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Question Number 110450 by mathmax by abdo last updated on 29/Aug/20
find ∫∫_([0,1]^2 )    ln(x^2  +3y^2 ) e^(−x^2 −3y^2 )  dxdy
$$\mathrm{find}\:\int\int_{\left[\mathrm{0},\mathrm{1}\right]^{\mathrm{2}} } \:\:\:\mathrm{ln}\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{3y}^{\mathrm{2}} \right)\:\mathrm{e}^{−\mathrm{x}^{\mathrm{2}} −\mathrm{3y}^{\mathrm{2}} } \:\mathrm{dxdy} \\ $$
Answered by mathmax by abdo last updated on 31/Aug/20
I =∫∫_([0,1])   ln(x^2  +3y^2 )e^(−x^2 −3y^2 ) dxdy  we consider the diffeomorphism  ϕ(r,θ) =(x,y) =(rcosθ ,(r/( (√3))) sinθ) =(ϕ_1  ,ϕ_2 )  M_j ϕ  = ((((∂ϕ_1 /∂r)                     (∂ϕ_1 /∂θ))),(((∂ϕ_2 /∂r)                      (∂ϕ_2 /∂θ))) )= (((cosθ            −rsinθ)),((((sinθ)/( (√3)))                  (r/( (√3))) cosθ)) )  ⇒det M_j =(r/( (√3))) cos^2 θ+(r/( (√3))) sin^2 θ =(r/( (√3)))  we have    { ((0≤x≤1)),((0≤y≤1)) :}   ⇒  { ((0≤x≤1         ⇒  0≤x^2  +3y^2  ≤4 ⇒0≤r≤2 ⇒)),((0≤(√3)y≤(√3))) :}  I =∫∫_(0≤r≤2  and  0≤θ≤(π/2))    ln(r^2 ) e^(−r^2 )   (r/( (√3)))  dr dθ  =(π/(2(√3)))  ∫_0 ^2   2r e^(−r^2 ) ln(r)dr  =(π/( (√3))) ∫_0 ^2   r e^(−r^2 ) ln(r) dr  by parts  ∫_0 ^2   r e^(−r^2 ) ln(r)dr =[−(1/2)(1−e^(−r^2 ) ) ln(r)]_0 ^2  +∫_0 ^2 (1/2)(1−e^(−r^2 ) )(dr/r)  =−(1/2)(1−e^(−4) )(ln4) +(1/2) ∫_0 ^2  ((1−e^(−r^2 ) )/r) dr  we have− e^(−r^2 )  =−Σ_(n=0) ^∞   (((−1)^n  r^(2n) )/(n!)) =−1+Σ_(n=1) ^∞  (((−1)^n  r^(2n) )/(n!)) ⇒  ((1−e^(−r^2 ) )/r) =Σ_(n=1) ^∞  (((−1)^n  r^(2n−1) )/(n!)) ⇒  ∫_0 ^2  ((1−e^(−r^2 ) )/r) dr =Σ_(n=1) ^∞  (((−1)^n )/(n!)) ∫_0 ^2  r^(2n−1)  dr  =Σ_(n=1) ^∞  (((−1)^n )/(n!))((1/(2n))(2)^(2n) ) =Σ_(n=1) ^∞  (((−4)^n )/(2n(n!))) ⇒  I =(e^(−4) −1)ln(2) +(1/4) Σ_(n=1) ^∞  (((−4)^n )/(n(n!)))
$$\mathrm{I}\:=\int\int_{\left[\mathrm{0},\mathrm{1}\right]} \:\:\mathrm{ln}\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{3y}^{\mathrm{2}} \right)\mathrm{e}^{−\mathrm{x}^{\mathrm{2}} −\mathrm{3y}^{\mathrm{2}} } \mathrm{dxdy}\:\:\mathrm{we}\:\mathrm{consider}\:\mathrm{the}\:\mathrm{diffeomorphism} \\ $$$$\varphi\left(\mathrm{r},\theta\right)\:=\left(\mathrm{x},\mathrm{y}\right)\:=\left(\mathrm{rcos}\theta\:,\frac{\mathrm{r}}{\:\sqrt{\mathrm{3}}}\:\mathrm{sin}\theta\right)\:=\left(\varphi_{\mathrm{1}} \:,\varphi_{\mathrm{2}} \right) \\ $$$$\mathrm{M}_{\mathrm{j}} \varphi\:\:=\begin{pmatrix}{\frac{\partial\varphi_{\mathrm{1}} }{\partial\mathrm{r}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\partial\varphi_{\mathrm{1}} }{\partial\theta}}\\{\frac{\partial\varphi_{\mathrm{2}} }{\partial\mathrm{r}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\partial\varphi_{\mathrm{2}} }{\partial\theta}}\end{pmatrix}=\begin{pmatrix}{\mathrm{cos}\theta\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{rsin}\theta}\\{\frac{\mathrm{sin}\theta}{\:\sqrt{\mathrm{3}}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{r}}{\:\sqrt{\mathrm{3}}}\:\mathrm{cos}\theta}\end{pmatrix} \\ $$$$\Rightarrow\mathrm{det}\:\mathrm{M}_{\mathrm{j}} =\frac{\mathrm{r}}{\:\sqrt{\mathrm{3}}}\:\mathrm{cos}^{\mathrm{2}} \theta+\frac{\mathrm{r}}{\:\sqrt{\mathrm{3}}}\:\mathrm{sin}^{\mathrm{2}} \theta\:=\frac{\mathrm{r}}{\:\sqrt{\mathrm{3}}}\:\:\mathrm{we}\:\mathrm{have}\: \\ $$$$\begin{cases}{\mathrm{0}\leqslant\mathrm{x}\leqslant\mathrm{1}}\\{\mathrm{0}\leqslant\mathrm{y}\leqslant\mathrm{1}}\end{cases}\:\:\:\Rightarrow\:\begin{cases}{\mathrm{0}\leqslant\mathrm{x}\leqslant\mathrm{1}\:\:\:\:\:\:\:\:\:\Rightarrow\:\:\mathrm{0}\leqslant\mathrm{x}^{\mathrm{2}} \:+\mathrm{3y}^{\mathrm{2}} \:\leqslant\mathrm{4}\:\Rightarrow\mathrm{0}\leqslant\mathrm{r}\leqslant\mathrm{2}\:\Rightarrow}\\{\mathrm{0}\leqslant\sqrt{\mathrm{3}}\mathrm{y}\leqslant\sqrt{\mathrm{3}}}\end{cases} \\ $$$$\mathrm{I}\:=\int\int_{\mathrm{0}\leqslant\mathrm{r}\leqslant\mathrm{2}\:\:\mathrm{and}\:\:\mathrm{0}\leqslant\theta\leqslant\frac{\pi}{\mathrm{2}}} \:\:\:\mathrm{ln}\left(\mathrm{r}^{\mathrm{2}} \right)\:\mathrm{e}^{−\mathrm{r}^{\mathrm{2}} } \:\:\frac{\mathrm{r}}{\:\sqrt{\mathrm{3}}}\:\:\mathrm{dr}\:\mathrm{d}\theta \\ $$$$=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{3}}}\:\:\int_{\mathrm{0}} ^{\mathrm{2}} \:\:\mathrm{2r}\:\mathrm{e}^{−\mathrm{r}^{\mathrm{2}} } \mathrm{ln}\left(\mathrm{r}\right)\mathrm{dr}\:\:=\frac{\pi}{\:\sqrt{\mathrm{3}}}\:\int_{\mathrm{0}} ^{\mathrm{2}} \:\:\mathrm{r}\:\mathrm{e}^{−\mathrm{r}^{\mathrm{2}} } \mathrm{ln}\left(\mathrm{r}\right)\:\mathrm{dr}\:\:\mathrm{by}\:\mathrm{parts} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}} \:\:\mathrm{r}\:\mathrm{e}^{−\mathrm{r}^{\mathrm{2}} } \mathrm{ln}\left(\mathrm{r}\right)\mathrm{dr}\:=\left[−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\mathrm{e}^{−\mathrm{r}^{\mathrm{2}} } \right)\:\mathrm{ln}\left(\mathrm{r}\right)\right]_{\mathrm{0}} ^{\mathrm{2}} \:+\int_{\mathrm{0}} ^{\mathrm{2}} \frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\mathrm{e}^{−\mathrm{r}^{\mathrm{2}} } \right)\frac{\mathrm{dr}}{\mathrm{r}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\mathrm{e}^{−\mathrm{4}} \right)\left(\mathrm{ln4}\right)\:+\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{2}} \:\frac{\mathrm{1}−\mathrm{e}^{−\mathrm{r}^{\mathrm{2}} } }{\mathrm{r}}\:\mathrm{dr} \\ $$$$\mathrm{we}\:\mathrm{have}−\:\mathrm{e}^{−\mathrm{r}^{\mathrm{2}} } \:=−\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{r}^{\mathrm{2n}} }{\mathrm{n}!}\:=−\mathrm{1}+\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{r}^{\mathrm{2n}} }{\mathrm{n}!}\:\Rightarrow \\ $$$$\frac{\mathrm{1}−\mathrm{e}^{−\mathrm{r}^{\mathrm{2}} } }{\mathrm{r}}\:=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{r}^{\mathrm{2n}−\mathrm{1}} }{\mathrm{n}!}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}} \:\frac{\mathrm{1}−\mathrm{e}^{−\mathrm{r}^{\mathrm{2}} } }{\mathrm{r}}\:\mathrm{dr}\:=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}!}\:\int_{\mathrm{0}} ^{\mathrm{2}} \:\mathrm{r}^{\mathrm{2n}−\mathrm{1}} \:\mathrm{dr} \\ $$$$=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}!}\left(\frac{\mathrm{1}}{\mathrm{2n}}\left(\mathrm{2}\right)^{\mathrm{2n}} \right)\:=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{4}\right)^{\mathrm{n}} }{\mathrm{2n}\left(\mathrm{n}!\right)}\:\Rightarrow \\ $$$$\mathrm{I}\:=\left(\mathrm{e}^{−\mathrm{4}} −\mathrm{1}\right)\mathrm{ln}\left(\mathrm{2}\right)\:+\frac{\mathrm{1}}{\mathrm{4}}\:\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{4}\right)^{\mathrm{n}} }{\mathrm{n}\left(\mathrm{n}!\right)} \\ $$$$ \\ $$

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