Question Number 89384 by nimnim last updated on 17/Apr/20
Commented by mathmax by abdo last updated on 17/Apr/20
![let try another way we consider the diffeomorphism (u,v)→(x,y) /x−y=u and x+y =v ⇒x =((u+v)/2) andy=((−u+v)/2) ⇒ϕ(u,v)=(ϕ_1 (u,v),ϕ_2 (u,v))=(x,y)=((u/2)+(v/2),−(u/2)+(v/2)) we have (1/2)≤x≤1 and (1/2)≤y≤1 ⇒1≤x+y≤2⇒v∈[1,2] −1≤−y≤−(1/2) ⇒−(1/2)≤x−y≤(1/2) ⇒u∈[−(1/2),(1/2)] M_j (ϕ)= ((((∂ϕ_1 /∂u) (∂ϕ_1 /∂v))),(((∂ϕ_2 /∂u) (∂ϕ_2 /∂v))) )= ((((1/2) (1/2))),((−(1/2) (1/2))) ) ∣detM_j (ϕ)∣ =(1/2) ∫∫_D f(x,y)dxdy =∫∫_w foϕ(u,v)∣J_ϕ ∣du dv ∫∫_(u∈[−(1/2),(1/2)]and v∈[1,2]) (u/v)×(1/2)du dv =(1/2) ∫_1 ^2 (∫_(−(1/2)) ^(1/2) udu)(dv/v) =(1/2)∫_1 ^2 [(u^2 /2)]_(−(1/2)) ^(1/2) (dv/v) =0](Q89444.png)
$${let}\:{try}\:{another}\:{way}\:\:{we}\:{consider}\:{the}\:{diffeomorphism} \\ $$$$\left({u},{v}\right)\rightarrow\left({x},{y}\right)\:/{x}−{y}={u}\:{and}\:{x}+{y}\:={v}\:\Rightarrow{x}\:=\frac{{u}+{v}}{\mathrm{2}}\:{andy}=\frac{−{u}+{v}}{\mathrm{2}} \\ $$$$\Rightarrow\varphi\left({u},{v}\right)=\left(\varphi_{\mathrm{1}} \left({u},{v}\right),\varphi_{\mathrm{2}} \left({u},{v}\right)\right)=\left({x},{y}\right)=\left(\frac{{u}}{\mathrm{2}}+\frac{{v}}{\mathrm{2}},−\frac{{u}}{\mathrm{2}}+\frac{{v}}{\mathrm{2}}\right) \\ $$$${we}\:{have}\:\frac{\mathrm{1}}{\mathrm{2}}\leqslant{x}\leqslant\mathrm{1}\:{and}\:\:\frac{\mathrm{1}}{\mathrm{2}}\leqslant{y}\leqslant\mathrm{1}\:\Rightarrow\mathrm{1}\leqslant{x}+{y}\leqslant\mathrm{2}\Rightarrow{v}\in\left[\mathrm{1},\mathrm{2}\right] \\ $$$$−\mathrm{1}\leqslant−{y}\leqslant−\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow−\frac{\mathrm{1}}{\mathrm{2}}\leqslant{x}−{y}\leqslant\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow{u}\in\left[−\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}}\right] \\ $$$${M}_{{j}} \left(\varphi\right)=\begin{pmatrix}{\frac{\partial\varphi_{\mathrm{1}} }{\partial{u}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\partial\varphi_{\mathrm{1}} }{\partial{v}}}\\{\frac{\partial\varphi_{\mathrm{2}} }{\partial{u}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\partial\varphi_{\mathrm{2}} }{\partial{v}}}\end{pmatrix}=\:\:\begin{pmatrix}{\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}}\\{−\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}}\end{pmatrix} \\ $$$$\mid{detM}_{{j}} \left(\varphi\right)\mid\:=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\int\int_{{D}} {f}\left({x},{y}\right){dxdy}\:=\int\int_{{w}} {fo}\varphi\left({u},{v}\right)\mid{J}_{\varphi} \mid{du}\:{dv} \\ $$$$\int\int_{{u}\in\left[−\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}}\right]{and}\:{v}\in\left[\mathrm{1},\mathrm{2}\right]} \:\:\:\:\frac{{u}}{{v}}×\frac{\mathrm{1}}{\mathrm{2}}{du}\:{dv} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{1}} ^{\mathrm{2}} \:\left(\int_{−\frac{\mathrm{1}}{\mathrm{2}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} {udu}\right)\frac{{dv}}{{v}}\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{1}} ^{\mathrm{2}} \left[\frac{{u}^{\mathrm{2}} }{\mathrm{2}}\right]_{−\frac{\mathrm{1}}{\mathrm{2}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{{dv}}{{v}}\:=\mathrm{0} \\ $$$$ \\ $$
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](Q89453.png)
$$\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \frac{{x}−{y}}{{x}+{y}}{dxdy} \\ $$$$=\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \left(\mathrm{1}−\frac{\mathrm{2}{y}}{{x}+{y}}\right){dxdy} \\ $$$$=\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \left\{\left[{x}−\mathrm{2}{aln}\mid{x}+{y}\mid\right]_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \right\}{dy} \\ $$$$=\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{2}{yln}\mid\frac{\frac{\mathrm{1}}{\mathrm{2}}+{y}}{\mathrm{1}+{y}}\mid\right){dy} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{2}\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} {yln}\mid\frac{\frac{\mathrm{1}}{\mathrm{2}}+{y}}{\mathrm{1}+{y}}\mid{dy} \\ $$$${y}>\mathrm{0}\rightarrow{ln}\mid\frac{\frac{\mathrm{1}}{\mathrm{2}}+{y}}{\mathrm{1}+{y}}\mid={ln}\left(\frac{\frac{\mathrm{1}}{\mathrm{2}}+{y}}{\mathrm{1}+{y}}\right) \\ $$$$\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} {yln}\left(\frac{\frac{\mathrm{1}}{\mathrm{2}}+{y}}{\mathrm{1}+{y}}\right){dy} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{y}^{\mathrm{2}} {ln}\left(\frac{\frac{\mathrm{1}}{\mathrm{2}}+{y}}{\mathrm{1}+{y}}\right)\mid_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{2}}\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \frac{{y}^{\mathrm{2}} }{\mathrm{2}{y}^{\mathrm{2}} +\mathrm{3}{y}+\mathrm{1}}{dy} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{8}}{ln}\left(\frac{\mathrm{512}}{\mathrm{243}}\right)−\frac{\mathrm{1}}{\mathrm{2}}\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \frac{{y}^{\mathrm{2}} }{\mathrm{2}{y}^{\mathrm{2}} +\mathrm{3}{y}+\mathrm{1}}{dy} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{8}}{ln}\left(\frac{\mathrm{512}}{\mathrm{243}}\right)−\frac{\mathrm{1}}{\mathrm{2}}\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \left[\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}{y}+\mathrm{1}\right)}−\frac{\mathrm{1}}{{y}+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}\right] \\ $$$$=−\frac{\mathrm{1}}{\mathrm{8}}{ln}\left(\frac{\mathrm{512}}{\mathrm{243}}\right)−\frac{\mathrm{1}}{\mathrm{8}}{ln}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\mathrm{4}}{\mathrm{3}}\right)−\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \frac{{x}−{y}}{{x}+{y}}\:{dxdy}=\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{2}×\frac{\mathrm{1}}{\mathrm{8}}=\mathrm{0} \\ $$
Commented by nimnim last updated on 17/Apr/20
$${Thank}\:{you}\:{Sir}. \\ $$