I am doing some independent learning on the Riemann Zeta function just out of curiosity. I have searched the site, and I know that there are plenty of questions on the Riemann Zeta function. However, this particular one has not been answered for me nor have I been able to find it easily on google. I teach high school math and adjunct at a local community college in lower level classes such as Trig and College Algebra. I say that so that you would keep in mind the level of mathematics that I have been exposed to and that you will word your answers accordingly (at an elementary level with little sophistication). I understand that it is claimed that the zeta function is only valid for Re(s)>1 which is why negative arguments require analytic continuation. But why is the function as it is written not defined for Re(s)<1? Is this because the series diverges when Re(s)<1? And if so, why may we claim that the function is undefined for those values just because it is divergent for those values? It will really take some convincing for me to believe that we can change a function just because it is divergent at some values.

  • $\begingroup$ As a perhaps more intuitive example of why we might want an extension of a function, the factorial function takes no value for non integers but its extension, the gamma function has some nice properties in-between integers and takes all the same values at integers. So the factorial is in a sense a subset of the gamma which only evaluates certain values. $\endgroup$ – samerivertwice May 14 '17 at 3:06
  • $\begingroup$ This isn't quite a direct answer, but I wrote a blog post some time ago that sought to explain some aspects of the continuation of the zeta function by looking at the easier topic of examining geometric series. $\endgroup$ – davidlowryduda May 14 '17 at 3:25
  • $\begingroup$ Yes, the series diverges for other values of $s$. That's why the series for the zeta function doesn't allow us to directly attribute a value to $\zeta(s)$ at those points. For example, when $s = 0$, you get $1 + 1 + 1 + 1 + \dots$. What number is that? The series diverges, so your guess is as good as mine. On the other hand, when $s = 2$, you get the series $1 + 1/2^2 + 1/3^2 + 1/4^2 + \dots$. It can be proved that this series converges to a finite sum (which turns out to be $\pi^2/6$), so by definition we say that $\zeta(2)$ is that sum. $\endgroup$ – user49640 May 14 '17 at 12:53
  • $\begingroup$ @user49640 $1 + 1 + 1 + ...$ is the hyperinteger $1 + 1 + 1 + ...$. It's a number alright! :) $\endgroup$ – tuespetre Jan 5 at 17:08

Let's begin by presenting a simple example of representing a function with a series. Let $f(z)=\frac{1}{1-z}$ for $z\ne 1$. Recall that $f(z)$ can be represented by the geometric series

$$f(z)=\sum_{n=0}^\infty z^n \tag 1$$

for $|z|<1$. So, although $f(z)$ exists for all $z\ne 1$, its representation as given in $(1)$ is valid only when $|z|<1$.

We can represent also represent $f(z)$ by the series

$$f(z)=-\sum_{n=1}^\infty \left(\frac{1}{z}\right)^n \tag 2$$

for $|z|>1$. So, we have two representations for the same function that are valid in distinct regions of the complex $z$-plane.

Now, suppose that we represent the function denoted $\zeta(s)$ by the series

$$\zeta(s)=\sum_{n=1}^\infty \frac{1}{n^s}$$

for $\text{Re}(s)>1$. We can easily extend the definition by writing

$$\underbrace{\sum_{n=1}^{\infty}\frac{1}{n^s}}_{\zeta(s)}=\underbrace{2\sum_{n=1}^{\infty}\frac{1}{(2n)^s}}_{2^{1-s}\zeta(s)} -\sum_{n=1}^{\infty} \frac{(-1)^n}{n^s}\tag 3$$

for $\text{Re}(s)>1$. Upon rearranging $(3)$ we find

$$\zeta(s)=(1-2^{1-s})^{-1}\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^s}\tag 4$$

But notice that the series on the right-hand side of $(4)$ converges for $\text{Re}(s)>0$. So, we have just developed another representation for $\zeta(s)$ that is valid in a larger region of the complex $s$-plane.

There are other series representations of the Riemann Zeta function, such as its Laurent series,

$$\zeta(s)=\frac1{s-1}+\sum_{n=1}^\infty \frac{(-1)^n\gamma_n}{n!}(s-1)^n$$

which converges for all $s\ne 1$.

And there are integral representation of $\zeta(s)$ such as

$$\zeta(s)=\frac1{\Gamma(s)}\int_0^\infty \frac{x^{s-1}}{e^x-1}\,dx$$

which converges for $\text{Re}(s)>1$ and (see 25.5.11 this reference)

$$\zeta(s)=\frac12 +\frac{1}{s-1}-2 \int_0^\infty \frac{\sin(s\arctan(x))}{(1+x^2)^{s/2}(e^{2\pi x}+1)}\,dx$$

which is valid for all $s\ne 1$.

