From 34c52947fb493edcaa9a01c27e4663a5f1f31bba Mon Sep 17 00:00:00 2001
From: unlockable <g@unlockableworld.com>
Date: Thu, 28 Sep 2023 13:23:35 +0800
Subject: [PATCH] =?UTF-8?q?=E6=B7=BB=E5=8A=A0Poisson=E7=A7=AF=E5=88=86?=
 =?UTF-8?q?=E8=AE=A1=E7=AE=97=E3=80=82?=
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---
 16多重积分.tex | 27 ++++++++++++++++++++++++++-
 1 file changed, 26 insertions(+), 1 deletion(-)

diff --git a/16多重积分.tex b/16多重积分.tex
index 3e59a75..6cea000 100644
--- a/16多重积分.tex
+++ b/16多重积分.tex
@@ -472,4 +472,29 @@
 这时
 \[\frac{\partial (x, y, z)}{\partial (r, \theta, \varphi)} = r^2 \sin \theta\]
 则
-\[\int \limits_\Omega f \dif \mu = \iiint \limits_{\tilde{\Omega}} f(r \sin \theta \cos \varphi, r \sin \theta \sin \varphi, r \cos \theta)\eqper\]
\ No newline at end of file
+\[\int \limits_\Omega f \dif \mu = \iiint \limits_{\tilde{\Omega}} f(r \sin \theta \cos \varphi, r \sin \theta \sin \varphi, r \cos \theta)\eqper\]
+
+\begin{example}
+    计算Poisson积分$I = \dint_{-\infty}^{+\infty} e^{-x^2} \dif x$。
+\end{example}
+
+\begin{proof}[解]
+    首先，我们回忆
+    \begin{align*}
+        V(R) & = \iint \limits_{x^2 + y^2 \leq R^2} e^{-(x^2 + y^2)} \dif x \dif y\\
+        & = \int_0^R \dif r \int_0^{2\pi}e^{-r^2} r \dif \theta\\
+        & = \pi (1 - e^{-R^2})
+    \end{align*}
+    再引入记号
+    \[I_R = \int_{-R}^{+R} e^{-x^2} \dif x, \quad D_R = [-R, R] \times [-R, R]\]
+    那么
+    \begin{align*}
+        I_R^2 & = \int_{-R}^{+R} e^{-x^2} \dif x \int_{-R}^{+R} e^{-y^2} \dif y\\
+        & = \iint \limits_{D_R} e^{-(x^2 + y^2)} \dif x \dif y
+    \end{align*}
+    利用区域可加性和积分的保号性，有
+    \[V(R) \leq I_R^2 \leq V(\sqrt{2}R)\]
+    由此得到
+    \[I^2 = \tolim{R}{+\infty} I_R^2 = \tolim{R}{+\infty} V(R) = \pi\]
+    因此$I = \sqrt{\pi}$。
+\end{proof}
\ No newline at end of file