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Contour Integration

April 12th, 2009 |

Author: Merrin

Contour integration is a powerful technique for evaluating many definite improper integrals. Contour integration occurs in the complex plane and is sort of the analogue of a line integral in two dimensions. Although contour integrals are complex we can manipulate them to gives some integration results for real functions. The following sections serve only as an introduction and emphasize the main points needed to evaluate integrals. Complex analysis is very rich theory and I recommend it if you get a chance to study it in detail at some later point. By now, you have already been exposed to a small amount of complex analysis. We will extend this a bit further so we will be able to carry out some simple contour integrals.

Definition 1. A complex function $latex f(z)=u(x,y)+iv(x,y)$ of the complex variable $latex z=x+iy$ is said to be analytic where it is differentiable.

To prove where a function is analytic we can use the so called Cauchy-Riemann conditions.

$\displaystyle \cfrac{\partial u}{\partial x} =\cfrac{\partial v}{\partial y} \quad \cfrac{\partial u}{\partial y} =-\cfrac{\partial v}{\partial x}$

Remember analytic is synonymous with differentiable.

Here we present the first major result of contour integration. If a function is analytic everywhere on and inside the contour of integration then that integral is zero. This is called Cauchy’s Theorem.

$\displaystyle \oint f(z)dz=0 \quad f(z)={\rm analytic}\, {\rm inside}\, {\rm and}\, \text{on } C$

This is the analogue of line integrals in a conservative field, but for complex numbers. Integrating around a closed path results in zero. We will find that for functions with singularities inside the contour the integral will not in general be zero. We need to have a way to recognize where a function is not analytic. In simple examples, we can locate non analytic locations where the function is divided by zero.

Example 1. Determine where the following function is not analytic.

$\cfrac{1}{(1+z^{2} )(1-z^{2} )}$

Solution 1. We are doing complex variables now so we have to take into account complex roots as well as real roots. If we factor the denominator further we are left with

$\cfrac{1}{(z+i)(z-i)(1-z)(1+z)}$

There are isolated singularities in this continuous function. The function is not analytic at $latex z=-i,i,1,-1$. After a partial fractions analysis we would have.

$\displaystyle \cfrac{1}{(z+i)(z-i)(1-z)(1+z)} =\cfrac{A}{z+i} +\cfrac{B}{z-i} +\cfrac{C}{1-z} +\cfrac{D}{1+z}$

Since the power of the singularity terms are negative one, we say the function has simple poles at these points. A double pole at $latex z=1$ would look like

$\cfrac{1}{(z-1)^{2} }$

Example 2. Show that $latex e^{z} $ is analytic in the entire complex plane.

Solution 2. We can write this complex form in the above form, and then test with the Cauchy-Riemann conditions.

$\displaystyle {e^{z} =e^{x+iy} =e^{x} (\cos y+i\sin y)=u(x,y)+iv(x,y)} \\ {\cfrac{\partial u}{\partial x} =e^{x} \cos y=\cfrac{\partial v}{\partial y} =e^{x} \cfrac{\partial \sin y}{\partial y} =e^{x} \cos y} \\ {\cfrac{\partial v}{\partial x} =e^{x} \sin y=-\cfrac{\partial u}{\partial y} =-e^{x} (-\sin y)=e^{x} \sin y}$

Since the resulting partial derivatives are satisfied everywhere in the complex plane, the function $latex e^{z} $ is analytic in the entire complex plane. Such functions that are analytic everywhere are also called entire.

It follows from Cauchy’s Theorem that

$\displaystyle \oint e^{z} dz =0\quad {\rm any}\, {\rm closed}\, {\rm curve}$

We will find that many complex integrals we will study are mainly determined by the parts of the complex function in the integrand which are not analytic such as isolated singularities. Around a singularity we can expand a function as a Laurent series which is more general than a Taylor series and contains terms of negative powers.

