Prerequisites for this post: previous post, and complex analysis. For this entire post, is a complex variable with .

## 1. The function

So there’s this thing called the Gamma function. Denoted , it is defined by

as long as . Here are its values at the first few integers:

Yep: the function is contrived so that the identity

always holds. (Proof: integration by parts.) Thus interpolates the factorial function a nice way. Moreover, this identity lets us extend to a meromorphic function on the entire complex plane.

We like the function because, unlike , we know what it’s doing. For example we actually know how to compute for any positive integer . (Contrast this with, say, ; we don’t even know if it’s an algebraic number, and it took until 1978 to prove that it was irrational). More to the point, we also know where all its zeros and poles are:

**Proposition 1**

Let be defined as above, extended to a meromorphic function on all of .

- The function has no zeros.
- It has a simple pole at , and these are the only poles.

The pole at should not be surprising: plug in to .

In any case, moral of story: is very friendly!

## 2. Functional Equation for Zeta

We will now do something really magical, due to Riemann. Pick an integer ; we set to derive the artificial equation

Replacing with and rearranging gives

Then, due to absolute convergence we can sum over ; this brings the Riemann zeta function into the right-hand side, to get

where

so that gives just the sum for . It turns out that this is special: a Jacobi theta function, which happens to satisfy

Also, for .

Using this and with some purely algebraic manipulations (namely, splitting the integral into two parts, and , and then using the property of the theta function), one can derive that

The right-hand side has two very useful properties:

- It is even around , meaning it remains the same when is replaced by .
- The integral is actually an entire function on all of (the integral converges for all because ).

So if we multiply both sides by , we get a symmetric entire function, called the function:

**Theorem 2**

The function

satisfies

and is an entire function on all of .

In particular we can use this to extend to a meromorphic function on the entire complex plane. We have

We can count zeros and poles from this. The ‘s don’t do anything. The ‘s have no zeros and just poles at non-positive integers.

Since has no zeros and no poles other than for , the only zeros it will have are for are at in order to cancel out the poles of . And so these are the so-called trivial zeros.

On the other hand in the strip we get very confused. More on that later.

## 3. Explicit Formula for Chebyshev Function

Recall that last post we had

according to Perron’s Formula. Write this as

Recall that if is a meromorphic function then so is

which has a simple pole for each zero/pole of ; the residue is for each simple zero (more generally for a zero of multiplicity ) and for each simple pole (more generally for a pole of order ). (See the argument principle).

Now we are going to do use the bestest theorem from complex analysis: Cauchy’s Residue Theorem. We replace the path with . Doing so picks up the following residues:

- Since has a pole at , has a simple pole there of residue , so we pick up a residue of corresponding to
This is the “main term”.

- has simple zeros at . So we pick up residues of , , and so on, or
- itself has a pole at , which gives us an additional term
It turns out that this equals , because why not.

- Finally, the hard-to-understand zeros in the strip . If is a zero, then it contributes a residue of . We only pick up the zeros with in our rectangle, so we get a term

In what follows, always denotes a zero in the critical strip, written .

Putting this all together, we obtain that

where is the integral along the three sides of the rectangle. With some effort, you can show that in fact ; but the speed of convergence depends on . To avoid this, we can take but leave intact, which gives the explicit formula:

**Theorem 3** **(Explicit Formula for the Chebyshev Function)**

For

where is some error term, with the property that as for any fixed .

(It’s necessary to use a truncated sum since the series actually converges only conditionally, and not absolutely.)

The error term is ugly enough that I haven’t included it in the formula, but if you want to know I’ll at least write it down: it is

where is the distance from to the nearest prime power (). Clearly, for any fixed , as . What I will say is that it’s not our main concern:

## 4. The Prime Number Theorem

Here’s the deal: we know , and want something as close as possible. We get the right off the bat, so we want everything else to be small.

The term is tiny, as is the constant. The can be handled as long as is big enough: the price we pay is that we introduce more zeros into the sum over . The sum of the zeros: well, what about the sum?

We know that for any zero , we have . But if , we’d be very upset, because now our sum has a term of size in it, which is bad since the thing we’re shooting for is . So what we’d like is to be able to force to be less than something, like . Then we’d have an error term around , which is not spectacular but better than .

In fact, we believe that , always — the Riemann Hypothesis. With some messing around with the value of , we would then get an error term of

which is pretty good, and actually near best possible (one can show that is not achievable).

Unfortunately, we can not even show . The most we know is that , which gives some pretty crummy bounds compared to what we think is true: using this bound, the best we can do is

which is worse than . That’s the current state of affairs.

That function is pretty mysterious.

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References: Kedlaya’s 18.785 notes and Hildebrand’s ANT notes.

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Seems the Riemann hypothesis could actually be false: https://figshare.com/articles/preprint/Untitled_Item/14776146.

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