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Question Number 89384 by nimnim last updated on 17/Apr/20

Commented by mathmax by abdo last updated on 17/Apr/20

$l e t t r y a n o t h e r w a y w e c o n s i d e r t h e d i f f e o m o r p h i s m$ $(u, v) \to (x, y) / x - y = u a n d x + y = v \Rightarrow x = \frac{u + v}{2} a n d y = \frac{- u + v}{2}$ $\Rightarrow φ (u, v) = (φ_{1} (u, v), φ_{2} (u, v)) = (x, y) = (\frac{u}{2} + \frac{v}{2}, - \frac{u}{2} + \frac{v}{2})$ $w e h a v e \frac{1}{2} ⩽ x ⩽ 1 a n d \frac{1}{2} ⩽ y ⩽ 1 \Rightarrow 1 ⩽ x + y ⩽ 2 \Rightarrow v \in [1, 2]$ $- 1 ⩽ - y ⩽ - \frac{1}{2} \Rightarrow - \frac{1}{2} ⩽ x - y ⩽ \frac{1}{2} \Rightarrow u \in [- \frac{1}{2}, \frac{1}{2}]$ $M_{j} (φ) = (\begin{matrix} \frac{\partial φ_{1}}{\partial u} \frac{\partial φ_{1}}{\partial v} \\ \frac{\partial φ_{2}}{\partial u} \frac{\partial φ_{2}}{\partial v} \end{matrix}) = (\begin{matrix} \frac{1}{2} \frac{1}{2} \\ - \frac{1}{2} \frac{1}{2} \end{matrix})$ $∣ {d e t M}_{j} (φ) ∣ = \frac{1}{2}$ $\int \int_{D} f (x, y) d x d y = \int \int_{w} f o φ (u, v) ∣ J_{φ} ∣ d u d v$ $\int \int_{u \in [- \frac{1}{2}, \frac{1}{2}] a n d v \in [1, 2]} \frac{u}{v} \times \frac{1}{2} d u d v$ $= \frac{1}{2} \int_{1}^{2} (\int_{- \frac{1}{2}}^{\frac{1}{2}} u d u) \frac{d v}{v} = \frac{1}{2} \int_{1}^{2} {[\frac{u^{2}}{2}]}_{- \frac{1}{2}}^{\frac{1}{2}} \frac{d v}{v} = 0$
Commented by Tony Lin last updated on 17/Apr/20
$∫_(1/2) ^1 ∫_(1/2) ^1 ((x−y)/(x+y))dxdy =∫_(1/2) ^1 ∫_(1/2) ^1 (1−((2y)/(x+y)))dxdy =∫_(1/2) ^1 {[x−2aln∣x+y∣]_(1/2) ^1 }dy =∫_(1/2) ^1 ((1/2)+2yln∣(((1/2)+y)/(1+y))∣)dy =(1/4)−2∫_(1/2) ^1 yln∣(((1/2)+y)/(1+y))∣dy y>0→ln∣(((1/2)+y)/(1+y))∣=ln((((1/2)+y)/(1+y))) ∫_(1/2) ^1 yln((((1/2)+y)/(1+y)))dy =(1/2)y^2 ln((((1/2)+y)/(1+y)))∣_(1/2) ^1 −(1/2)∫_(1/2) ^1 (y^2 /(2y^2 +3y+1))dy =−(1/8)ln(((512)/(243)))−(1/2)∫_(1/2) ^1 (y^2 /(2y^2 +3y+1))dy =−(1/8)ln(((512)/(243)))−(1/2)∫_(1/2) ^1 [(1/(2(2y+1)))−(1/(y+1))+(1/2)] =−(1/8)ln(((512)/(243)))−(1/8)ln((3/2))+(1/2)ln((4/3))−(1/8) =−(1/8) ∫_(1/2) ^1 ∫_(1/2) ^1 ((x−y)/(x+y)) dxdy=(1/4)−2×(1/8)=0$
$\int_{\frac{1}{2}}^{1} \int_{\frac{1}{2}}^{1} \frac{x - y}{x + y} d x d y$ $= \int_{\frac{1}{2}}^{1} \int_{\frac{1}{2}}^{1} (1 - \frac{2 y}{x + y}) d x d y$ $= \int_{\frac{1}{2}}^{1} {{[x - 2 a l n ∣ x + y ∣]}_{\frac{1}{2}}^{1}} d y$ $= \int_{\frac{1}{2}}^{1} (\frac{1}{2} + 2 y l n ∣ \frac{\frac{1}{2} + y}{1 + y} ∣) d y$ $= \frac{1}{4} - 2 \int_{\frac{1}{2}}^{1} y l n ∣ \frac{\frac{1}{2} + y}{1 + y} ∣ d y$ $y > 0 \to l n ∣ \frac{\frac{1}{2} + y}{1 + y} ∣= l n (\frac{\frac{1}{2} + y}{1 + y})$ $\int_{\frac{1}{2}}^{1} y l n (\frac{\frac{1}{2} + y}{1 + y}) d y$ $= \frac{1}{2} y^{2} l n (\frac{\frac{1}{2} + y}{1 + y}) ∣_{\frac{1}{2}}^{1} - \frac{1}{2} \int_{\frac{1}{2}}^{1} \frac{y^{2}}{2 y^{2} + 3 y + 1} d y$ $= - \frac{1}{8} l n (\frac{512}{243}) - \frac{1}{2} \int_{\frac{1}{2}}^{1} \frac{y^{2}}{2 y^{2} + 3 y + 1} d y$ $= - \frac{1}{8} l n (\frac{512}{243}) - \frac{1}{2} \int_{\frac{1}{2}}^{1} [\frac{1}{2 (2 y + 1)} - \frac{1}{y + 1} + \frac{1}{2}]$ $= - \frac{1}{8} l n (\frac{512}{243}) - \frac{1}{8} l n (\frac{3}{2}) + \frac{1}{2} l n (\frac{4}{3}) - \frac{1}{8}$ $= - \frac{1}{8}$ $\int_{\frac{1}{2}}^{1} \int_{\frac{1}{2}}^{1} \frac{x - y}{x + y} d x d y = \frac{1}{4} - 2 \times \frac{1}{8} = 0$
Commented by nimnim last updated on 17/Apr/20

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