Jordan's lemma中函数需满足的条件

hbghlyj

没看出证明哪里用到meromorphic似乎只用到$f$有界+连续
mathworld

for "nice" functions which satisfy $\lim_{R\to\infty}|f(Re^{i\theta})|=0$

wikipedia
complex.pdf page 65

Lemma 11.7. (Jordan's Lemma): Let $f: \mathbb{H} \rightarrow \mathbb{C}_{\infty}$ be a meromorphic function on the upper-half plane $\mathbb{H}=\{z \in \mathbb{C}: \Im(z)>0\}$. Suppose that $f(z) \rightarrow 0$ as $z \rightarrow \infty$ in $\mathbb{H}$. Then if $\gamma_R(t)=R e^{i t}$ for $t \in[0, \pi]$ we have
\[
\int_{\gamma_R} f(z) e^{i \alpha z} d z \rightarrow 0
\]
as $R \rightarrow \infty$ for all $\alpha \in \mathbb{R}_{>0}$.

Proof. Suppose that $\epsilon>0$ is given. Then by assumption we may find an $S$ such that for $|z|>S$ we have $|f(z)|<\epsilon$. Thus if $R>S$ and $z=\gamma_R(t)$, it follows that
\[
\left|f(z) e^{i \alpha z}\right|=|f(z)|e^{\Re(i\alpha z)}=|f(z)|e^{-\alpha R\sin(t)}< \epsilon e^{-\alpha R \sin (t)} .
\]
But now applying Lemma 11.6 to the function $g(t)=\sin (t)$ with $x=0$ and $y=\pi / 2$ we see that $\sin (t) \geq \frac{2}{\pi} t$ for $t \in[0, \pi / 2]$. Similarly we have $\sin (\pi-t) \geq 2(\pi-t) / \pi$ for $t \in[\pi / 2, \pi]$. Thus we have
\[
\left|f(z) e^{i \alpha z}\right| \leq\left\{\begin{array}l
\epsilon . e^{-2 \alpha R t / \pi} &, t \in[0, \pi / 2] \\
\epsilon . e^{-2 \alpha R(\pi-t) / \pi} &, t \in[\pi / 2, \pi]
\end{array}\right.
\]
But then it follows that
\[
\left|\int_{\gamma_R} f(z) e^{i \alpha z} d z\right| \leq 2 \int_0^{\pi / 2} \epsilon R . e^{-2 \alpha R t / \pi} d t=\epsilon . \pi \frac{1-e^{-\alpha R}}{\alpha}<\epsilon . \pi / \alpha,
\]
Thus since $\pi / \alpha>0$ is independent of $R$, it follows that $\int_{\gamma_R} f(z) e^{i \alpha z} d z \rightarrow 0$ as $R \rightarrow \infty$ as required.

hbghlyj

在I_6.pdf提到
$\int_{-\infty}^\infty\frac{e^{iz}}z$不可积

1. Integrate[Exp[I z]/z, {z, -Infinity, Infinity}]

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但是它的主值是$\pi$
$$PV\int_{-\infty}^\infty\frac{e^{iz}}z=\pi$$

1. Integrate[Exp[I z]/z, {z, -Infinity, Infinity}, PrincipalValue -> True]

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用semicircle contour$$\operatorname*{Res}_{z=0}\frac{e^{iz}}z=1$$

1. Residue[Exp[I z]/z, {z, 0}]

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$\frac{e^{iz}}z$在半圆弧$R$上的积分$=i (\pi -2 \text{Si}(R))$

1. Assuming[R > 0,
2. Integrate[
3.   Exp[I R Exp[Pi I t]]/(R Exp[Pi I t]) R Pi I Exp[Pi I t], {t, 0, 1}]]

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因为$\text{Si}(R)\to\frac\pi2$,这个表达式的极限为0

hbghlyj

complex.pdf page67

Lemma 11.11. Let $f: U \rightarrow \mathbb{C}$ be a meromorphic function with a simple pole at $a \in U$ and let $\gamma_\epsilon:[\alpha, \beta] \rightarrow \mathbb{C}$ be the path $\gamma_\epsilon(t)=a+\epsilon e^{i t}$, then
\[
\lim _{\epsilon \rightarrow 0} \int_{\gamma_\epsilon} f(z) d z=\operatorname{Res}_a(f) \cdot(\beta-\alpha) i .
\]

Proof. Since $f$ has a simple pole at $a$, we may write
\[
f(z)=\frac{c}{z-a}+g(z)
\]
Where $g(z)$ is holomorphic near $z$ and $c=\operatorname{Res}_a(f)$ (indeed $c /(z-a)$ is just the principal part of $f$ at $a$). But now as $g$ is holomorphic at $a$, it is continuous at $a$, and so bounded. Let $M, r>0$ be such that $|g(z)|<M$ for all $z \in B(a, r)$. Then if $0<\epsilon<r$ we have
\[
\left|\int_{\gamma_\epsilon} g(z) d z\right| \leq \ell\left(\gamma_\epsilon\right) M=(\beta-\alpha) \epsilon \cdot M,
\]
which clearly tends to zero as $\epsilon \rightarrow 0$. On the other hand, we have
\[
\int_{\gamma_\epsilon} \frac{c}{z-a} d z=\int_\alpha^\beta \frac{c}{\epsilon e^{i t}} i \epsilon e^{i t} d t=\int_\alpha^\beta(i c) d t=i c(\beta-\alpha)
\]
Since $\int_{\gamma_\epsilon} f(z) d z=\int_{\gamma_\epsilon} c /(z-a) d z+\int_{\gamma_\epsilon} g(z) d z$ the result follows.

hbghlyj

$\frac{\exp (i z)}{\sqrt{z}}$负实轴branch cut
但在$(0,\infty)$可积
$$\int_0^{\infty }\frac{\exp (i z)}{\sqrt{z}}\, dz=e^{i\pi\over4}\sqrt{\pi}$$

1. Assuming[0 < k < 1, Integrate[Exp[I z]/z^k, {z, 0, Infinity}]]

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对于$0<k<1$有
$$\int_0^{\infty } \frac{\exp (i z)}{z^k} \, dz=i (-i)^k \Gamma (1-k)$$根据Gamma函数的定义

hbghlyj

Integral transform page 93

The LT of $δ(x−a)$ is $e^{−pa}$. By inversion formula

$\delta(x-\mathrm{a})=\frac{1}{2 \pi \mathrm{i}} \int_{\sigma-\mathrm{i} \infty}^{\sigma+\mathrm{i} \infty} \mathrm{e}^{-p a} \mathrm{e}^{p x} \mathrm{~d} p=\frac{1}{2 \pi \mathrm{i}} \int_{\sigma-\mathrm{i} \infty}^{\sigma+\mathrm{i} \infty} \mathrm{e}^{p(x-\mathrm{a})} \mathrm{d} p$

Now in this contour integral (in the $p$-plane),
If $x<a$, you can close the contour with a semicircle on the right, as $x-a<0$ gives exponential decay.¹² By Cauchy's theorem, the integral is zero.

---

¹² NB you can't use Jordan directly to show decay of the arc-integral because the maximum on the arc of $\left|\mathrm{e}^{p(x-a)}\right|$ does not tend to zero. You have to use the rapid oscillation property as well, as in the R-L Lemma. (You might like to try this, by splitting the range of integration as we did when we showed that a LT vanishes at $+\infty$.) But the result is true.