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hbghlyj
Posted at 2023-2-24 07:15:15
We want to find the image of the region $\cases{0\leq y\leq 1\\y\leq x\leq \frac{1}{y}}$ under the change of variable $\cases{u=xy\\v=\frac yx}$.
Rewrite the inequalities in terms of $u,v$:$$0\le x\leq\frac{1}{y}\xRightarrow{\times y}0 \leq u \leq 1$$
$$0\le y\leq x\xRightarrow{\div x}0\le v \leq 1$$
$\therefore$ 在新变量 $u,v$ 下的边界为 $[0,1]\times[0,1]$. 和1楼所给结果相同.
1楼公式漏了)号, 应该是
$$I=\int_0^1\int_y^{\frac 1y} \frac {y^3}x e^{y^2(x^2+x^{-2})}\rmd x\rmd y$$
The Jacobian of the transformation $\begin{cases}u = xy \\ v = \frac{y}{x}\end{cases}$ is
$$
\mathbf{J} = \begin{pmatrix}
\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\
\frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}
\end{pmatrix}
= \begin{pmatrix}
y & x \\
-\frac{y}{x^2} & \frac{1}{x}
\end{pmatrix}
$$The determinant of the Jacobian is:
$$
\det(\mathbf{J}) = y \cdot \frac{1}{x} - x \cdot \left(-\frac{y}{x^2}\right) = \frac{2y}{x}=2v
$$
The Jacobian of the inverse transformation is:
$$
\det(\mathbf{J})^{-1}=\frac1{2v}
$$
Substituting $\cases{x = \sqrt{uv}\\y = \sqrt{u\over v}}$ into the integrand and simplify:
$$\frac{y^3}{x} e^{y^2(x^2+x^{-2})}=uv^2e^{u^2+v^2} $$
Now the integral becomes
\begin{align*}\int_0^1\int_0^1uv^2e^{u^2+v^2} \times\frac1{2v}\rmd u\rmd v&=\frac12\int_0^1\int_0^1e^{u^2+v^2}uv\rmd u\rmd v\end{align*}The double integral is separable:
\begin{align*}\frac12\int_0^1\int_0^1e^{u^2+v^2}uv\rmd u\rmd v&=\frac12\left(\int_0^1e^{u^2}u\rmd u\right)\left(\int_0^1e^{v^2}v\rmd v\right)\\&=\frac12\left(\int_0^1e^{u^2}u\rmd u\right)^2\\&=\frac12\left(\frac12e^{u^2}|_0^1\right)^2
\end{align*}
$\text{Final result}=\frac{(e-1)^2}{8}=$
终于筭对了 |
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