Definition 2. Laurent series are written as

$\displaystyle f(z)=\sum _{k=-\infty }^{\infty }a_{k} (z-z_{0} )$
$f(z)=\cfrac{a_{-n} }{(z-a)^{n} } +...+\cfrac{a_{-1} }{z-a} +a_{0} +a_{1} (z-a)+...$

Now we will try a contour integral around each piece in the Laurent series. We will choose the Laurent series to be centered at $latex a = 0$ and the contour to be a counterclockwise unit circle in the complex plane. If we can integrate each piece then we can sum them to find the contour integral of the function. Since there are so many singularities we expect to get a nonzero value for the integral.

Example 3. Suppose complex function is written as a Laurent series centered about $latex z=0$. Show that the counter clockwise contour integral of f(z) along the path of a unit circle centered at zero in the complex plane is zero except for the integral of the term $latex a_{-1} \cfrac{1}{z} $

Solution 3. We can write the contour integral as

$\displaystyle {\oint f(z)dz }={\int _{0}^{2\pi }\sum _{k=-\infty }^{\infty }a_{k} z^{k} dz= \sum _{k=-\infty }^{\infty }a_{k} \int _{0}^{2\pi }\cfrac{1}{(\rho e^{i\theta } )^{-k} } \rho ie^{i\theta } d\theta }$

Something remarkable happened although there were poles of all orders in the integral the only piece that contributes is the simple pole corresponding to the term

$\displaystyle \oint \cfrac{a_{-1} }{z} dz=2\pi ia_{-1}$

The integral of the function only depends on the coefficient of the Laurent series of power negative one. This coefficient is called the residue. It is a matter of practical algebra and differentiation to find this coefficient for any function. So it appears we can do any contour integral if the function has a Laurent series inside the contour.

There are other formulas to find which are shifting of the power of poles. For example, writint

$\displaystyle \cfrac{1}{z-a} \sum _{n=-\infty }^{\infty }a_{n} (z-a)^{n} = \sum _{n=-\infty }^{\infty }a_{n} (z-a)^{n-1}$

The residue of the new series is just $latex a_{0}$, shifted by one. Thus we have the Cauchy integral formula

$\displaystyle f(a)=\cfrac{1}{2\pi i} \oint \cfrac{f(z)}{z-a} dz$

With more shifts, we just find the coefficients of the Taylor series part of the Laurent series

$\displaystyle f^{(n)} (a)=\cfrac{n!}{2\pi i} \oint \cfrac{f(z)}{(z-a)^{n+1} } dz$

The remarkable things about these formulas depend on integration far away from the point we are probing an answer for.

There may be more than one isolated singularity inside the contour integral. In this case, we have to make separate Laurent expansions around these points with poles. The contour integration is then related to the sum of the residues from each Laurent series. This is known as the \textbf{Residue Theorem} and is what is used to calculate some of the advanced integrals we are after.

$\displaystyle \oint f(z)dz =2\pi i(a_{-1} +b_{-1} +...)=2\pi i\sum _{k=1}^{n}Res(f(z),z_{k} )$

We say the contour integral is equal to $latex 2\pi i$times the sum of the residues within the closed contour. Now we direct our attention to calculate the residues for any general function.

Calculating residues is a necessary step of certain contour integrations when the integral is not zero. We have a residue formula for evaluating residues but there are several other methods we will also demonstrate. A series expansion can identify the coefficient which is the residue. Polynomial division or partial fractions for a rational function may also reveal the residues. Direct calculation of residues can always be done through differentiation using the following formula.

$\displaystyle \text{Res}(f(z),z_{0} )=\cfrac{1}{(n-1)!} \mathop{\lim }\limits_{z\to z_{0} } \cfrac{d^{n-1} }{dz^{n-1} } [f(z)(z-z_{0} )^{n} ]$

It is worth emphasizing that for the case of n=1 the formula reduces to

$\displaystyle \text{Res}(f(z),z_{0} )=\mathop{\lim }\limits_{z\to z_{0} } f(z)(z-z_{0} )$

The residue formula for a simple pole has a simple form with no derivatives and just a limit.

