Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 66470 by mathmax by abdo last updated on 15/Aug/19

calculate ∫_0 ^∞ (dx/((x^n +8)^3 )) withn>1

calculate∫0∞dx(xn+8)3withn>1

Commented by~ À ® @ 237 ~ last updated on 16/Aug/19

let state ∀ a>0 f(a)=∫_0 ^∞ (dx/((x^n +a)^2 )) f(a)=∫_0 ^∞ (dx/(a^2 [((x/a^(1/n) ))^n +1]^2 )) when changing x=a^(1/n) u we get f(a)= a^((1/n) −2) ∫_0 ^∞ (du/((u^n +1)^2 )) now change v=(1/(u^n +1)) ⇒ u=((1/v) −1)^(1/n) ⇒ du=(1/n).(((−1)/v^2 ))((1/v) −1)^((1/n) −1) dv then f(a)=(1/n). a^((1/n)−2) ∫_0 ^1 v^2 .((1/v^2 ))((1/v) −1)^((1/n) −1) dv = (a^((1/n)−2) /n) ∫_0 ^1 v^(1−(1/n)) (1−v)^((1/n)−1) dv = (a^((1/n) −2) /n) B(2−(1/n) , (1/n)) = (a^((1/n) −2) /n) .((Γ(2−(1/n))Γ((1/n)))/(Γ(2))) as Γ(x+1)=xΓ(x) we have Γ(2−(1/n))=(1−(1/n))Γ(1−(1/n)) and Γ(2)=Γ(1)=1 So f(a)=(a^((1/n) −2) /n) .(1−(1/n))Γ(1−(1/n))Γ((1/n)) as Γ(1−z)Γ(z)=(π/(sin(πz))) we finally get f(a)= (1/n)(1−(1/n))a^((1/n) −2) .(π/(sin((π/n)))) Now we can deduce f(8) , f(3), f(n)

letstate∀a>0f(a)=∫0∞dx(xn+a)2 f(a)=∫0∞dxa2[(xa1n)n+1]2 whenchangingx=a1nuweget f(a)=a1n−2∫0∞du(un+1)2 nowchangev=1un+1⇒u=(1v−1)1n⇒du=1n.(−1v2)(1v−1)1n−1dv then f(a)=1n.a1n−2∫01v2.(1v2)(1v−1)1n−1dv =a1n−2n∫01v1−1n(1−v)1n−1dv =a1n−2nB(2−1n,1n) =a1n−2n.Γ(2−1n)Γ(1n)Γ(2) asΓ(x+1)=xΓ(x)wehaveΓ(2−1n)=(1−1n)Γ(1−1n)andΓ(2)=Γ(1)=1 Sof(a)=a1n−2n.(1−1n)Γ(1−1n)Γ(1n) asΓ(1−z)Γ(z)=πsin(πz)wefinallyget f(a)=1n(1−1n)a1n−2.πsin(πn) Nowwecandeducef(8),f(3),f(n)

Commented by~ À ® @ 237 ~ last updated on 16/Aug/19

if ∀ p>2 g_p (a)=∫_0 ^∞ (dx/((x^n +a)^p )) start as the previous and reach to g_p (a)=(a^((1/n) −p) /n) ∫_(0 ) ^1 v^p .((1/v^2 ))((1/v)−1)^((1/n)−1) dv = (a^((1/n) −p) /n) ∫_0 ^1 v^(p−2+1−(1/n)) (1−v)^((1/n)−1) dv =(a^((1/n)−p) /n) B(p−(1/n), (1/n)) =(a^((1/n)−p) /n) .((Γ(p−(1/n))Γ((1/n)))/(Γ(p))) as Γ(p+z)=z(z+1).......(z+p−1)Γ(z) we have Γ(p−(1/n))=Γ(p−1+(1−(1/n)))=Π_(k=0) ^(p−2) [(1−(1/n))+k] Γ(1−(1/n)) Now g_p (a)=(a^((1/n) −p) /(n.(p−1)!)) .(Π_(k=0) ^(p−2) [(1−(1/n))+k]) Γ(1−(1/n))Γ((1/n)) =(a^((1/n)−p) /(n.(p−1)!)) .(π/(sin((π/n)))) .Π_(k=0) ^(p−2) [(1−(1/n))+k] now we can deduce g_3 (8) ,∙.......

