$x e^{a x+b y+c z}$球体积分

hbghlyj

For real constants $a,b,c$, show that$$\int_{R} x e^{a x+b y+c z} \mathrm{~d} V=\frac{4 \pi a}{q^{5}}\left(\left(3+q^{2}\right) \sinh q-3q \cosh q\right)$$where $R$ is the region $x^2+y^2+z^2\leqslant 1$ and $q=\sqrt{a^2+b^2+c^2}$.

Czhang271828

math.stackexchange.com/questions/3526310/triple-integral-iiint-x ... unit-sphere-x2y2z2-l

一般都旋转坐标轴换元吧

hbghlyj

设$p=\frac{ax+by+cz}q$, 那么在$p=$定值的截面上, 函数$ (ax+by+cz) e^{a x+b y+c z}$为定值.
乘以$p=$定值的截面面积然后做一元积分:
\begin{align}\int_{R} (ax+by+cz) e^{a x+b y+c z} \mathrm{~d} V\nonumber&=\int_{R} pq e^{pq} \mathrm{~d} V\\\nonumber&=\int_{-1}^1 pq e^{pq}π(1-p^2)dp\nonumber\\&=\frac{4\pi}{q^3} \left(\left(3+q^2\right) \sinh q-3 q \cosh q\right)\label1\end{align}
根据1楼公式有
\[\int_{R} ax e^{a x+b y+c z} \mathrm{~d} V=\frac{4 \pi a^2}{q^{5}}\left(\left(3+q^{2}\right) \sinh q-3q \cosh q\right)\]同理\[\int_{R} by e^{a x+b y+c z} \mathrm{~d} V=\frac{4 \pi b^2}{q^{5}}\left(\left(3+q^{2}\right) \sinh q-3q \cosh q\right)\]和\[\int_{R} cz e^{a x+b y+c z} \mathrm{~d} V=\frac{4 \pi c^2}{q^{5}}\left(\left(3+q^{2}\right) \sinh q-3q \cosh q\right)\]
加起来和\eqref{1}相同✅

hbghlyj

设$q=\sqrt{a^2+b^2}$.
\[\int_{-1}^1 2 \sqrt{1-p^2} \cdot pq e^{p q} dp=2 \pi  I_2(q)\]
其中$I_2$是$n=2$的BesselI函数.
\[\int_R x e^{a x+b y} \, dxdy=\frac{2 \pi a}{q^2} I_2(q)\]其中$R$是区域$x^2+y^2\leqslant 1$.

---

验证:

1. a=RandomReal[];b=RandomReal[];q=Sqrt[a^2+b^2];
2. NIntegrate[x Exp[a x+b y],{x,y}\[Element]Disk[]]
3. 2Pi a/q^2 BesselI[2,q]

复制代码

hbghlyj

\[\int_{-1}^1 \frac{4 \pi}3\left(1-p^2\right)^{3/2}\cdot p q e^{pq} dp=\frac{4 \pi ^2}q I_3(q)\]其中$I_3$是$n=3$的BesselI函数.
\[\int_R x e^{a x+b y+cz+dt} \, dxdydzdt=\frac{4 \pi^2 a}{q^3} I_3(q)\]其中$R$是区域$x^2+y^2+z^2+t^2\leqslant 1$.

---

验证:

1. {a,b,c,d}=Table[RandomReal[],4];
2. q=Sqrt[a^2+b^2+c^2+d^2];
3. NIntegrate[x Exp[a x+b y+c z+d t],{x,y,z,t}\[Element]Ball[4]]
4. 4Pi^2 a/q^3 BesselI[3,q]

复制代码

hbghlyj

hbghlyj 发表于 2023-1-28 00:38
二维
\[\int_R x e^{a x+b y} \, dxdy=\frac{2 \pi a}{q^2} I_2(q)\]

hbghlyj 发表于 2023-1-28 01:01
四维\[\int_R x e^{a x+b y+cz+dt} \, dxdydzdt=\frac{4 \pi^2 a}{q^3} I_3(q)\]

用2#方法可以证明吗