let B_n = ∫∫_([0,n[^2 ) ((arctan(x^2 +3y^2 ))/(√(x^2 +3y^2 )))dxdy calculate lim_(n→+∞) (B


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Question Number 104774 by mathmax by abdo last updated on 23/Jul/20

$let B_{n} = \int \int_{[0, n [^{2}} \frac{\arctan (x^{2} + {3 y}^{2})}{\sqrt{x^{2} + {3 y}^{2}}} dxdy$ $calculate \lim_{n \to + \infty} \frac{B_{n}}{n}$
	Answered by mathmax by abdo last updated on 26/Jul/20
	$we do the changement { ((x =rcosθ)),((y =(r/(√3))sinθ)) :} we hsve 0≤x<n and 0≤y <n ⇒0 ≤x^2 +3y^2 <4n^2 ⇒ 0≤r^2 <4n^2 ⇒0≤r<2n ⇒B_n =∫_0 ^(2n) ∫_0 ^(π/2) ((arctan(r^2 ))/r) r dr dθ =(π/2) ∫_0 ^(2n) arctan(r^2 )dr by psrts ∫ arctan(r^2 )dr =r arctan(r^2 )−∫ r×((2r)/(1+r^4 ))dr =r arctan(r^2 )−2 ∫ (r^2 /(1+r^4 ))dr and ∫ (r^2 /(1+r^4 )) dr =∫ (1/((1/r^2 ) +r^2 ))dr =(1/2)∫ ((1−(1/r^2 )+1+(1/r^2 ))/(r^2 +(1/r^2 )))dr =(1/2) ∫ (((1−(1/r^2 ))dr)/((r+(1/r))^2 −2))(→u =r+(1/r))+(1/2)∫ (((1+(1/r^2 ))dr)/((r−(1/r))^2 +2))(v=r−(1/r)) =(1/2)∫ (du/(u^2 −2)) +(1/2)∫ (dv/(v^2 +2))(→v =(√2)α) =(1/(4(√2)))∫((1/(u−(√2)))−(1/(u+(√2))))du +(1/2)∫ (((√2)dα)/(2(1+α^2 ))) =(1/(4(√2)))ln∣((u−(√2))/(u+(√2)))∣ +(1/(2(√2))) arctan((v/(√2))) +c =(1/(4(√2)))ln∣((r+(1/r)−(√2))/(r+(1/r)+(√2)))∣+(1/(2(√2))) arctan((1/(√2))(r−(1/r))) +c$
	$we do the changement {\begin{cases} x = rcos θ \\ y = \frac{r}{\sqrt{3}} \sin θ \end{cases}$ $we hsve 0 ⩽ x < n and 0 ⩽ y < n \Rightarrow 0 ⩽ x^{2} + {3 y}^{2} < {4 n}^{2} \Rightarrow$ $0 ⩽ r^{2} < {4 n}^{2} \Rightarrow 0 ⩽ r < 2 n \Rightarrow B_{n} = \int_{0}^{2 n} \int_{0}^{\frac{π}{2}} \frac{\arctan (r^{2})}{r} r dr d θ$ $= \frac{π}{2} \int_{0}^{2 n} \arctan (r^{2}) dr by psrts$ $\int \arctan (r^{2}) dr = r \arctan (r^{2}) - \int r \times \frac{2 r}{1 + r^{4}} dr$ $= r \arctan (r^{2}) - 2 \int \frac{r^{2}}{1 + r^{4}} dr and$ $\int \frac{r^{2}}{1 + r^{4}} dr = \int \frac{1}{\frac{1}{r^{2}} + r^{2}} dr = \frac{1}{2} \int \frac{1 - \frac{1}{r^{2}} + 1 + \frac{1}{r^{2}}}{r^{2} + \frac{1}{r^{2}}} dr$ $= \frac{1}{2} \int \frac{(1 - \frac{1}{r^{2}}) dr}{{(r + \frac{1}{r})}^{2} - 2} (\to u = r + \frac{1}{r}) + \frac{1}{2} \int \frac{(1 + \frac{1}{r^{2}}) dr}{{(r - \frac{1}{r})}^{2} + 2} (v = r - \frac{1}{r})$ $= \frac{1}{2} \int \frac{du}{u^{2} - 2} + \frac{1}{2} \int \frac{dv}{v^{2} + 2} (\to v = \sqrt{2} α)$ $= \frac{1}{4 \sqrt{2}} \int (\frac{1}{u - \sqrt{2}} - \frac{1}{u + \sqrt{2}}) du + \frac{1}{2} \int \frac{\sqrt{2} d α}{2 (1 + α^{2})}$ $= \frac{1}{4 \sqrt{2}} \ln ∣ \frac{u - \sqrt{2}}{u + \sqrt{2}} ∣ + \frac{1}{2 \sqrt{2}} \arctan (\frac{v}{\sqrt{2}}) + c$ $= \frac{1}{4 \sqrt{2}} \ln ∣ \frac{r + \frac{1}{r} - \sqrt{2}}{r + \frac{1}{r} + \sqrt{2}} ∣ + \frac{1}{2 \sqrt{2}} \arctan (\frac{1}{\sqrt{2}} (r - \frac{1}{r})) + c$
	Commented by mathmax by abdo last updated on 26/Jul/20
	$=(1/(2(√2))){ ln((√((r+(1/r)−(√2))/(r+(1/r)+(√2))))) +arctan((1/(√2))(r−(1/r))) +c ⇒ ∫_0 ^(2n) arctan(r^2 )dr =[r arctan(r^2 )]_0 ^(2n) −(1/(√2))[{ln((√((r^2 +1−(√2)r)/(r^2 +1+(√2)r)))) +arctan((1/(√2))(r−(1/r)))]_0 ^(2n) =2n arctan(4n^2 )−(1/(√2)){ln((√((4n^2 +1−2(√2)n)/(4n^2 +1+2(√2)n))))+arctan((1/(√2))(2n−(1/(2n)))=u_n +(π/2)} ⇒B_n =(π/2)u_n ⇒(B_n /n) =(π/2)(u_n /n) =π arctan(4n^2 ) −(1/(n(√2))){ln(√((4n^2 +1−2(√2)n)/(4n^2 +1+2(√2)n)))+arctan((1/(√2))(2n−(1/n))} ⇒lim_(n→+∞) (B_n /n) =π×(π/2) =(π^2 /2)$
	$= \frac{1}{2 \sqrt{2}} {\ln (\sqrt{\frac{r + \frac{1}{r} - \sqrt{2}}{r + \frac{1}{r} + \sqrt{2}}}) + \arctan (\frac{1}{\sqrt{2}} (r - \frac{1}{r})) + c \Rightarrow$ $\int_{0}^{2 n} \arctan (r^{2}) dr = {[r \arctan (r^{2})]}_{0}^{2 n} - \frac{1}{\sqrt{2}} [{\ln (\sqrt{\frac{r^{2} + 1 - \sqrt{2} r}{r^{2} + 1 + \sqrt{2} r}})$ ${+ \arctan (\frac{1}{\sqrt{2}} (r - \frac{1}{r}))]}_{0}^{2 n}$ $= 2 n \arctan ({4 n}^{2}) - \frac{1}{\sqrt{2}} {\ln (\sqrt{\frac{{4 n}^{2} + 1 - 2 \sqrt{2} n}{{4 n}^{2} + 1 + 2 \sqrt{2} n}}) + \arctan (\frac{1}{\sqrt{2}} (2 n - \frac{1}{2 n}) = u_{n}$ $+ \frac{π}{2}} \Rightarrow B_{n} = \frac{π}{2} u_{n} \Rightarrow \frac{B_{n}}{n} = \frac{π}{2} \frac{u_{n}}{n}$ $= π \arctan ({4 n}^{2}) - \frac{1}{n \sqrt{2}} {\ln \sqrt{\frac{{4 n}^{2} + 1 - 2 \sqrt{2} n}{{4 n}^{2} + 1 + 2 \sqrt{2} n}} + \arctan (\frac{1}{\sqrt{2}} (2 n - \frac{1}{n})}$ $\Rightarrow \lim_{n \to + \infty} \frac{B_{n}}{n} = π \times \frac{π}{2} = \frac{π^{2}}{2}$
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