on the 1-line, there is a \(\sigma {\lt}1\) (depending on \(T\)), so that the box \([\sigma ,1] \times _{ℂ} [-T,T]\) is free of zeros of \(\zeta \).
For any \(T{\gt}0\), there is a constant \(\sigma {\lt}1\) so that
for all \(|t| {\lt} T\) and \(\sigma ' \ge \sigma \).
Assume not. Then there is a sequence \(|t_n| \le T\) and \(\sigma _n \to 1\) so that \(\zeta (\sigma _n + it_n) = 0\). By compactness, there is a subsequence \(t_{n_k} \to t_0\) along which \(\zeta (\sigma _{n_k} + it_{n_k}) = 0\). If \(t_0\ne 0\), use the continuity of \(\zeta \) to get that \(\zeta (1 + it_0) = 0\); this is a contradiction. If \(t_0=0\), \(\zeta \) blows up near \(1\), so can’t be zero nearby.
We now prove that there’s an absolute constant \(\sigma _0\) so that \(\zeta '/\zeta \) is holomorphic on a rectangle \([\sigma _2,2] \times _{ℂ} [-3,3] \setminus \{ 1\} \).
There is a \(\sigma _2 {\lt} 1\) so that the function
is holomorphic on \(\{ \sigma _2 \le \Re s \le 2, |\Im s| \le 3 \} \setminus \{ 1\} \).
The derivative of \(\zeta \) is holomorphic away from \(s=1\); the denominator \(\zeta (s)\) is nonzero in this range by Lemma 82.
There is an \(A{\gt}0\) so that for all \(T{\gt}3\), the function \( \frac{\zeta '}{\zeta }(s) \) is holomorphic on \(\{ 1-A/\log ^9 T \le \Re s \le 2, |\Im s|\le T \} \setminus \{ 1\} \).
The derivative of \(\zeta \) is holomorphic away from \(s=1\); the denominator \(\zeta (s)\) is nonzero in this range by Lemma 79.
It would perhaps (?) be better to refactor this entire file so that we’re not using explicit constants but instead systematically using big Oh notation... The punchline would be:
There is an \(A{\gt}0\) so that for \(1-A/\log ^9 |t| \le \sigma {\lt} 1\) and \(|t|\to \infty \),
(Same statement but using big-Oh and filters.)
Same as above.
3.5 Proof of Medium PNT
The approach here is completely standard. We follow the use of \(\mathcal{M}(\widetilde{1_{\epsilon }})\) as in [Kontorovich 2015].
The (second) Chebyshev Psi function is defined as
where \(\Lambda (n)\) is the von Mangoldt function.
It has already been established that zeta doesn’t vanish on the 1 line, and has a pole at \(s=1\) of order 1. We also have the following.
We have that, for \(\Re (s){\gt}1\),
Already in Mathlib.
The main object of study is the following inverse Mellin-type transform, which will turn out to be a smoothed Chebyshev function.
Fix \(\epsilon {\gt}0\), and a bumpfunction supported in \([1/2,2]\). Then we define the smoothed Chebyshev function \(\psi _{\epsilon }\) from \(\mathbb {R}_{{\gt}0}\) to \(\mathbb {C}\) by
where we’ll take \(\sigma = 1 + 1 / \log X\).
Fix a nonnegative, continuously differentiable function \(F\) on \(\mathbb {R}\) with support in \([1/2,2]\), and total mass one, \(\int _{(0,\infty )} F(x)/x dx = 1\). Then for any \(\epsilon {\gt}0\), and \(\sigma \in (1, 2]\), the function
is integrable on \(\mathbb {R}\).
By Lemma 53 the integrand is \(O(1/t^2)\) as \(t\rightarrow \infty \) and hence the function is integrable.
Fix a nonnegative, continuously differentiable function \(F\) on \(\mathbb {R}\) with support in \([1/2,2]\), and total mass one, \(\int _{(0,\infty )} F(x)/x dx = 1\). Then for any \(\epsilon {\gt}0\) and \(\sigma \in (1,2]\), the function \(x \mapsto \sum _{n=1}^\infty \frac{\Lambda (n)}{n^{\sigma +it}} \mathcal{M}(\widetilde{1_{\epsilon }})(\sigma +it) x^{\sigma +it}\) is equal to \(\sum _{n=1}^\infty \int _{(0,\infty )} \frac{\Lambda (n)}{n^{\sigma +it}} \mathcal{M}(\widetilde{1_{\epsilon }})(\sigma +it) x^{\sigma +it}\).
Interchange of summation and integration.
Inserting the Dirichlet series expansion of the log derivative of zeta, we get the following.
We have that
We have that
We have enough decay (thanks to quadratic decay of \(\mathcal{M}(\widetilde{1_{\epsilon }})\)) to justify the interchange of summation and integration. We then get
and apply the Mellin inversion formula (Theorem 5).
The smoothed Chebyshev function is close to the actual Chebyshev function.
