PHYS20672 Summary 5

The Residue Theorem

image of mappings

  1. If we expand a function $f(z)$ about the point $z=a$ in a Laurent series, valid for $0<|z-a| < R$: $$f(z)=\sum_{n=1}\frac{b_n}{(z-a)^n} +\sum_{n=0}a_n(z-a)^n $$ 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 $z=a$.
  2. Hence for a curve $C$ which encloses a number of poles of $f(z)$, the residue theorem states that $$\oint_C f(z)\d z=2\pi i \text{(Sum of residues at all poles inside $C$)}$$
  3. The residue at each pole is the term $b_1$ in the Laurent expansion in the vicinity of that pole; this expansion will be different for each pole.
  4. For a function $f(z)$ with a simple pole at $z=a$, we can use one of the following formulae for the residue: ($g(z)$ and $h(z)$ represent functions which are analytic about $a$, with $g(a)$ finite and $h(a)=0$): $$b_1=\lim_{z\to a}\bigl((z-a) f(z)\bigr);\quad \text{for } f(z)=g(z)/h(z),\; b_1=g(a)/h'(a);\quad \text{for } f(z)=g(z)/(z-a),\; b_1=g(a).$$
  5. For a function $f(z)$ with a pole of order $n$ at $z=a$, $$b_1=\lim_{z\to a}\frac 1 {(n-1)!}\frac{\d^{n-1} }{\d z^{n-1}}\bigl((z-a)^n f(z)\bigr).$$
  6. 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.

Spiegel 7.1-3

Riley 18.14; Boas 14.5, 14.6 ; Arfken 7.2

Real integrals via contour integration

image of mappings

  1. The contours above are examples of contours which can be used to calculate real integrals
  2. 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 $$ I=\lim_{R\to\infty}I_1= \lim_{R\to\infty}\left(\oint_C f(z)\d z-I_2\right) =2\pi i\text{(Sum of residues at all poles inside $C$)}$$
  3. If $I=\int_{-\infty}^\infty f(x)\e^{ikx}\d x$, and (i) $k>0$, (ii) $f(z)$ has a finite number of poles in the upper half plane and (iii) $|f(R\e^{i\theta})|\to 0$ as $R\to \infty$, 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.
  4. 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 $\lim_{\epsilon\to 0}\int_{-\infty}^{a-\epsilon} f(x)\d x+\int_{a+\epsilon}^{\infty} f(x)\d x$. 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.
  5. 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 $f(x)\ne f(x\e^{2\pi i})$.
  6. If $I=\int_{0}^{2\pi} F(\sin\theta,\cos\theta)\d \theta$, the integral may be able to be evaluated through the variable change $z=\e^{i\theta}$ which maps the real $\theta$ integral onto a contour integral round the unit circle.
  7. Integrands of the form $f(z) \tan z$, $f(z)/ \sin z$ etc may be used to evaluate series of the form $\sum_i \pm f(x_i)$ where $x_i$ are the zeros of $\sin x$ or $\cos x$. The relevant contour integral is a large circle whose radius $R$ is such that the circle crosses the real axis half way between a pair of poles. Then provided $f(z)$ falls off faster than $1/R$ as $R\to\infty$, the contour integral will tend to zero.

Spiegel 7.4-8

Riley 18.16-19; Boas 14.7; Arfken 7.2