Lecture 34
Example 1: \(f(z)=\dfrac{z+i}{z^2+9}\). \(f\) is analytic on \(\C\setminus\{\pm 3i\}\).
\(\pm 3i \) are isolated singularities.
Near \(z=3i\), write \[f(z) = \frac{\phi(z)}{z-3i} \] where \(\phi(z) = \dfrac{z+i}{z+3i}\) is analytic and non-zero near \(z=3i\).
Example 1 (cont): Theorem 2 (Lec 33) \(\Rightarrow\) simple pole at \(z=3i\), and \((**) \Rightarrow\)
\(\underset{z=3i}{\res} \,f (z) =\phi(3i)\) \(= \dfrac{2}{3}\).
📝 Do for \(z=-3i\).
Example 2: \(f(z)=\dfrac{z^3+2z}{(z-i)^3}\). \(f\) is analytic on \(\C\setminus\{ i \}\).
Near \(z=i\), we have \(f(z) = \dfrac{\phi(z)}{(z-i)^3} \) where \(\phi(z) = z^3+2z\) is analytic and \(\phi(i)= i \neq 0\).
Theorem 1 (Lec 33) \(\Rightarrow\) pole of order 3 at \(z=i\), so
\(\underset{z=i}{\res} \,f(z)=\dfrac{\phi^{(2)}(i)}{2!}\) \(= 3i\).
Suppose that \(f\) analytic at \(z_0\): \(f\) has a zero of order \(m\) at \(z_0\) if \[ \left\{ \begin{array}{lr} f^{\left(j\right)}(z_0) = 0 & j=0, 1, \ldots , m-1\\ f^{\left(m\right)}(z_0) \neq 0 & \end{array} \right.\]
E. g. \(f(z)=(z-i)^4(z-4)\) has a zero of order \(4\) at \(i\) and a simple zero at \(4\).
Theorem 1: \(f\) analytic at \(z_0\) has a zero of order \(m\) at \(z_0\) if and only if \[f(z)=(z-z_0)^mg(z)\] with \(g\) analytic and \(g(z_0)\neq 0\).
Theorem 2: Suppose \(p\) and \(q\) are analytic at \(z_0\), \(p(z_0)\ne 0,\) \(q\) has a zero of order \(m\) at \(z_0\). Then \(p/q\) has a pole of order \(m\) at \(z_0\).
Example: \(p(z)=1\) and \(q(z)=z(e^z-1)\). \(p/q\) has an isolated singularity at \(0\). \(p\) is analytic and nonzero.
\(q(0)=0, \;\quad\quad\)
\(q'(0)= 0, \quad\;\;\) 📝
\(q''(0)= 2\neq 0\). 📝
Theorem 2 \(\Rightarrow\) has a pole of order \(2\) at \(0\).
Theorem 3: Let \(p\), \(q\) be analytic at \(z_0\). If \(p(z_0)\neq 0\), \(q(z_0)=0\), \(q'(z_0)\neq 0\). Then \(p/q\) has a simple at \(z_0\), and \[\underset{z=z_0}{\res}\frac{p(z)}{q(z)} = \frac{p(z_0)}{q'(z_0)}\]
Note: There exist higher order analogues but they are messy.
Recall: \[{\small\int_{-\infty}^{\infty} f(x)\,dx = \lim_{M_1\ra -\infty}\int_{M_1}^{0} f(x)\,dx + \lim_{M_2\ra \infty}\int_{0}^{M_2} f(x)\,dx}\] if both of the limits exist.
Remark: We can replace \(0\) by any fixed \(c\in \R\).
You cannot in general replace RHS by \[\lim_{M\ra -\infty}\int_{-M}^{M} f(x)\,dx .\]
If you do this anyhow, it defines the Cauchy Principal value (PV) integral.
E. g. \(\ds\int_{-\infty}^{\infty}x \,dx\) \(=\ds\lim_{M_1\ra -\infty}\int_{M_1}^{0} x\,dx + \lim_{M_2\ra \infty}\int_{0}^{M_2} x\,dx\)
\(\qquad\qquad\quad = \ds\lim_{M_1\ra -\infty} \frac{-M_1^2}{2} + \lim_{M_2\ra \infty} \frac{M_2^2}{2}\)
\(\qquad\qquad\quad = \) undefined, but
\(\ds\text{PV} \int_{-\infty}^{\infty}x\, dx = \lim_{M\ra \infty}\int_{-M}^{M} x\,dx\) \(=\ds \lim_{M\ra \infty}\left[\frac{M^2}{2}-\frac{M^2}{2}\right]\) \(=\ds \lim_{M\ra \infty}\left[0\right]\) \(=0\).
When is \(\text{PV}\ds \int_{-\infty}^{\infty} f = \int_{-\infty}^{\infty}f\,\)? E. g. for \(f\) even (or for \(f\geq 0\)).
If \(f\) is even, i. e. \(f(x) = f(-x)\) \(\forall x\in \R\), then
\(\ds\int_{0}^{\infty} f(x)\, dx = \frac{1}{2} \int_{-\infty}^{\infty} f(x)\, dx\) \(=\ds \frac{1}{2}\text{PV} \int_{-\infty}^{\infty} f(x)\, dx\)
and all of these converge or diverge together.