Question Number 148249 by mnjuly1970 last updated on 26/Jul/21

Commented byKamel last updated on 26/Jul/21

Ω_n =∫_0 ^(+∞) ((cos(πx))/((x^2 +1^2 )(x^2 +2^2 )....(x^2 +n^2 )))dx (1/((x^2 +1^2 )(x^2 +2^2 )....(x^2 +n^2 )))=Σ_(k=1) ^n (a_k /((x^2 +k^2 )))...(E) X=x^2 , (E)×(x^2 +k^2 ), X→−k^2 a_k =(1/((1^2 −k^2 )(2^2 −k^2 )...((k−1)^2 −k^2 )((k+1)^2 −k^2 )((k+2)^2 −k^2 )...(n^2 −k^2 ))) =(((−1)^(k−1) k!(2k))/((k−1)!(2k−1)!(n−k)!(2k+1)(2k+2)(2k+3)...(n+k))) =((2(−1)^(k−1) k^2 )/((n+k)!(n−k)!)) Ω_n =2Σ_(k=1) ^n ((k^2 (−1)^(k−1) )/((n+k)!(n−k)!))∫_0 ^(+∞) ((cos(πx))/(x^2 +k^2 ))dx ∫_0 ^(+∞) ((cos(πx))/(x^2 +k^2 ))dx=(π/(2k))e^(−kπ) ∴ Ω_n =πΣ_(k=1) ^n (((−1)^(k−1) ke^(−kπ) )/((n+k)!(n−k)!)) ∴ ∫_0 ^(+∞) ((cos(πx))/((x^2 +1^2 )(x^2 +2^2 )....(x^2 +n^2 )))dx=πΣ_(k=1) ^n (((−1)^(k−1) ke^(−kπ) )/((n+k)!(n−k)!))

Answered by Olaf_Thorendsen last updated on 26/Jul/21

Ω = ∫_0 ^∞ ((cos(πx))/(Π_(k=1) ^n (x^2 +k^2 ))) dx Ω = (1/2)∫_(−∞) ^∞ ((cos(πx))/(Π_(k=1) ^n (x^2 +k^2 ))) dx Let f(x) = ((cos(πx))/(Π_(k=1) ^n (x^2 +k^2 ))) dx The function f is even : Ω = (1/2)Re∫_(−∞) ^∞ (e^(iπx) /(Π_(k=1) ^n (x^2 +k^2 ))) dx = (1/2)∫_(−∞) ^∞ (e^(iπx) /(Π_(k=1) ^n (x^2 +k^2 ))) dx Ω = 2iπΣ_(k=1) ^n Res(f(x),ik) Res(f(x),ik) = lim_(x→ik) ((e^(iπx) /((x+ik)Π_(p=1_(p≠k) ) ^n (x^2 +p^2 )))) Res(f(x),ik) = (e^(−kπ) /((2ik)Π_(p=1_(p≠k) ) ^n (p^2 −k^2 ))) Ω = πΣ_(k=1) ^n (e^(−kπ) /(kΠ_(p=1_(p≠k) ) ^n (p^2 −k^2 ))) ...to be continued...

Answered by mathmax by abdo last updated on 26/Jul/21

Ψ_n =∫_0 ^∞ ((cos(πx))/((x^2 +1^2 )(x^2 +2^2 ).....(x^2 +n^2 )))dx ⇒ 2Ψ_n =Re(∫_(−∞) ^(+∞) (e^(iπx) /((x^2 +1^2 )(x^2 +2^2 )....(x^2 +n^2 )))dx) let ϕ(z)=(e^(iπz) /((z^2 +1^2 )(z^2 +2^2 )....(z^2 +n^2 ))) ⇒ ϕ(z)=(e^(iπz) /((z−i)(z+i)(z+2i)(z−2i)....(z−ni)(z+ni))) the poles of ϕ are +^− ki rdsidus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπΣ_(k=1) ^n Res(ϕ,ki) ϕ(z)=(e^(iπz) /((z−i)(z−2i)....(z−ni)(z+i)(z+2i)...(z+ni))) =(e^(iπz) /((z−i)(z−2i)....(z−(k−1)i)(z−ki)(z−(k+1)i)...(z−ni)Π_(k=1) ^n (z+ki))) ⇒Res(ϕ,ki)=(e^(iπ(ki)) /((ki−i)(ki−2i)....(i)(−i)...(ki−ni)Π_(k=1) ^n (2ki))) =(e^(−πk) /(i^(k−1) (k−1)!(−i)^(n−k) (n−k)!(2i)^n n!)) =(e^(−kπ) /((−1)^(n−k) i^(n−1) (i)^n 2^n (k−1)!(n−k)!n!)) =i(e^(−kπ) /((−1)^k 2^n (k−1)!(n−k)!n!)) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =((2iπ)/(n!×2^n ))iΣ_(k=1) ^n (((−1)^k e^(−kπ) )/((k−1)!(n−k)!)) =−(π/(n! 2^(n−1) ))Σ_(k=1) ^n (((−1)^k e^(−kπ) )/((k−1)!(n−k)!)) =(π/(n!2^(n−1) ))Σ_(k=1) ^n (((−1)^(k−1) e^(−kπ) )/((k−1)!(n−k)!))=2Ψ_n ⇒ Ψ_n =(π/(2^n n!))Σ_(k=1) ^n (((−1)^(k−1) e^(−kπ) )/((k−1)!(n−k)!))