We have that
Returning to the definition of \(\psi _{\epsilon }\), fix a large \(T\) to be chosen later, and set \(\sigma _0 = 1 + 1 / log X\), \(\sigma _1 = 1- A/ \log T^9\), and \(\sigma _2{\lt}\sigma _1\) a constant. Pull contours (via rectangles!) to go from \(\sigma _0-i\infty \) up to \(\sigma _0-iT\), then over to \(\sigma _1-iT\), up to \(\sigma _1-3i\), over to \(\sigma _2-3i\), up to \(\sigma _2+3i\), back over to \(\sigma _1+3i\), up to \(\sigma _1+iT\), over to \(\sigma _0+iT\), and finally up to \(\sigma _0+i\infty \).
| | I₉ +-----+ | I₈ | I₇ | | | +-----------+ | I₆ I₅| --σ₂----------σ₁--1-σ₀---- | | I₄ +-----------+ | | I₃| | | I₂ +---+ | | I₁ |
In the process, we will pick up the residue at \(s=1\). We will do this in several stages. Here the interval integrals are defined as follows:
For \(\sigma _0 {\gt} 1\), there exists a constant \(C {\gt} 0\) such that
Write as Dirichlet series and estimate trivially using Theorem 19.
The integrand
is integrable on the contour \(\sigma _0 + t i\) for \(t \in \mathbb {R}\) and \(\sigma _0 {\gt} 1\).
The \(\zeta '(s)/\zeta (s)\) term is bounded, as is \(X^s\), and the smoothing function \(\mathcal{M}(\widetilde{1_{\epsilon }})(s)\) decays like \(1/|s|^2\) by Theorem 53. Actually, we already know that \(\mathcal{M}(\widetilde{1_{\epsilon }})(s)\) is integrable from Theorem 86, so we should just need to bound the rest.
Let \(g : \mathbb {C}\to \mathbb {C}\) be a holomorphic function on a rectangle, then \(g\) is bounded above on the rectangle.
Use the compactness of the rectangle and the fact that holomorphic functions are continuous.
We have that
Pull rectangle contours and evaluate the pole at \(s=1\).
Next pull contours to another box.
We have that
Mimic the proof of Lemma 22.
We insert this information in \(\psi _{\epsilon }\). We add and subtract the integral over the box \([1-\delta ,2] \times _{ℂ} [-T,T]\), which we evaluate as follows
For all \(\epsilon {\gt} 0\) sufficiently close to \(0\), the rectangle integral over \([1-\delta ,2] \times _{ℂ} [-T,T]\) of the integrand in \(\psi _{\epsilon }\) is
where the implicit constant is independent of \(X\).
Unfold the definitions and apply Lemma 54.
It remains to estimate all of the integrals.
This auxiliary lemma is useful for what follows.
Given \(k{\gt}0\), there exists \(C{\gt}0\) so that for all \(T{\gt}3\),
Elementary. Use ‘isLittleO_log_rpow_rpow_atTop‘ in Mathlib.
We have that
Same with \(I_9\).
Unfold the definitions and apply the triangle inequality.
By Theorem 88 (once fixed!!), \(\zeta '/\zeta (\sigma _0 + t i)\) is bounded by \(\zeta '/\zeta (\sigma _0)\), and Theorem 17 gives \(\ll 1/(\sigma _0-1)\) for the latter. This gives:
where we used Theorem 53. Continuing the calculation, we have
where we used that \(\sigma _0=1+1/\log X\), and \(X^{\sigma _0} = X\cdot X^{1/\log X}=e \cdot X\).
We have that
Same with \(I_8\).
Unfold the definitions and apply the triangle inequality.
where we used Theorems 53 and 81, and the fact that \(X^\sigma \le X^{\sigma _0} = X\cdot X^{1/\log X}=e \cdot X\). Since \(T{\gt}3\), we have \(\log T^9 \leq C''' T\).
We have that
Same with \(I_7\).
Unfold the definitions and apply the triangle inequality.
where we used Theorems 53 and 80. Now we estimate \(X^{\sigma _1} = X \cdot X^{-A/ \log T^9}\), and the integral is absolutely bounded.
We have that
Same with \(I_6\).
The analysis of \(I_4\) is similar to that of \(I_2\), (in Lemma 94) but even easier. Let \(C\) be the sup of \(-\zeta '/\zeta \) on the curve \(\sigma _2 + 3 i\) to \(1+ 3i\) (this curve is compact, and away from the pole at \(s=1\)). Apply Theorem 53 to get the bound \(1/(\epsilon |s|^2)\), which is bounded by \(C'/\epsilon \). And \(X^s\) is bounded by \(X^{\sigma _1} = X \cdot X^{-A/ \log T^9}\). Putting these together gives the result.
We have that
Here \(\zeta '/\zeta \) is absolutely bounded on the compact interval \(\sigma _2 + i [-3,3]\), and \(X^s\) is bounded by \(X^{\sigma _2}\). Using Theorem 53 gives the bound \(1/(\epsilon |s|^2)\), which is bounded by \(C'/\epsilon \). Putting these together gives the result.
3.6 MediumPNT
We have
Evaluate the integrals.
3.7 Strong PNT
This section has been removed.