  • $\begingroup$ Ok, this makes sense. Does this "extension of the function" only work for infinite series? For example, could we extend f(x)=lnx for values of x that are not defined in this function? $\endgroup$ – MathGuy May 14 '17 at 16:10
  • $\begingroup$ @TyeCampbell Well, what do you mean about the logarithm function? $\log(z)=\log(|z|)+i\arg(z)$ is a continuous and analytic function if we delete from (cut) the plane a contour that adjoins $z=0$ and $z=\infty$. Do you mean a series representation for $\log(z)$ such as $\log(z)=\sum_{n=1}^\infty \frac{(z-1)^n}{n}$ for $|z-1|<1$? $\endgroup$ – Mark Viola May 14 '17 at 21:22
  • $\begingroup$ I can't seem to verify your integral representation of $\zeta(s)$. I get $\zeta (s)=\frac{2^{s-1}}{s-1}-2^s\int\limits_0^\infty\frac{\sin\left(s \tan^{-1}(x)\right)}{\left(1+x^2\right)^{s/2} \left(e^{\pi x}+1\right)}\,dx$. $\endgroup$ – Steven Clark Jan 26 at 16:05
  • $\begingroup$ @StevenClark Indeed. Much appreciated. I've edited the result and provided a reference. $\endgroup$ – Mark Viola Jan 26 at 17:27

You are not "changing" the function. You are extending the function, i.e. making a new function with a larger domain that agrees with the original function on the old domain. More precisely, you are extending the function $\sum_{n=1}^\infty n^{-s}$ whose domain is the half plane $\Re(s)>1$ to a function with domain $\mathbb C \setminus\{1\}$ that agrees with $\sum_{n=1}^\infty n^{-s}$ for all values of $s$ with $\Re (s) >1.$ It's common practice to call both $\sum_{n=1}^\infty n^{-s}$ and its analytic continuation to $\mathbb C\setminus\{1\}$ (which is a very special type of extension) by the same name $\zeta(s)$ even though they're different functions. Remember a function actually has two parts: (1) Its domain and (2) a "rule" for producing an output from any number in the domain. Don't forget about the first part.

As for whether function is somehow "defined" for $\Re (s) \le 1$ by the sum $\sum_{n=1}^\infty n^{-s},$ I take exception on two fronts. First, what do you imagine it would be defined $as$? Infinity perhaps? This would be complicated since the sum might oscillate and not limit to infinity in a meaningful way. Maybe you just imagine "is divergent" is a definition? It might help to go back to an old textbook exercise where you're asked "what's the domain of $\sqrt{x}"$ and you dutifully answer "$x\ge 0$". Why? Because (forgetting for a moment that we know about complex numbers) we couldn't get a number from the rule $\sqrt{x}$ for $x<0.$ We're in sort of the same situation with zeta expression... we can't get a number for $\sum_{n=1}^\infty n^{-s}$ for $\Re(s) \le 1$ so this expression's (largest possible) domain is $\Re(s) > 1.$

On the other hand, perhaps you didn't mean "defined" and what you really meant was "determined". So the expression $\sum_{n=1}^\infty n^{-s}$ somehow determines uniquely that the values for $\Re(s)\le1$ are divergent, or whatever you want to call their state of being. However, this is much too strong to be true. Take our example of $\sqrt{x}$ with "domain" $x\ge0.$ Can we extend this to a function on all the reals? Of course we can! In fact we can define the function to be absolutely anything we want for $x<0.$ For instance, we could just define the extension to be zero for $x<0.$ This is a perfectly valid extension of the function whose rule is $\sqrt{x}$ and whose domain is $x\ge 0$ to a function whose domain is all the real numbers. Notice that I haven't abused notation and called the new function $\sqrt{x}$ as I mentioned above is done routinely with the zeta function.

Now, you might object "but I already know how to define a square root for all real numbers. $\sqrt{-2}=2i$, not zero." Yes, this is another valid extension. You still object "but this one's better... the other was kinda arbitrary." And I think it's better and more natural too. In fact I think it's the best and most natural definition you could ask for. It's so good I'm inclined to abuse notation and call this extension $\sqrt{x}$ and consider $\sqrt{x}$ to henceforth have domain $\mathbb R$ rather than $x>0.$

So what I've hoped to show is that extending/modifying a function is pretty much an arbitrary process, but that some extensions are nicer than others. This is exactly the same thing that's going on with the zeta function. The function whose rule is $\sum_{n=1}^\infty n^{-s}$ and whose domain is $\Re(s)>1$ has a ton of extensions to larger domains, but there is a special and natural one called the analytic continuation that extends the domain to $\mathbb C\setminus\{1\}.$ It agrees with $\sum_{n=1}^\infty n^{-s}$ for $\Re(s)>1$ and also is defined outside that domain, e.g. $\zeta(0) = -1/2$ and $\zeta(-1) = -1/12.$ And like last paragraph we like this new extended function so much that we use $\zeta(s)$ to denote it, even though anybody writing an exposition on the zeta function will first write down $\zeta(s) = \sum_{n=1}^\infty n^{-s},$ referring to a restricted version of the zeta function function with domain $\Re(s)>1.$ This is because it's a nice explicit expression that serves as a prototype for the "true" analytically continued zeta function. Just like the restricted $\sqrt{x}$ for $x\ge 0$ is a easier-to-understand prototype for the extended square root function where the domain includes negative numbers (at the conceptual cost of needing to introduce complex numbers).

