Prerequisites for this post: previous post, and complex analysis. For this entire post, $s$ is a complex variable with $s = \sigma + it$.

1. The $\Gamma$ function

So there’s this thing called the Gamma function. Denoted $\Gamma(s)$, it is defined by $$\Gamma(s) = \int_0^{\infty} x^{s-1} e^{-x} dx$$ as long as $\sigma > 0$. Here are its values at the first few integers:

$$\begin{aligned} \Gamma(1) &= 1 \\ \Gamma(2) &= 1 \\ \Gamma(3) &= 2 \\ \Gamma(4) &= 6 \\ \Gamma(5) &= 24. \end{aligned}$$

Yep: the $\Gamma$ function is contrived so that the identity $$\Gamma(s+1) = s\Gamma(s)$$ always holds. (Proof: integration by parts.) Thus $\Gamma$ interpolates the factorial function a nice way. Moreover, this identity lets us extend $\Gamma$ to a meromorphic function on the entire complex plane.

We like the $\Gamma$ function because, unlike $\zeta$, we know what it’s doing. For example we actually know how to compute $\Gamma(n)$ for any positive integer $n$. (Contrast this with, say, $\zeta(3)$; 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 $\Gamma$ be defined as above, extended to a meromorphic function on all of $\mathbb C$.

• The function $\Gamma$ has no zeros.
• It has a simple pole at $s= 0, -1, -2, \dots$, and these are the only poles.

The pole at $s=0$ should not be surprising: plug in $s=0$ to $\Gamma(s+1)=s\Gamma(s)$.

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

2. Functional Equation for Zeta

We will now do something really magical, due to Riemann. Pick an integer $n$; we set $t = \pi n^2x$ to derive the artificial equation $$\Gamma(s) = (n^2\pi)^{s} \int_0^{\infty} x^{s-1} e^{-n^2\pi x} dx.$$ Replacing $s$ with $\frac{1}{2} s$ and rearranging gives $$\pi^{-s/2} \cdot \Gamma(s/2) n^{-s} = \int_0^\infty x^{s/2 - 1} e^{-n^2\pi x} dx \qquad \sigma > 0.$$ Then, due to absolute convergence we can sum over $n \ge 1$; this brings the Riemann zeta function into the right-hand side, to get

$$\pi^{-s/2} \cdot \Gamma(s/2) \zeta(s) = \int_0^\infty x^{s/2 - 1} \frac{\theta(x)-1}{2} dx \qquad \sigma > 1$$

where $$\theta(x) = \sum_{n=-\infty}^{\infty} e^{-n^2\pi x};$$ so that $\frac{\theta(x)-1}{2}$ gives just the sum for $n \ge 1$. It turns out that this $\theta$ is special: a Jacobi theta function, which happens to satisfy $$\theta(x^{-1}) = \sqrt{x} \cdot \theta(x) \quad \forall x > 0.$$ Also, $\omega(x) = O(e^{-\pi x})$ for $x > 0$.

Using this and with some purely algebraic manipulations (namely, splitting the integral into two parts, $0 \le x \le 1$ and $x \ge 1$, and then using the property of the theta function), one can derive that

$$\pi^{-s/2}\Gamma(s/2)\zeta(s) = \frac1{s-1} - \frac 1s + \int_1^\infty \frac{\theta(x)-1}{2} \cdot \left( x^{s/2-1} + x^{(1-s)/2-1} \right) dx.$$

The right-hand side has two very useful properties:

• It is even around $\frac{1}{2}$, meaning it remains the same when $s$ is replaced by $1-s$.
• The integral is actually an entire function on all of $\mathbb C$ (the integral converges for all $s$ because $\theta(x) = O(e^{-\pi x})$).

So if we multiply both sides by $\frac{1}{2} s(s-1)$, we get a symmetric entire function, called the $\xi$ function:

Theorem 2. The function $$\xi(s) = \frac{1}{2} s(s-1) \pi^{-s/2} \Gamma(s/2) \zeta(s)$$ satisfies $$\xi(s) = \xi(1-s)$$ and is an entire function on all of $\mathbb C$.

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

$$\pi^{-\frac{1}{2} s} \Gamma\left( \frac s2 \right) \zeta(s) = \pi^{-\frac{1}{2} (1-s)} \Gamma\left( \frac{1-s}{2} \right) \zeta(1-s).$$

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

Since $\zeta$ has no zeros and no poles other than $s = 1$ for $\sigma > 0$, the only zeros it will have are for $\sigma \le 0$ are at $s = -2, -4, -6, \dots$ in order to cancel out the poles of $\Gamma(s/2)$. And so these are the so-called trivial zeros.

On the other hand in the strip $0 < \sigma < 1$ we get very confused. More on that later.

