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)) ■