So there's nothing really mysterious going on here. We can extend functions to larger domains (or even modify them on their 'natural domain') in any way we want... it's just not guaranteed to be a productive enterprise. However certain extensions are natural and it is productive. The analytic continuation of the zeta function is a good example of this.

However, explaining why the analytic continuation is a good extension that we like so much would take too much more time and this answer is already long enough.


A function f(z) of a complex variable z, which is defined in a region R can be analytically continued for points beyond the region by using a Taylor's- Laurent's expansion. Example: If f(z) is defined (say) within a unit circle centered around z=0, and f(z) is analytic at all points inside this circle, then it can be 'continued' in the following manner: Take a point 'a' where 'a' is a complex number within the unit circle also let us assume 'a' to be within the unit circle but very close to the boundary of the circle. Then consider the Taylor's series :

f(a) + (z-a).f'(a) + f''(a)/2! . (z-a)^2 . + f'''(a)/3! . (z-a)^3 +...

When this series is summed for a particular value z near a, it will normally converge and the sum represents the value f(z), even though z is outside the circle. When I said normally converge, I mean one can draw another circle of radius r (small) with 'a' as center,then z can be any point within this new circle. But in certain cases (for some r) the new circle may encounter a pole then the series will not converge. If this happens then one must choose a smaller circle around 'a', of smaller radius r. One can then skirt around the pole if it is an isolated pole. See Figure. Once f(z) is known then the same series can be used to find all the derivatives of f(z) at z, viz. f'(z), f''(z), f'''(z),.... We then choose this z as our new 'a' and then draw another circle around this new 'a' and use another Taylor's expansion to find out f(z) for a new 'z' beyond the new 'a'. So you see if there is a point P inside the unit circle and another point Q (not a pole) anywhere outside the unit circle.Then we can draw a series of overlapping circles each of whose centers lie on a certain path connecting P to Q. Then a succession of Taylor's expansions starting from P will finally give the value f(Q).

It has been proved that such an 'analytic continuation' is unique and independent of the path (provided you don't cross a branch cut).

One can continue the zeta function from a point Rez > 1 to any other point to the left of the vertical Re(z)=1 line and the zeta function will be uniquely determined.

It is assured that even if one uses contour integrals one will get the same answer for the value of zeta(z) for any particular z as one would have got by Taylor's expansions.

See Figure Below Analytic Continuation Figure

  • $\begingroup$ Please use MathJax. $\endgroup$ – José Carlos Santos Jul 9 '17 at 13:32
  • $\begingroup$ And see Mark Viola's answer, already explaining the analytic continuation with the simplest example : $\sum_{n=0}^\infty z^n$ $\endgroup$ – reuns Jul 9 '17 at 22:19

When the real part is more than one, the total length of the pieces $|n^{-k}|$ is finite. So there is a well-defined sum.
When the real part is between zero and one, the pieces shrink to zero, but the total length of the pieces is infinite. As a result, you can rearrange the terms and form any sum you like.
For example, $$1-\frac12+\frac13-\frac14+...=\ln 2\\ (1+\frac13-\frac12)+(\frac15+\frac17-\frac14)+(\frac19+\frac1{11}-\frac16)+...>(1+\frac13-\frac12)=5/6$$

When $k<0$, so the pieces are $n^{-k}$ with a positive exponent, the pieces get bigger and bigger, so as you say can't settle to anything.

A real function can change its form as you cross a point. There is a function that is zero if $x\leq0$, but $\exp(-1/x^2)$ when $x>0$ This function is quite smooth at zero.

A complex function can't change suddenly. When a complex function is differentiable, there is only one way to extend it. That is second-year calculus.

We have $f(x)=\sum n^{-k}$ defined for $\Re(k)>1$ and $g(x)$ defined everywhere. $f(x)=g(x)$ when $\Re(k)>1$, and there is only one way to extend it. So we call the sum $g(x)$, even though the sum doesn't converge.

  • $\begingroup$ So this is definitely not how it goes when the real part of $s$ is nonpositive. When the real part is positive and less than or equal to $1$, I think the details of conditional convergence depend sensitively on the imaginary part of $s$. $\endgroup$ – Ian May 14 '17 at 3:29

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