- Suppose that we expand a function $f(z)$ about the point $z=a$
in a Laurent series, valid for $0<|z-a| < R$:
$$f(z)=\sum_{n=0}^\infty a_n(z-a)^n +
\sum_{n=1}^\infty\frac{b_n}{(z-a)^n}\,.$$ Then, for a curve $C$
lying in the region $0<|z-a| < R$, $$\oint_C f(z)\,\d z=2\pi i\,
b_1.$$ The coefficient $b_1$ is called the
*residue*of $f(z)$ at $z=a$; in the lectures it is often denoted $\res(a)$. - Hence for a curve $C$ which encloses a number of poles of $f(z)$, the residue theorem states that $$\oint_C f(z)\,\d z=2\pi i \times[\text{sum of residues at all poles inside $C$}].$$
- The residue at each pole is the term $b_1$ in the Laurent expansion in the vicinity of that pole; note that this expansion will be different for each pole!
- For a function $f(z)$ with a
*simple pole*at $z=a$, we can often use one of the following formulas for the residue: $$b_1=\lim_{z\to a}\bigl[(z-a) f(z)\bigr];\qquad \text{for } f(z)=\frac{g(z)}{(z-a)},\;\; b_1=g(a);\qquad \text{for } f(z)=\frac{g(z)}{h(z)},\;\; b_1=\frac{g(a)}{h'(a)};$$ where $g(z)$ and $h(z)$ represent functions which are analytic at $a$, with $g(a)\ne0$ (and finite) and $h(a)=0$. Note that the first of the above formulas is always applicable and that the others follow easily from it. - For a function $f(z)$ with a pole of order $p$ at $z=a$, then in principle we can find $b_1$ from $$b_1=\frac1{(p-1)!}\lim_{z\to a}\left\{\frac{\d^{p-1} }{\d z^{p-1}}\left[(z-a)^p f(z)\right]\right\}.$$ This formula can be tedious to apply if $p$ is large, but it has a nice self-correcting feature: if you overestimate the order of the pole, the formula still gives the right answer for $b_1$. You should think about why that is.
- For higher-order poles, if $f(z)$ can be written as a ratio of analytic functions it can be simpler to use the Taylor expansions for these and hence construct the Laurent series up to $b_1$ explicitly. It is therefore worth knowing the first few terms of the Taylor expansions of familiar functions such as $\e^z$, $\sin z$, $\tan z$, $\ln(1+z)$, etc.

Spiegel 7.1-3

Riley 18.14; **Boas 14.5, 14.6** ; Arfken 7.2

- The contours above are examples of contours which can be used to calculate real integrals
- If $I=\int_{-\infty}^\infty f(x)\d x$, and $f(z)$ has a finite number of poles none of which are on the real axis, and $|f(R\e^{i\theta})|\to 0$ faster than $1/R$ as $R\to \infty$, then we can use contour $C$ shown in (A). With $R$ large enough so that all poles in the upper half plane are within $C$, $\lim_{R\to\infty}I_2=0$. Hence $$ I=\lim_{R\to\infty}I_1= \lim_{R\to\infty}\left(\oint_C f(z)\,\d z-I_2\right) =2\pi i\times\left[\text{sum of residues at all poles inside $C$}\right]$$
- If $I=\int_{-\infty}^\infty f(x)\e^{ikx}\,\d x$, and (i) $k>0$, (ii) $f(z)$ is meromorphic in the upper half plane and (iii) $|f(z)|\to 0$ as $|z|\to \infty$ in the upper half plane, then we can also use contour $C$ shown in (A). The vanishing of $I_2$ as $R\to \infty$ in this case is called Jordan's lemma. If $k<0$, but the other conditions of Jordan's lemma hold for the lower half plane, we can close the contour in the lower half plane instead.
- If in either case above there is a pole on the real axis at $z=a$, we define the principal value of the integral to be $$\lim_{\epsilon\to 0}\left[\int_{-\infty}^{a-\epsilon} f(x)\,\d x+\int_{a+\epsilon}^{\infty} f(x)\,\d x\right];\;\text{the limiting value is denoted by}\;\mathrm{P}\!\int_{-\infty}^\infty f(x)\,\d x\quad\text{or}\quad\Pint{-\infty}\infty{f(x)\,\d x}.$$ To evaluate it we use the contour shown in (B). If the pole is simple or of odd order we have $\lim_{\epsilon\to 0}I_3=-i\pi\,b_1$ where $b_1$ is the residue of the pole at $z=a$. If the pole is of even order the principal value of the integral will not exist.
- If the integrand has a branch point at $z=0$ we may be able to evaluate $I=\int_{0}^\infty f(x)\,\d x$ through the use of the contour shown in (C), where the branch cut is taken along the positive real axis. The contributions $I_1$ and $I_3$ will not cancel because $f(x)\ne f(x\e^{2\pi i})$.
- If $I=\int_{0}^{2\pi} F(\sin\theta,\cos\theta)\,\d \theta$, you may be able to be evaluate the integral through the variable change $z=\e^{i\theta}$, which turns the real $\theta$ integral into a contour integral around the unit circle $\abs z=1$.
- Integrands of the form $f(z) \cot z$, $f(z)/ \sin z$, etc., may be used to evaluate series of the form $\sum_i \pm f(x_i)$ where $x_i$ are the zeros of $\tan x$ or $\sin x$. The most convenient contour to use is normally a large square of side $L$, each of whose vertical sides crosses the real axis half way between a pair of poles. Then provided $f(z)$ falls off faster than $1/L$ as $L\to\infty$, the contour integral will tend to zero.

Spiegel 7.4-8

**Riley 18.16-19**; **Boas 14.7**; Arfken 7.2