To calculate a contour integral, use theorems rather than direct calculation

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To calculate a contour integral, use theorems rather than direct calculation

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[QUICK DESCRIPTION]
Suppose that $f$ is a holomorphic function defined on a domain $D\subset\C$, $C$ is a (sufficiently nice) closed curve in $D$, and you want to calculate the contour integral $\int_Cf(z)dz$. In theory you could convert this into a real integral (or pair of real integrals) by working directly from the definition of a contour integral. But this is almost never the right thing to do.

[PREREQUISITES]
The definition of a contour integral, [[w:Cauchy's theorem]], [[w:Cauchy's integral formula]], [[w:Cauchy's residue theorem]]

[EXAMPLE]
Let $C$ be a contour defined by the function $\gamma:[0,1]\rightarrow\C$, $\gamma(t)=e^{2\pi it}+e^{4\pi it}$. What is the contour integral $\int_Cz^3dz$?

One way of tackling this question is to use the definition of a contour integral:
[maths]\int_Cz^3dz=\int_0^1(e^{2\pi it}+e^{4\pi it})^3(2\pi ie^{2\pi it}+4\pi ie^{4\pi it})dt[/maths]
Clearly when the integrand is expanded out it will be a linear combination of positive integer powers of $e^{2\pi it}$. Since $\int_0^1e^{2\pi int}dt=0$ for every positive integer $n$, the whole integral is $0$.
Another way of tackling the question is to use a bit of theory: the function $z\mapsto z^3$ is holomorphic on the whole of $\C$, and $\gamma$ is a closed curve, so $\int_Cz^3dz=0$ by Cauchy's theorem.

[EXAMPLE]
Let $C$ be the unit circle -- that is, the contour defined by the path $\gamma:t\mapsto e^{2\pi it}$. What is $\int_C\frac{e^{z^2dz}}{1-2z}$?

If we plunge straight in and use the definition of the contour integral, we get
[maths]\int_C\frac{e^{z^2dz}}{1-2z}=\int_0^{2\pi}\frac{e^{e^{4\pi it}}2\pi ie^{2\pi it}dt}{1-2e^{2\pi it}},[/maths]
which looks horrific. However, if we rewrite the integral as $-\frac 12\int_C\frac{e^{z^2}dz}{z-1/2}$, then Cauchy's integral formula tells us immediately that the answer is $-\pi ie^{1/4}$.

[GENERAL DISCUSSION]
A clear indication that using the definition of the contour integral is not the right thing to do (besides the fact that in the above two examples the answer follows immediately from a little bit of theory) is that if you continuously deform $C$ while avoiding any singularities of $f$, then the integral does not change. Thus, for a specific holomorphic function $f$, a path integral is telling you something about topological properties of $C$ in the domain $D$ (or, if you allow $f$ to have a few singularities, the topology of the domain $D$ with these singularities removed). To put it another way, the path integral $\int_Cf(z)dz$ is really a function of an ''equivalence class'' of closed paths $C$, and we are not really understanding it properly if we calculate it for the particular member of the equivalence class that we happen to be presented with.

There is in fact a case for saying that there is only one integral that you ever need to calculate by hand. It is the integral $\int_C\frac{dz}z$, where $C$ is the unit circle. If you do that directly you get $2\pi i$. For every $n$ other than $-1$, the integral $\int_Cz^ndz$ is zero, [add]since the function $z^n$ has the antiderivative $z^{n+1}/(n+1)$ on $\C\setminus\{0\}${{, and the fundamental theorem of calculus therefore tells us that its integral round any closed curve is zero. This can be thought of as the $(-1)$st result in complex analysis: it is helpful in developing the theory but can be proved just using real analysis}}[/add]. Given the result that $\int_C\frac{dz}z=2\pi i$, all the rest of complex analysis can be developed, culminating in the residue theorem, which one then uses to calculate integrals round closed curves. (One may want more sophisticated versions of the residue theorem if e.g. the curves wind more than once round some of the singularities of $f$.)

[EXAMPLE]
Another example, which actually illustrates ''three'' Tricki tricks, is the standard proof of [[w:Liouville's theorem]]. Let us suppose that $f$ is a bounded holomorphic function defined on the whole of $\C$. We would like to prove that $f$ is constant. So let $u$ and $v$ be complex numbers and let us try to prove that $f(u)=f(v)$. The first idea (not a Tricki trick, but perhaps it could be turned into one) is to use Cauchy's integral formula to write $f(u)=\int_C\frac{f(z)dz}{z-u}$ and $f(v)=\int_C\frac{f(z)dz}{z-v}$. For this to be valid, let $C$ be a circle that contains both $u$ and $v$ inside it. Subtracting, we find that
[maths]f(u)-f(v)=\int_Cf(z)\left(\frac 1{z-u}-\frac 1{z-v}\right)dz=\int_C\frac{f(z)(u-v)dz}{(z-u)(z-v)}.[/maths]

The next idea is one of the main themes of this article: think of the integral on the right-hand side as a function not of $C$ itself but of the ''equivalence class'' of all such curves $C$, where we think of two curves as equivalent if one can be deformed continuously to the other without crossing $u$ or $v$ (where the function $f(z)/(z-u)(z-v)$ has singularities). More formally, the integral does not change if you replace $C$ by another closed curve that is homotopic (or even homologous) to $C$ in the domain $\C\setminus\{u,v\}$.

That was Tricki trick number 1. Trick number 2 is to bound a path integral $\int_Cg(z)dz$ by the length of $C$ times the supremum of $|g|$ on $C$. This idea is discussed in the article [[Bound your integral by its base times its height]]. In our case, if $C$ is a circle of radius $R$, then one can easily use this to obtain an upper bound for the modulus of the right-hand side that tends to zero as $R$ goes to infinity (at a rate proportional to $R^{-1}$, since $C$ has length $2\pi R$ and the integrand, because $f$ is bounded, has a supremum on $C$ that tends to zero at a rate of $R^{-2}$ or faster).

This tells us that there are equivalent curves for which the integral is arbitrarily small. Since the integral is the same for all equivalent curves, it must be zero for all of them. This reasoning is an example of the trick discussed in the article [[To prove that two real numbers are equal, prove that neither is greater than the other]]. (Here we use an important special case of this technique: if a number has modulus less than $\delta$ for every $\delta>0$ then it must be $0$.)

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