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I'm having trouble with this double integral:

$$\int_0^2\int_0^{2-x} \exp\left(\frac{x−y}{x+y}\right)\text dy\,\text dx$$

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closed as off-topic by user223391, graydad, abiessu, Jonas Meyer, user147263 May 27 '15 at 3:21

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  • $\begingroup$ this is first one. $\endgroup$ – SHM May 26 '15 at 17:44
  • $\begingroup$ $dxdy$? or $dydx$? $\endgroup$ – 3SAT May 26 '15 at 17:46
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    $\begingroup$ first of all i would recommend a change of variables $x-y=v, \quad x+y=u$ with jacobian $1/2$ $\endgroup$ – tired May 26 '15 at 17:50
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    $\begingroup$ It is not appropriate to phrase questions posted here in a way suitable for assigning homework. That is probably the reason why some people are voting to close the question. They ought to have told you that, but around here that usually doesn't happen. $\endgroup$ – Michael Hardy May 26 '15 at 17:55
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    $\begingroup$ homework? i am graduated in computer software engineering :D just remembering these stuff for fun. $\endgroup$ – SHM May 26 '15 at 18:02
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The integral is one over the $2$-simplex $\Delta_2(2) = \{ (x,y ) : x, y \ge 0, x+ y \le 2 \}$.

One standard trick to deal with integral over $d$-simplex of the form $$\Delta_d(L) = \{ (x_1, x_2, \ldots, x_d ) : x_i \ge 0, \sum_{i=1}^d x_i \le L \}$$ is convert it to one over the $d$-cuboid $[0,L] \times [0,1]^{d-1}$ through following change of variables

$$\begin{align} \lambda &= x_1 + x_2 + \cdots + x_d\\ \lambda\mu_1 &= x_1 + x_2 + \cdots + x_{d-1}\\ \lambda\mu_1\mu_2 &= x_1 + x_2 + \cdots + x_{d-2}\\ &\;\;\vdots\\ \lambda\mu_1\mu_2\cdots\mu_{d-1} &= x_1 \end{align} $$ Under such change of variables, an integral of $(x_1,\ldots,x_d)$ over $\Delta_d(L)$ becomes an integral of $(\lambda,\mu_1,\ldots,\mu_{d-1})$ over $[0,L] \times [0,1]^{d-1}$.

For the integral at hand, let $\lambda = x + y$ and $x = \mu\lambda$. The area element can be rewritten as

$$dx \wedge dy = dx \wedge d(x+y) = d(\mu \lambda) \wedge d\lambda = \lambda d\mu \wedge d\lambda$$

So the integral becomes

$$\int_{\Delta_2(2)} \exp\left(\frac{x-y}{x+y}\right)dxdy = \int_0^2 \int_0^1 e^{2\mu - 1} \lambda d\mu d\lambda = \left(\int_0^2 \lambda d\lambda \right)\left(\int_0^1 e^{2\mu-1} d\mu\right)\\ = 2 \times \frac{1}{2e}(e^2 - 1) = 2\sinh(1) \approx 2.35040238728760291376476370119120163 $$

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    $\begingroup$ thanks for the clear answer. $\endgroup$ – SHM May 26 '15 at 18:31
  • $\begingroup$ Did I understand this correctly: For $\lambda$ alone, we treat it as a variable but for $\mu_1 \lambda$ we treat $\lambda$ as a constant. And if we had $\mu_2 \mu_1 \lambda$ then $\mu_1 \lambda$ would be treated as constants? $\endgroup$ – grdgfgr May 26 '15 at 23:19
  • $\begingroup$ @grdgfgr Are you referring to the $d(\mu\lambda)\wedge d\lambda = \lambda d\mu\wedge d\lambda$ part? In that case, the answer is no, both $\lambda$ and $\mu$ are variables all the time. The full derivation of above should be $$d(\mu\lambda)\wedge d\lambda = (\lambda d\mu + \mu d\lambda) \wedge d\lambda = \lambda d\mu\wedge d\lambda + \mu d\lambda \wedge d\lambda = \lambda d\mu\wedge d\lambda $$ The last equality is true because wedge product between 1-forms are anti-commutative. i.e for any function $f, g$, $df \wedge dg = - dg \wedge df$. In particular $d\lambda \wedge d\lambda = 0$. $\endgroup$ – achille hui May 27 '15 at 4:25
  • $\begingroup$ Remember, if you exchange two rows (or columns) of a matrix, the determinant get an extra minus sign. Just think of above manipulation of differential forms as a efficient ways for computing the Jacobian. $\endgroup$ – achille hui May 27 '15 at 4:26

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