Question Number 103312 by bemath last updated on 14/Jul/20

∫_0 ^∞  (1/((1+x^2 )^6 )) dx ?

Answered by Dwaipayan Shikari last updated on 14/Jul/20

∫_0 ^(π/2) (1/(sec^(12) θ))sec^2 θdθ=∫_0 ^(π/2) cos^(10) θdθ....continue  ∫_0 ^(π/2) cos^(10) θdθ=((63π)/(512))

Commented byAr Brandon last updated on 14/Jul/20

Cool!  ∫_0 ^(π/2) cos^(10) θdθ=(1/2)∙(3/4)∙(5/6)∙(7/8)∙(9/(10))∙(π/2)=((63π)/(512))  Walli′s method

Commented byDwaipayan Shikari last updated on 14/Jul/20

Thanking you

Answered by mathmax by abdo last updated on 14/Jul/20

I =∫_0 ^∞   (dx/((1+x^2 )^6 ))  ⇒2I =∫_(−∞) ^(+∞)  (dx/((x^2  +1)^6 ))  let ϕ(z) =(1/((z^2 +1)^6 ))  we have lim_(z→+∞) ∣zf(z)∣ =0  and f(z) =(1/((z−i)^6 (z+i)^6 ))  so ∫_(−∞) ^(+∞)  ϕ(z)dz =2iπ Res(ϕ,i)  Res (ϕ,i) =lim_(z→i)     (1/((6−1)!)){(z−i)^6 ϕ(z)}^((5))   =lim_(z→i)     (1/(5!)){(z+i)^(−6) }^((5))   =(1/(5!))(−6){(z+i)^(−7) }^((4))   =lim_(z→i) (1/(5!))(−6)(−7){(z+i)^(−8) }^((3))

Commented byabdomathmax last updated on 14/Jul/20

Res(ϕ,i) =lim_(z→i) ((42)/(5!))(−8){(z+i)^(−9) }^((2))   =lim_(z→i)   ((42)/(5!))(−8)(−9){(z+i)^(−10) }^((1))   =lim_(z→i)   ((42×72)/(5!))(−10)(z+i)^(−11)   =−((42×720)/(5!(2i)^(11) )) =−((42×720)/(5! ×2^(11) (−i))) =((42×720)/(5! ×2^(11) i))  ⇒∫_(−∞) ^(+∞)  ϕ(z)dz =2iπ ×((42×720)/(5!×2^(11) i ))  =((84×720)/(5!×2^(11) ))π =2I ⇒  I =((84×720π)/(2^(12)  ×5!))  rest simplification..

Answered by OlafThorendsen last updated on 14/Jul/20

I = ∫_0 ^∞ (dx/((1+x^2 )^6 ))  x = tanu  dx = (1+tan^2 u)du  I = ∫_0 ^(π/2) ((1+tan^2 u)/((1+tan^2 u)^6 ))du  I = ∫_0 ^(π/2) (du/((1+tan^2 u)^5 ))  I = ∫_0 ^(π/2) (cos^2 u)^5 du  I = ∫_0 ^(π/2) cos^(10) udu  I = ∫_0 ^(π/2) (((e^(iu) +e^(−iu) )/2))^(10) du  I = (1/2^(10) )∫_0 ^(π/2) Σ_(k=0) ^(10) C_(10) ^k e^(iku) e^(−iu(10−k)) du  I = (1/2^(10) )∫_0 ^(π/2) Σ_(k=0) ^(10) C_(10) ^k e^(2iu(k−5)) du  I = (1/2^(10) )[C_(10) ^5 u+2Σ_(k=0) ^(k=4) C_(10) ^k ((sin(2u(k−5)))/(2(k−5)))]_0 ^(π/2)   I = (1/2^(10) )[C_(10) ^5 (π/2)+Σ_(k=0) ^(k=4) C_(10) ^k ((sin(2π(k−5)))/((k−5)))−0]  I = ((πC_(10) ^5 )/2^(11) )   I = ((63)/(512))π (≈0,387...)

Answered by bramlex last updated on 15/Jul/20

consider I_n =∫_0 ^∞ (dx/((1+x^2 )^n )) , n≥1  by parts  { ((u=(1/((1+x^2 )^n )))),((dv = 1.dx)) :}  I_n  = (x/((1+x^2 )^n ))∣_0 ^∞ +∫_0 ^∞   ((2nx)/((1+x^2 )^(n+1) )) dx  I_n  = 0+2n∫_0 ^∞  (((x^2 +1)−1)/((1+x^2 )^(n+1) )) dx  I_n  = 2n ∫_0 ^∞ ((1/((1+x^2 )^n ))  −(1/((1+x^2 )^(n+1) )))dx  I_n  = 2n(I_n −I_(n+1) ) ,solving for  I_(n+1)  yields I_(n+1) =((2n−1)/(2n))×I_n   so then we have I_1 = ∫_0 ^∞  (dx/((1+x^2 )^1 ))  I_1 = arctan x ∣_0 ^∞  = (π/2).  repeated use of the recurrence  gives us I_n = ((1.3.5...(2n−3))/(2.4.6...(2n−2)))I_1   finally we take n=6  to conclude that   ∫_0 ^∞ (dx/((1+x^2 )^6 )) = ((1.3.5.7.9)/(2.4.6.8.10))×(π/2)  = ((63π)/(512)) ■