We have:

$\zeta (n)=1/1^n+1/2^n+\dots$(*)

Multiply by $1/2^n$:

$\zeta (n)/2^n=1/2^n+1/4^n+\dots$

Subtract this twice from (*):

$\zeta (n)[(1-2/2^n)]=1/1^n-1/2^n+1/3^n-1/4^n+\dots$(**)

whenever (*) converges. We now define $\zeta (n)$ by (**) even when (*) doesn't converge. In other words we define:

$\zeta (n)=1(1/1^n-1/2^n+\dots )/(1-2/2^n)$ (***)

This still doesn't converge for $n=0$. It does however now converge for $0<n\le 1$ (which it didn't originally when we defined $\zeta (n)$ by (*)). So we may decide to define $\zeta (0)$ as $\underset{n\to 0+}{lim}\zeta (n)$ where $\zeta (n)$ is given by (***) - we define $\zeta (0)$ this way to make $\zeta (n)$ continuous at the origin. If you now work out $\zeta (0)$ this way it should come to 1/2.

Well another interesting result concerning the zeta function and the gamma function is:

$\zeta (n)=\zeta (1-n)\Gamma (1-n)2^n{\pi }^{n-1}\sin (\pi n/2)$

I'll prove this if you like, but you do need residues.

Ah, thanks. I thought I'd make an error somewhere...

Now about the reflection formula for $\zeta (n)$:

$\zeta (n)=\zeta (1-n)\Gamma (1-n)2^n{\pi }^{n-1}\sin (n\pi /2)$

I'll sketch the proof, and then fill in on any details if requested. We start with the formula Brad proved in a different discussion:

$\zeta (s)=1/\Gamma (s){\int }_0^{\infty }{x}^{s-1}/(e^x-1)\mathrm{dx}$

Now consider the integral:

$I(s)={\int }_C{z}^{s-1}/(e^z-1)$

where the contour $C$ starts at $\infty$on the positive real axis, encircles the origin once (anti-clockwise) and doesn't loop round $2n\pi i$ for any non-zero integral $n$. (Why is this well defined? Well the integrand is analytic except at $x=2n\pi i$, so the only pole it encloses is at 0, so Cauchy's 2nd theorem shows that all such contours give the same value for the integral.)

Define the contour $C_n$ to go from $\infty$on the positive real axis to $(2n+1)\pi$ then round the squares with corners $(2n+1)\pi (\pm 1\pm i)$ and then back to $\infty$along the positive real axis.

Between $C$ and $C_n$, $z^{s-1}/(e^z-1)$ has poles at $\pm 2i\pi$, ..., $\pm 2in\pi$. We can therefore apply Cauchy's 2nd integral theorem to the region between $C$ and $C_n$ and you get:

$I(s)={\int }_{C_n}{z}^{s-1}/(e^z-1)\mathrm{dz}+4\pi i{e}^{i\pi s}\sin (\pi s/2)\sum _{m=1}^n(2m\pi {)}^{s-1}$

Please write back if you need help obtaining this. The key is to work out the residues at each of the points listed above.

Letting $\mathrm{Re}s<0$ and $n\to \infty$you notice that the integral round $C_n$ tends to 0 (since $1/(e^z-1)$ is bounded on the contours $C_n$ and $z^{s-1}<K|z{|}^{\mathrm{Re}(s-1)}$ for some $K>0$). So:

$I(s)=4\pi i{e}^{i\pi s}\sin (\pi s/2)\sum _{m=1}^{\infty }(2m\pi {)}^{s-1}$

And:

$I(s)=4\pi i{e}^{i\pi s}\sin (\pi s/2)(2\pi {)}^{s-1}\zeta (1-s)$

Sorry, I wrote a reply to this, but forgot to send.

If you're not too familiar with residues, the derivation of the reflection formula really isn't the best introduction, since it is pretty complicated. Here is the gist of complex integration. The following definitions are not totally rigorous and I've left out lots of details. Please ask if there's anything you aren't sure of.

As you know an analytic (or holomorphic) function is one in which $\underset{h\to 0}{lim}[f(z+h)-f(z)]/h$ exists for each $z$. Note that $h$ is a complex number and can tend to 0 in any direction. The limit must be independent of the direction in which $h$ approaches 0, in order for $f$ to be analytic at $z$. So the condition of analyticity is actually a very strong one. However lots of useful functions are analytic and certainly all functions that I know arising physically are analytic, so it's not nearly as restrictive as it looks.

Now integration. Firstly, we can integrate functions $f:R\to C$ no problem. Simply write $f(x)=g(x)+ih(x)$ where $g$, $h(z)$ are real. Then define:

$\int f(x)\mathrm{dx}=\int g(x)\mathrm{dx}+i\int h(x)\mathrm{dx}$

Now how about functions $f:C\to C$? Well, we can define integration of $f$ along a path $P$ in the Argand diagram in a fairly intuitive way.