if∀p>2gp(a)=∫0∞dx(xn+a)p startasthepreviousandreachto gp(a)=a1n−pn∫01vp.(1v2)(1v−1)1n−1dv =a1n−pn∫01vp−2+1−1n(1−v)1n−1dv =a1n−pnB(p−1n,1n) =a1n−pn.Γ(p−1n)Γ(1n)Γ(p) asΓ(p+z)=z(z+1).......(z+p−1)Γ(z)wehaveΓ(p−1n)=Γ(p−1+(1−1n))=∏p−2k=0[(1−1n)+k]Γ(1−1n) Now gp(a)=a1n−pn.(p−1)!.(∏p−2k=0[(1−1n)+k])Γ(1−1n)Γ(1n) =a1n−pn.(p−1)!.πsin(πn).∏p−2k=0[(1−1n)+k] nowwecandeduceg3(8),⋅.......

Commented bymathmax by abdo last updated on 16/Aug/19

thank you sir.

thankyousir.

Commented bymathmax by abdo last updated on 16/Aug/19

let g(a) =∫_0 ^∞ (dx/((a+x^n )^2 )) with a>0 we have proved that g(a) =((π(1−(1/n))a^((1/n)−2) )/(nsin((π/n)))) also we have g^′ (a) =−∫_0 ^∞ ((2(a+x^n ))/((a+x^n )^4 ))dx =−2 ∫_0 ^∞ (dx/((a+x^n )^3 )) ⇒ ∫_0 ^∞ (dx/((a+x^n )^3 )) =−(1/2)g^′ (a) we have g^′ (a) =(π/(nsin((π/n))))(1−(1/n))((1/n)−2)a^((1/n)−3) ⇒ ∫_0 ^∞ (dx/((a+x^n )^3 )) =((2π)/(nsin((π/n))))(1−(1/n))(2−(1/n))a^((1/n)−3) a=8 ⇒∫_0 ^∞ (dx/((8+x^n )^3 )) =((2π)/(nsin((π/n))))(1−(1/n))(2−(1/n))8^((1/n)−3)

letg(a)=∫0∞dx(a+xn)2witha>0wehaveprovedthat g(a)=π(1−1n)a1n−2nsin(πn)alsowehave g′(a)=−∫0∞2(a+xn)(a+xn)4dx=−2∫0∞dx(a+xn)3⇒ ∫0∞dx(a+xn)3=−12g′(a)wehave g′(a)=πnsin(πn)(1−1n)(1n−2)a1n−3⇒ ∫0∞dx(a+xn)3=2πnsin(πn)(1−1n)(2−1n)a1n−3 a=8⇒∫0∞dx(8+xn)3=2πnsin(πn)(1−1n)(2−1n)81n−3

Commented bymathmax by abdo last updated on 16/Aug/19

error at final line ∫_0 ^∞ (dx/((a+x^n )^3 )) =(π/(2nsin((π/n))))(1−(1/n))(2−(1/n))a^((1/n)−3) a=8 ⇒∫_0 ^∞ (dx/((8+x^n )^3 )) =(π/(2nsin((π/n))))(1−(1/n))(2−(1/n))8^((1/n)−3)

erroratfinalline∫0∞dx(a+xn)3=π2nsin(πn)(1−1n)(2−1n)a1n−3 a=8⇒∫0∞dx(8+xn)3=π2nsin(πn)(1−1n)(2−1n)81n−3

Terms of Service

Privacy Policy

Contact: info@tinkutara.com