Example 4. Calculate the residue at z=i of

$\cfrac{1}{z(1+z^{2} )(1-z^{2} )}$

Solution 4. At the point z=i there is a simple pole so we can use the residue formula for a simple pole. We get a simple plug in answer.

$\displaystyle Res\left(\cfrac{1}{z(1+z^{2} )(1-z^{2} )} ,i\right)=\mathop{\lim }\limits_{z\to i} \cfrac{(z-i)}{z(z+i)(z-i)(1-z^{2} )} =\cfrac{1}{i(2i)(1-i^{2} )} =-\cfrac{1}{4}$

Example 5. Find the residue at z=0 of the following function

$f(z)=\cfrac{1}{z^{3} (1-z)(1+z)}$

Solution 5. Here there is a triple pole so we use the residue formula with two derivatives.

$\displaystyle {Res(f(z),0)} = {\cfrac{1}{(2)!} \mathop{\lim }\limits_{z\to 0} \cfrac{d^{2} }{dz^{2} } [\cfrac{1}{z^{3} (1-z)(1+z)} z^{3} ]=\mathop{\lim }\limits_{z\to 0} \cfrac{1}{2!} \left(\cfrac{1}{1-z^{2} } \right)^{{'} {'} } } \\ = {\mathop{\lim }\limits_{z\to 0} \left[\cfrac{1}{2} \cfrac{2z}{(1-z^{2} )^{2} } \right]^{{'} } =\mathop{\lim }\limits_{z\to 0} \cfrac{1}{(1-z^{2} )^{2} } +\cfrac{2z^{2} }{(1-z^{2} )^{3} } =1}$

Another way to calculate the residues is when we recognize functions that have a known Laurent of Taylor series. Carry out a Laurent series and find the coefficient of the reciprocal of z-a.

Example 6. Find the residue of

$\cfrac{\sin z}{z^{5} }$

Solution 6. This problem is well suited to calculation by a series expansion.

$\displaystyle \cfrac{\sin z}{z^{5} } =\cfrac{z-\cfrac{1}{3!} z^{3} +\cfrac{1}{5!} z^{5} -\cfrac{1}{7!} z^{7} }{z^{5} } =\cfrac{1}{z^{4} } -\cfrac{1}{6z^{2} } +\cfrac{1}{120} -...$

This series has no residue, so the residue is equal to zero.

Example 7. Find the residue of

$\cfrac{\sin z}{z^{4} }$

Solution 7. Do a series expansion.

$\displaystyle \cfrac{z-\cfrac{1}{6} z^{3} +\cfrac{1}{120} z^{5} }{z^{4} } =\cfrac{1}{z^{3} } -\cfrac{1}{6z} +\cfrac{z}{120} -...$

The residue is $latex -1/6$.

Example 8. Find the residue at z=0 of

$\displaystyle \cfrac{\cos z}{(1-z^{2} )z^{3} }$

Solution 8. We will use a series method to determine the residue

$\displaystyle {\cfrac{(1-\cfrac{1}{2} z^{2} +\cfrac{1}{4} z^{4} -...)}{(1-z^{2} )z^{3} } } = {\cfrac{(1-\cfrac{1}{2} z^{2} +\cfrac{1}{4} z^{4} -...)(1+z^{2} +z^{4} +z^{6} +...)}{z^{3}}}$

Multiply out the numerator to order $latex O(z^4)$ and you will find that the residue is equal to $latex 1/2$.

Example 9. We can also calculate residues by partial fractions. Find the residue at z = 4 of the following function

$\cfrac{9z^{2} +8z-2}{(z-4)^{3} }$

Solution 9. We know how to write out the partial fractions for this function but there is a trick, to avoid polynomial division.

$\displaystyle {u}= {z-4} \\ {\cfrac{3z^{2} +4z-5}{(z-4)^{3} } } = {\cfrac{3(u+4)^{2} +8(u+4)-2}{u^{3} } =\cfrac{3u^{2}+32u+78}{u^{3} } }$

So the residue is just equal to 3. Now that we have all the pieces for calculating contour integral we will move on to some specific classes of real integrals that can be done with complex variables.

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