${\int }_{P}f(z)\mathrm{dz}$

is defined by breaking the path $P$ into tiny pieces and then summing $f(z)\delta z$. Specifically, if we break up the path $P$ into pieces which start/end at points $z_0$, $z_1$, ..., $z_N$ (where $z_i$, $z_{i+1}$ are close points on the path) then we define:

${\int }_{P}f(z)\mathrm{dz}=lim((z_1-z_0)f(z_0)+(z_1-z_0)f(z_1)+...+(z_N-z_{N-1})f(z_{N-1}))$

where the limit is taken as the number of pieces tends to infinity, and the sizes of the pieces of the path, i.e. $|z_1-z_0|$, ..., $|z_N-z_{N-1}|$, all tend to 0.

An entirely equivalent definition is:

${\int }_{P}f(z)\mathrm{dz}=\int f(P(t))P\text{'}(t)\mathrm{dt}$

where $P(T)$ is a parametrisation of the path $P$ in the complex plane (where $T$ is real) and the RHS is defined by a normal integral. (The integrand is a function from $R$ to $C$ so is defined as above.)

You should be able to see why these two definitions are equivalent intuitively. (Proving this formally isn't entirely trivial though.)

Most of the properties you know concerning ordinary integration (e.g. the rule for changing variables) carry over to the complex case.

We can also define an integral round a contour - that is a closed, non-intersecting shape in the plane (e.g. a circle or a rectangle). Just consider it as a path which starts and ends at one point, then use the definition above. By convention you always go round anti-clockwise. Now the massive theorem behind complex analysis is Cauchy's first theorem. If $f$ is analytic then if $C$ is a contour then:

${\int }_{C}f(z)=0$

In other words if you integrate an analytic function round any closed path, you get 0. You can prove this using Stoke's theorem (I don't know whether you're familiar with this or not). If not, we can do it manually, though it won't be totally rigorous without going into much more detail.

There are some other important properties: 1) If you join the ends of paths $P_1$, $P_2$ to get one path $P$ then:

${\int }_{P}f(z)\mathrm{dz}={\int }_{P_1}f(z)\mathrm{dz}+{\int }_{P_2}f(z)\mathrm{dz}$

2) If paths $P_1$, $P_2$ start and end at the same point then ${\int }_{P_1}f(z)\mathrm{dz}={\int }_{P_2}f(z)\mathrm{dz}$ where $f$ is analytic on and between the two paths $P_1$, $P_2$. (To prove this, you need Cauchy's first theorem.)

3) A contour $C\text{'}$ encloses a contour $C$. Suppose $f$ is not necessarily analytic (so the integrals round the contours aren't necessarily 0) but is analytic on and between the two contours. Then the integral of $f$ around each of the contours is the same. (This is another application of Cauchy's first theorem.)

4) Suppose the contour $C$ encloses contours $C_1$, ..., $C_n$ which are all disjoint. Suppose further that $f$ is analytic everywhere in $C$ apart from in $C_1$, ..., $C_n$. Then:

${\int }_{C}f(z)\mathrm{dz}={\int }_{C_1}f(z)\mathrm{dz}+...+{\int }_{C_n}f(z)\mathrm{dz}$

Please ask for hints if you get stuck.

Anyway, that was all extremely quick and there are probably numerous points you'll want to raise. Then we can get onto writing complex functions as Taylor series ( $\sum _{i=0}^{\infty }{a}_i(z-z_0{)}^i$) and Laurent series ( $\sum _{-\infty }^{\infty }{a}_i(z-z_0{)}^i$. We can always write analytic functions as the former, whereas if a function is analytic everywhere apart from at $z=z_0$ (where traditionally it blows up) we have to revert to a Laurent series. The residue at $z=z_0$ of a function with the above Laurent series is defined as $a_{-1}$. Then we can use the properties given above in the list to prove that if a function $f$ is analytic in a contour $C$ except at $z_1$, ..., $z_n$ then:

${\int }_{C}f(z)\mathrm{dz}=2\pi i\sum _{i=1}^n{r}_i$

where $r_i$ is the residue at $z_i$. This is Cauchy's residue theorem.

(If you want to have a go at this now, see if you can show that:

${\int }_{C}f(z)/(z-z_0)\mathrm{dz}=2\pi if(z_0)$

where $f$ is analytic and $C$ encloses $z_0$.

Nice site, Arun.

(BTW I made an obvious mistake with the first definition of the integral along a path of $f:C\to C$.) The point about the second definition of the integral of $f:C\to C$ is that we have defined it in terms of the integral of a function which is $R\to C$. You are right the formula in definition 2) does also hold for any function $f:R\to C$ so you can think of it as a generalisation.

