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Question Number 199236 by SANOGO last updated on 30/Oct/23

calcul ∫_0 ^(+oo) (1/(x^α +x^β ))dx a>o

calcul∫0+oo1xα+xβdxa>o

Answered by witcher3 last updated on 30/Oct/23

α≥β x^a +x^β =x^β (1+x^(a−𝛃) )∼x^β ,a−β≥0 integrable if β<1 x^a +x^β =x^a (1+x^(β−α) )∼x^a integrable if a>1,in ∞ f(a,b)=∫_0 ^∞ (dx/(x^a +x^b )), a>1>b ∫_0 ^∞ ((x^(−a) dx)/(1+x^(b−a) ))dx x^(b−a) =t⇒x=t^(1/(b−a)) =∫_∞ ^0 (t^(a/(a−b)) /(1+t)).(t^((1/(b−a))−1) /((b−a)))dt =(1/(a−b))∫_0 ^∞ (t^(((a−1)/(a−b))−1) /(1+t))dt β(x,y)=∫_0 ^∞ (t^(x−1) /((1+t)^(x+y) ))dx f(a,b)=(1/(a−b)).β(((a−1)/(a−b)),1−((a−1)/(a−b)))=(1/(a−b)).(π/(sin(π((a−1)/(a−b))))) f(a,b)=(π/((a−b)sin(((π(a−1))/(a−b)))));a>1>b

α⩾βxa+xβ=xβ(1+xa−β)∼xβ,a−β⩾0integrableifβ<1xa+xβ=xa(1+xβ−α)∼xaintegrableifa>1,in∞f(a,b)=∫0∞dxxa+xb,a>1>b∫0∞x−adx1+xb−adxxb−a=t⇒x=t1b−a=∫∞0taa−b1+t.t1b−a−1(b−a)dt=1a−b∫0∞ta−1a−b−11+tdtβ(x,y)=∫0∞tx−1(1+t)x+ydxf(a,b)=1a−b.β(a−1a−b,1−a−1a−b)=1a−b.πsin(πa−1a−b)f(a,b)=π(a−b)sin(π(a−1)a−b);a>1>b

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