find ∫∫_([0,1]^2 ) ln(x^2 +3y^2 ) e^(−x^2 −3y^2 ) dxdy


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Question Number 110450 by mathmax by abdo last updated on 29/Aug/20

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