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

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

Commented bymathmax 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)