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Question Number 36946 by maxmathsup by imad last updated on 07/Jun/18
calculateϕ(λ)=  ∫_0 ^π        ((cos(t))/(1−2λ cost +λ^2 )) dt
calculateφ(λ)=0πcos(t)12λcost+λ2dt
Commented by math khazana by abdo last updated on 08/Jun/18
changement tan((t/2)) =x give  ϕ(λ) = ∫_0 ^∞  (((1−x^2 )/(1+x^2 ))/(1−2λ((1−x^2 )/(1+x^2 )))) ((2dx)/(1+x^2 ))  = 2 ∫_0 ^∞      ((1−x^2 )/((1+x^2 ){1+x^2  −2λ +2λx^2 }))dx  =2 ∫_0 ^∞      ((1−x^2 )/((1+x^2 )((2λ+1)x^2 +1−2λ)))dx let  decompose F(x)=((1−x^2 )/((1+x^2 ){(2λ+1)x^2  +1−2λ)))  F(x)= ((ax+b)/(x^2  +1))  +((cx+d)/((2λ+1)x^2  +1−2λ))  F(−x)=F(x)⇒((−ax+b)/(x^2  +1)) +((−cx +d)/((2λ+1)x^2  +1−2λ)) =F(x)  ⇒a=c=0 ⇒F(x)=(b/(x^2  +1)) + (d/((2λ+1)x^2  +1−2λ))  lim_(x→+∞) x^2 F(x)=((−1)/(2λ+1)) = b +(d/(2λ+1)) ⇒  −1=(2λ+1)b +d ⇒d=−1−(2λ+1)b  F(x)= (b/(x^2  +1)) −((1+(2λ+1)b)/((2λ+1)x^2  +1−2λ))  F(o)= (1/(1−2λ)) = b −((1+(2λ+1)b)/(1−2λ)) ⇒  1=(1−2λ)b −1−(2λ+1)b =−4λb −1 ⇒  2=−4λb ⇒b=((−1)/(2λ))  ( λ≠0 and λ≠(1/2))  ∫_0 ^(+∞) F(x)dx =((−1)/(2λ))∫_0 ^∞  {(1/(x^2  +1)) −((1−((2λ+1)/(2λ)))/((2λ+1)x^2  +1−2λ))}dx  =((−1)/(2λ)) ∫_0 ^∞    (dx/(1+x^2 )) −(1/(2λ)) .((−1)/(2λ))∫_0 ^∞     (dx/((2λ+1)x^2  +1−2λ))  =−(1/(2λ)) (π/2) +(1/(4λ^2 )) ∫_0 ^∞     (dx/((2λ+1)x^2  +1−2λ)) but  ∫_0 ^∞      (dx/((2λ+1)x^2  +1−2λ))  =(1/(2λ+1))∫_0 ^∞      (dx/(x^2  +((1−2λ)/(1+2λ))))  if ((1−2λ)/(1+2λ)) >0 we use the chang.x=(√((1−2λ)/(1+2λ)))u  =(1/(2λ+1))∫_0 ^∞     (1/(((1−2λ)/(1+2λ))(1+x^2 ))) ((√(1−2λ))/( (√(1+2λ)))) du  =(1/(2λ+1))  ((1+2λ)/(1−2λ)) ((√(1−2λ))/( (√(1+2λ)))) (π/2)  = (π/(2(2λ+1))) (√((1+2λ)/(1−2λ)))=(π/2)  (1/( (√(1−4λ^2 )))) ⇒  ∫_0 ^∞  F(x)dx= −(π/(4λ)) +(1/(4λ^2 )) (π/2) (1/( (√(1−4λ^2 ))))  =−(π/(4λ)) + (π/(8λ^2 (√(1−4λ^2 ))))  ϕ(λ) =2 ∫_0 ^∞ F(x)dx ⇒  ϕ(λ) =  (π/(4λ^2 (√(1−4λ^2 )))) −(π/(2λ)) .
changementtan(t2)=xgiveφ(λ)=01x21+x212λ1x21+x22dx1+x2=201x2(1+x2){1+x22λ+2λx2}dx=201x2(1+x2)((2λ+1)x2+12λ)dxletdecomposeF(x)=1x2(1+x2){(2λ+1)x2+12λ)F(x)=ax+bx2+1+cx+d(2λ+1)x2+12λF(x)=F(x)ax+bx2+1+cx+d(2λ+1)x2+12λ=F(x)a=c=0F(x)=bx2+1+d(2λ+1)x2+12λlimx+x2F(x)=12λ+1=b+d2λ+11=(2λ+1)b+dd=1(2λ+1)bF(x)=bx2+11+(2λ+1)b(2λ+1)x2+12λF(o)=112λ=b1+(2λ+1)b12λ1=(12λ)b1(2λ+1)b=4λb12=4λbb=12λ(λ0andλ12)0+F(x)dx=12λ0{1x2+112λ+12λ(2λ+1)x2+12λ}dx=12λ0dx1+x212λ.12λ0dx(2λ+1)x2+12λ=12λπ2+14λ20dx(2λ+1)x2+12λbut0dx(2λ+1)x2+12λ=12λ+10dxx2+12λ1+2λif12λ1+2λ>0weusethechang.x=12λ1+2λu=12λ+10112λ1+2λ(1+x2)12λ1+2λdu=12λ+11+2λ12λ12λ1+2λπ2=π2(2λ+1)1+2λ12λ=π2114λ20F(x)dx=π4λ+14λ2π2114λ2=π4λ+π8λ214λ2φ(λ)=20F(x)dxφ(λ)=π4λ214λ2π2λ.
Commented by math khazana by abdo last updated on 08/Jun/18
if ((1−2λ)/(1+2λ))<0  we get ∫_0 ^∞      (dx/(x^2  +((1−2λ)/(1+2λ))))  =∫_0 ^∞    (dx/(x^2   −((√(−((1−2λ)/(1+2λ)))))^2 ))   and we use the  decompositon ...
if12λ1+2λ<0weget0dxx2+12λ1+2λ=0dxx2(12λ1+2λ)2andweusethedecompositon

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