积分换元

业余的业余

画出积分 $$I=\int_0^1\int_y^{\cfrac 1y} \cfrac {y^3}x e^{y^2(x^2+x^{-2}}\mathrm{d}x\mathrm{d}y$$
的积分域， 描述其在新变量 $u=xy$ 和 $v=\cfrac yx$ 下的边界，从而对 $I$ 求值。

折腾好久，终于看到一线曙光。主要是积分区域的变换有点烧脑，后来发现是 $[0,1]\times[0,1]$ (不知正确与否), 最后算出来值为  $\cfrac {(e-1)^2}4$。 请高手确认下区间转化是否正确。

如果没有换元，直接算，要求 $\int \cfrac {e^x}x \mathrm{d} x$ 型的积分，刚好看到紧下一楼的讨论，没有初等形式。

战巡

回复 1# 业余的业余


区间没错，但最后结果应该是$\frac{(e-1)^2}{8}$，我估计你可能雅可比式出了问题

业余的业余

回复 2# 战巡

谢谢！

hbghlyj

$\begin{rcases}0\leq y\leq 1\\y\leq x\leq \frac{1}{y}\end{rcases}\implies$import graph;
size(8cm);
path p=graph(new real (real y) {return 1/y;},1/3,1)--(0,0)--(0,3);
p=xscale(-1)*rotate(90)*p;
draw(p,linewidth(2)+linecap(0));
fill(p--cycle,gray(0.8));
xaxis("$x$",xmax=3);
yaxis("$y$",ymax=1.5);
label("1",(1,0),S);
label("1",(0,1),W);
draw((0,1)--(1,1)--(1,0), dashed+linewidth(0.5));

hbghlyj

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
$$I=\int_0^1\int_y^{\frac 1y} \frac {y^3}x e^{y^2(x^2+x^{-2})}\rmd x\rmd y$$import graph;

unitsize(4cm);

filldraw(box((0,0),(1,1)), fillpen=paleyellow);

xaxis("$u$", xmax=1.2, Arrow);
yaxis("$v$", ymax=1.2, Arrow);
label("1",E,S);
label("1",N,W);

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*}
$\text{Final result}=\frac{(e-1)^2}{8}=$real result = 1/8 * (exp(1) - 1) ^2;
label((0,0),string(result));
终于筭对了