3. Explicit Formula for Chebyshev Function

Recall that last post we had $$\psi(x) = \frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} -\frac{\zeta'(s)}{\zeta(s)} \cdot \frac{x^s}{s} ds$$ according to Perron’s Formula. Write this as

$$\psi(x) = \frac{1}{2\pi i} \lim_{T \rightarrow \infty} \left( \int_{c-iT}^{c+iT} \frac{\zeta'(s)}{\zeta(s)} \cdot -\frac{x^s}{s} ds \right)$$

Recall that if $f$ is a meromorphic function then so is $$\frac{1}{2\pi i} \cdot \frac{f'}{f}$$ which has a simple pole for each zero/pole of $f$; the residue is $+1$ for each simple zero (more generally $+m$ for a zero of multiplicity $m$) and $-1$ for each simple pole (more generally $-m$ for a pole of order $m$). (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 $c-iT \rightarrow c+iT$ with $c-iT \rightarrow -U-iT \rightarrow -U + iT \rightarrow c+iT$. Doing so picks up the following residues:

• Since $\zeta$ has a pole at $s=1$, $\zeta'/\zeta$ has a simple pole there of residue $-1$, so we pick up a residue of $+x$ corresponding to

  $$(-1) \cdot -x^1/1 = +x.$$ This is the “main term”.

• $\zeta$ has simple zeros at $s=-2,-4,-6,\dots$. So we pick up residues of $-\frac{x^2}{2}$, $-\frac{x^4}{4}$, and so on, or

  $$- \sum_{n \ge 1} \frac{x^{2n}}{2n} = -\frac{1}{2} \log(1-x^{-2}).$$

• $x^s/s$ itself has a pole at $s=0$, which gives us an additional term

  $$-\frac{\zeta'(0)}{\zeta(0)}.$$ It turns out that this equals $\log(2\pi)$, because why not.

• Finally, the hard-to-understand zeros in the strip $0 < \sigma < 1$. If $\rho = \beta+i\gamma$ is a zero, then it contributes a residue of $-\frac{x^\rho}{\rho}$. We only pick up the zeros with $\left\lvert \gamma \right\rvert < T$ in our rectangle, so we get a term

  $$\sum_{\substack{\rho, \left\lvert \gamma \right\rvert < T}} \frac{x^\rho}{\rho}$$ In what follows, $\rho$ always denotes a zero in the critical strip, written $\rho = \beta + i \gamma$.

Putting this all together, we obtain that

$$\psi(x) = \frac{1}{2\pi i} \lim_{U \rightarrow \infty} \lim_{T \rightarrow \infty} \left( x - \frac{1}{2} \log(1-x^{-2}) - \log(2\pi) - \sum_{\substack{\rho, \left\lvert \gamma \right\rvert < T}} \frac{x^\rho}{\rho} + I(x,U,T) \right)$$

where $I(x,U,T)$ is the integral along the three sides of the rectangle. With some effort, you can show that in fact $I(x,U,T) \rightarrow 0$; but the speed of convergence depends on $x$. To avoid this, we can take $U \rightarrow \infty$ but leave $T$ intact, which gives the explicit formula:

Theorem 3 (Explicit Formula for the Chebyshev Function)

For $x \ge 2$

$$\psi(x) = x - \sum_{\rho, \left\lvert \gamma \right\rvert < T} \frac{x^\rho}{\rho} - \log(2\pi) - \frac{1}{2}\log(1-x^{-2}) + R(x,T)$$

where $R(x,T)$ is some error term, with the property that $R(x,T) \rightarrow 0$ as $T \rightarrow \infty$ for any fixed $x$.

(It’s necessary to use a truncated sum since the series $\sum_{\rho} \frac{1}{\left\lvert \rho \right\rvert}$ 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

$$R(x,T) = O\left( \frac{x \log(xT)^2}{T} + \log x \min \left\{ 1, \frac{x}{T\left<x\right>} \right\} \right)$$

where $\left<x\right>$ is the distance from $x$ to the nearest prime power ($\ne x$). Clearly, for any fixed $x$, $R(x,T) \rightarrow 0$ as $T \rightarrow \infty$. What I will say is that it’s not our main concern:

4. The Prime Number Theorem

Here’s the deal: we know $\psi(x) \approx x$, and want something as close as possible. We get the $x$ right off the bat, so we want everything else to be small.

The term $-\frac{1}{2}\log(1-x^{-2})$ is tiny, as is the constant. The $R(x,T)$ can be handled as long as $T$ is big enough: the price we pay is that we introduce more zeros into the sum over $\rho$. The sum of the zeros: well, what about the sum?

We know that for any zero $\rho = \beta + i \gamma$, we have $\beta < 1$. But if $\beta = 1 - \varepsilon$, we’d be very upset, because now our sum has a term of size $\left\lvert x^\rho \right\rvert = x^\beta = x^{1-\varepsilon}$ in it, which is bad since the thing we’re shooting for is $\psi(x) \approx x$. So what we’d like is to be able to force $\beta$ to be less than something, like $0.9$. Then we’d have an error term around $x^{0.9}$, which is not spectacular but better than $x^{1-\varepsilon}$.

In fact, we believe that $\beta = \frac12$, always – the Riemann Hypothesis. With some messing around with the value of $T$, we would then get an error term of $$O\left( x^{\frac12 + \varepsilon} \right)$$ which is pretty good, and actually near best possible (one can show that $O(\sqrt x)$ is not achievable).

Unfortunately, we can not even show $\beta < 0.999$. The most we know is that $\beta < 1 - \frac{c}{\log \gamma}$, which gives some pretty crummy bounds compared to what we think is true: using this bound, the best we can do is $$\psi(x) = x + O\left( x \exp\left( -c\sqrt{\log x} \right) \right)$$ which is worse than $O(x^{0.999})$. That’s the current state of affairs.

That $\zeta$ function is pretty mysterious.

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