Just to summarise the definitions of various integrals; there are at least three types of function which we can integrate (provided they behave nicely enough). They are $f:R\to R$, $f:R\to C$, $f:C\to C$.

1) $f:R\to R$

These functions are integrated under the standard calculus definition. The integral is defined as the limit of a Riemann sum.

2) $f:R\to C$

These can be integrated by an obvious extension of 1). Either you can write some definition involving something similar to the limit of a Riemann sum (you need to define the limit of a complex sequence but this isn't hard) or you can just separate the real and imaginary parts of the integrand (like I did in my first message) and define:

${\int }_a^{b}f(x)\mathrm{dx}={\int }_a^b\mathrm{Re}f(x)\mathrm{dx}+i{\int }_a^b\mathrm{Im}f(x)\mathrm{dx}$

3) $f:C\to C$

We are now considering $f$ to be a mapping from a plane to a plane (i.e. the function takes one point on Argand diagram and sends it to another point on the Argand diagram). With integrals 1), 2) we integrate $f(x)$ where $x$ is real so the path $x$ takes is totally restricted, whereas with 3) we must specify the path (knowing the endpoints alone isn't good enough - unless the function happens to be analytic everywhere in which case by property 2) in my first message the integral is independent of the path). Then there are two ways of defining the integral of $f$ - either we can divide up the path into little pieces and sum $f(z)\delta z$ and take the limit. Or we can define it in terms of 2) by:

${\int }_{P}f(z)\mathrm{dz}={\int }_a^{b}f(P(T))P\text{'}(T)\mathrm{dT}$ (*)

Note that $P(T)$ is a function from $R\to C$ such that $P(a)$ is one end of the path $P$ and $P(b)$ is the other end, and as $T$ varies in $[a,b]$, $P(T)$ traces out the path $P$. In order to show the integral is well defined we need to prove that for any differentiable parametrisation $P(T)$ of the path $P$ (where $T$ is real) the RHS of (*) gives the same result.

"Cauchy's theorem makes great intuitional sense. If we impicture adding up areas over the real and imaginary parts of the curve, it is relatively intuitional to see the area's multiplied by the 'top' and 'bottom' (real part) of the closed curve cancel out, and similarly, the areas multiplyed by the 'left' and 'right' (imaginary part) should cancel out as well.«par/> I don't quite get what you mean here. I'm not sure how areas come into it.

"For the fourth one, I'm not sure what disjoint means. Does it mean that the path's all share common endpoints? If it does, then that makes sense to me. I'll have a try proving the last theorem, and post a bit later."

Disjoint means they don't overlap at all (e.g. two disjoint sets are disjoint if they don't have any common members). I didn't really know what the term meant either before university.

Now onto non-analytic functions. Another example. Take $f(z)=\left|z\right|$

In other words $f(z)$ is the distance on the Argand diagram between the origin and $z$.

We show that this isn't analytic anywhere. For it to be analytic at $z_0$ we require that:

$(f(z_0+h)-f(z_0))/h$ (*)

converges as $h\to 0$ in any direction.

Now there are several ways $h$ can tend to 0. For example $h$ could tend to 0 along the real line. In this case since $f$ is always real, (*) is always real so the limit will be real. On the other hand $h$ could tend to 0 along the imaginary line (like $\lambda i$ where $\lambda$ is real and tending to 0) in which case (*) would be imaginary, so the limit would be imaginary. If you let $h$ tend to 0 in yet a different way, you would probably get another different answer. So in other words (*) tends to different values depending on the direction $h$ tends to 0 in.

(Alternatively, show that the Cauchy-Riemann conditions aren't satisfied. The Cauchy-Riemann conditions are necessary for $f$ to be analytic but not quite sufficient. To prove $f$ is analytic we need to not only prove CR is satisfied but also show that $f$ is locally linear - meaning that at each point $f$ is a linear mapping to first order. Alternatively you can simply show that $(f(z+h)-f(z))/h$ converges to one value however $h\to 0$.)

By the way, there's a theorem saying that any non-constant analytic function is unbounded, so in fact there are plenty of simple examples of non-analytic functions.

Now if you try integrating $f(z)=\left|z\right|$ round a closed path in the complex plane then you probably won't get 0 - because Cauchy's theorem doesn't hold for non-analytic functions.

The other obvious example of non-analytic functions are ones which blow up at a point. For instance the function $f(z)=1/z$ blows up at the origin. If you take a unit circle around the origin then we can parametrise this by $P(T)=e^{it}$ where $T$ is in $[0,2\pi ]$. By the second definition we get:

${\int }_{C}f(z)\mathrm{dz}={\int }_0^{2\pi }f(e^{iT})i{e}^{iT}\mathrm{dT}=2\pi i$

${\int }_{\Gamma }z*\mathrm{dz}=2\times$ area bounded by $\Gamma$

for simple closed curve $\Gamma$.