Question Number 100850 by Dwaipayan Shikari last updated on 28/Jun/20

∫_0 ^(102) (x−1)(x−2).....(x−100)×((1/(x−1))+(1/(x−2))+...+(1/(x−100)))dx

Commented byDwaipayan Shikari last updated on 28/Jun/20

∫_0 ^(102) (d/dx)((x−1)(x−2).....(x−100))dx ∫_0 ^(102) d((x−1)(x−2).....(x−100)) [(x−1)(x−2)....(x−100)]_0 ^(102) =101!−100!

Answered by maths mind last updated on 28/Jun/20

∫_0 ^(102) ((Γ(x))/(Γ(x−100)))(Ψ(x)−Ψ(x−100))dx =∫_0 ^(102) ((Γ(x))/(Γ(x−100)))Ψ(x)−∫_0 ^(102) ((Γ(x)Ψ(x−100))/(Γ(x−100))) =∫_0 ^(102) ((Γ′(x)Γ(x−100)−Γ′(x−100)Γ(x))/(Γ(x−100)^2 ))dx =∫_0 ^(102) d(((Γ(x))/(Γ(x−100)))) =[((Γ(x))/(Γ(x−100)))]_0 ^(102) =((Γ(102))/(Γ(2)))−lim_(a→0) ((Γ(a))/(Γ(a−100))) Γ(a)=(a−1)......(a−100)Γ(a−100) ⇒((Γ(a))/(Γ(a−100)))=(100−a)....(1−a) ⇒lim_(a→0) ((Γ(a))/(Γ(a−100)))=Γ(101) we get Γ(102)−Γ(101)=100Γ(101)

Commented bymaths mind last updated on 28/Jun/20

we can get ∫_0 ^(k+2) Π_(j=0) ^k (x−j).(Σ_(j=0) ^k (1/(x−j)))dx =kΓ(k+1)

Answered by mathmax by abdo last updated on 28/Jun/20

let p(x) =Π_(k=1) ^(100) (x−k) ⇒((p^′ (x))/(p(x))) =Σ_(k=1) ^(100) (1/(x−k)) ⇒ p(x)Σ_(k=1) ^(100) (1/(x−k)) =p^′ (x) ⇒∫_0 ^(102) (x−1)(x−2)...(x−100)×((1/(x−1))+(1/(x−2))+...+(1/(x−100)))dx =∫_0 ^(102) p^′ (x)dx =[p(x)]_0 ^(102) =p(102)−p(0) =101×100×...×2−1×2×3×...100 =(101)!−100!