Question Number 100590 by Rohit@Thakur last updated on 27/Jun/20

∫_0 ^∞ (dx/((1+x^(18) )^2 ))

Answered by mathmax by abdo last updated on 27/Jun/20

let f(t) =∫_0 ^∞ (dx/(t+x^(18) )) with t>0 we have (df/dt)(t) =−∫_0 ^∞ (dx/((t+x^(18) )^2 )) ⇒∫_0 ^∞ (dx/((t+x^(18) )^2 )) =−f^′ (t) and ∫_0 ^∞ (dx/((1+x^(18) )^2 )) =−f^′ (1) f(t) =∫_0 ^∞ (dx/(t(1+(x^(18) /t)))) =(1/t)∫_0 ^∞ (dx/(1+((x/α))^(18) )) with α =t^(1/(18)) =_((x/α)=u) (1/t)∫_0 ^∞ ((αdu)/(1+u^(18) )) =(α/t) ∫_0 ^∞ (du/(1+u^(18) )) =_(u =z^(1/(18)) ) t^((1/(18))−1) ∫_0 ^∞ (z^((1/(18))−1) /(18(1+z)))dz =(t^((1/(18))−1) /(18)) ∫_0 ^∞ (z^((1/(18))−1) /(1+z))dz =(t^((1/(18))−1) /(18))×(π/(sin((π/(18))))) =(π/(18sin((π/(18)))))×t^(−((17)/(18))) ⇒f^′ (t) =−((17)/(18)) t^(−((17)/(18))−1) ×(π/(18sin((π/(18))))) ⇒f^′ (1) =−((17)/(18))×(π/(18 sin((π/(18))))) ⇒ ∫_0 ^∞ (dx/((1+x^(18) )^2 )) =−f^′ (1) =((17π)/(18^2 sin((π/(18)))))

Commented byAr Brandon last updated on 27/Jun/20

Understood, impressionnant ! 😅

Commented by1549442205 last updated on 27/Jun/20

Great method !But how do you know ∫_0 ^∞ (z^((1/(18))−1) /(z+1))dz=(π/(18sin(π/(18)))) Please,can you show me?

Commented byCoronavirus last updated on 27/Jun/20

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Commented byabdomathmax last updated on 27/Jun/20

i have proved this relation take alook in the platform...i have used residus for that...

Commented bymaths mind last updated on 27/Jun/20

∫_0 ^1 t^(x−1) (1−t)^(y−1) dt=β(x,y)..E (t/(1−t))=u⇒t=(u/(1+u))⇒dt=(du/((1+u)^2 )) E⇔∫_0 ^(+∞) (u^(x−1) /((1+u)^(x−1) )).(1/((1+u)^(y−1) )).(du/((1+u)^2 )) =∫_0 ^(+∞) (u^(x−1) /((1+u)^(x+y) ))du x=(1/(18)),y=((17)/(18)) ⇒β((1/(18)),1−(1/(18)))=(π/(sin((π/(18))))) β(a,b)=((Γ(a)Γ(b))/(Γ(a+b))),if a+b=1⇒Γ(a+b)=Γ(1)=1, Γ(a)Γ(1−a)=(π/(sin(πa)))....

Commented by1549442205 last updated on 27/Jun/20

Thankyou,sir a lot

Commented bymathmax by abdo last updated on 27/Jun/20

let prove that ∫_0 ^∞ (t^(a−1) /(1+t))dt =(π/(sin(πa))) with 0<a<1 changement t=tan^2 θ give ∫_0 ^∞ (t^(a−1) /(1+t))dt =∫_0 ^(π/2) ((tan^(2a−2) )/(1+tan^2 θ))×2tanθ(1+tan^2 θ)dθ =2 ∫_0 ^(π/2) tan^(2a−1) θ dθ =2∫_0 ^(π/2) ((sin^(2a−1) θ)/(cos^(2a−1) θ))dθ =2 ∫_0 ^(π/2) sin^(2a−1) θ cos^(1−2a) θ dθ we know 2 ∫_0 ^(π/2) cos^(2p−1) θ sin^(2q−1) θ =B(p,q) =((Γ(p).Γ(q))/(Γ(p+q))) 2q−1 =1−2a ⇒2q =2−2a ⇒q =1−a ⇒ 2∫_0 ^(π/2) sin^(2a−1) θ cos^(1−2a) θ dθ =2 ∫_0 ^(π/2) sin^(2a−1) θ cos^(2(1−a)−1) θ dθ =B(a,1−a) =((Γ(a).Γ(1−a))/(Γ(a+1−a))) =Γ(a).Γ(1−a) =(π/(sin(πa))) (compoliment formulae) the value is proved.

Commented bymathmax by abdo last updated on 27/Jun/20

error of typo (compliment formulae)

Commented byRohit@Thakur last updated on 28/Jun/20

thank you sir

Answered by smridha last updated on 27/Jun/20

let x^9 =tan𝛉 so we get (1/9)∫_0 ^(𝛑/2) (sin𝛉)^(−(8/9)) .(cos𝛉)^((26)/9) d𝛉 =(1/(18)).((𝚪((1/(18))).Γ(((35)/(18))))/(𝚪(2)))=((17)/(324)).𝚪((1/(18))).Γ(1−(1/(18))) =((17𝛑)/(324)).(1/(sin((𝛑/(18))))).

Commented byRohit@Thakur last updated on 28/Jun/20

plz provide the detailed solution

Commented bysmridha last updated on 28/Jun/20

∫_0 ^∞ (dx/((1+x^(18) )^2 )) let x^9 =tan(𝛉) so we get (1/9)∫_0 ^(𝛑/2) sec^(−4) (𝛉).tan^((1/9)−1) (𝛉)sec^2 (𝛉)d𝛉 =(1/9)∫_0 ^(𝛑/2) (((sin𝛉)^(−(8/9)) )/((cos𝛉)^(−(8/9)) )).cos^2 (𝛉)d𝛉 =(1/9)∫_0 ^(𝛑/2) (sin(𝛉))^(−(8/9)) (cos(𝛉))^((26)/9) =(1/9).(1/2).((𝚪(((−(8/9)+1)/2)).𝚪(((((26)/9)+1)/2)))/(𝚪(((−(8/9)+((26)/9)+2)/2))))=(1/(18))((𝚪((1/(18))).𝚪(((35)/(18))))/(𝚪(2))) =(1/(18))((𝚪((1/(18))).𝚪(((17)/(18))+1))/(1!))=((17)/(18^2 )).𝚪((1/(18))).Γ(((17)/(18))) [𝚪(n)=(n−1)! and 𝚪(n+1)=n𝚪(n)] =((17)/(324))𝚪((1/(18))).𝚪(1−(1/(18)))=((17𝛑)/(324)).(1/(sin((𝛑/(18))))) [𝚪(m).𝚪(1−m)=(𝛑/(sin(m𝛑))) where m<1 this is called reflection formula for Gamma f^n ]

Commented bysmridha last updated on 28/Jun/20

thus you posted something like this!!it should be needed to know the basic formulism at first.