I have trouble in doing a substitution. That is one step in a whole demonstration. I need to compute this integral : $\epsilon << 1 $
$$ \\ \int_{ [0,1-\epsilon] } \int_{ [0,1-\epsilon] } \frac 1 {1-xy} \, dx dy$$ And for that I need to do the substitution :
$$ x = u-v \, ; \, y = u + v $$
Can you help me to do it? I'm not asking for the whole calculus, I have the answer under the eyes. My problem is that it is one of the first multi-integral I'm computing and I struggle to understand what is depending of what, what do I need to write like a function of the rest etc...
For example,
- I know that I need to compute the determinant so this I ve done : it gives 2.
- I know that I need to find the lines that bound my submanifold, and so I have written :
$$ \left\{ \begin{array}{c} y = 0 \\ x = 0 \\ y = 1 - \epsilon \\ x = 1 - \epsilon \end{array} \right. \implies \left\{ \begin{array}{c} u= -v \\ u = v \\ u+v = 1 - \epsilon \\ u-v = 1 - \epsilon \end{array} \right. $$
- I know that I need to find the boundaries for u and v : $$ \left\{ \begin{array}{c} 0 < x < 1 - \epsilon \\ 0 < y < 1 - \epsilon \end{array} \right. \implies \left\{ \begin{array}{c} 0 < u < 1 - \epsilon \\ -1 + \epsilon < v < 1 - \epsilon \end{array} \right. $$
I would be so greatfull if you can help me figure out what I need to do in order to have a general method to deal with problems like this. thank you!
P.S. : please, so many times, people told me just to do a graph... I have done it !!!! I just need some help at first, and I'm so sorry guys to ask you this, but it's really necessary for me... I tried for 2 days now :(
On the $uv$-plane, we can graph the intersection of the following inequalities
$$ \begin{align} u-v > 0 \\ u-v < 1 - \epsilon \\ u+v > 0 \\ u+v < 1 - \epsilon \end{align} $$
Here is what the intersection looks like.
When $u \in \left[0, \dfrac{1-\epsilon}{2}\right]$, it follows then that $v \in [-u,u]$. Otherwise, when $u \in \left[\dfrac{1-\epsilon}{2}, 1-\epsilon\right]$, it follows that $v \in [(u+\epsilon) - 1, 1 - (u+\epsilon)]$. Setting up the integrals:
$$I = \int_{0}^{(1-\epsilon)/2} \int_{-u}^{u} \frac{2}{1-(u^2-v^2)} \, dvdu + \int_{(1-\epsilon)/2}^{1-\epsilon} \int_{(u+\epsilon)-1}^{1-(u+\epsilon)} \frac{2}{1-(u^2-v^2)} \, dvdu$$