find the value of ∫_(−∞) ^(+∞) ((cosx)/((x^2 +1)^n ))dx (n fromN and n≥1)


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Question Number 163109 by mathmax by abdo last updated on 03/Jan/22

$find the value of \int_{- \infty}^{+ \infty} \frac{cosx}{{(x^{2} + 1)}^{n}} dx (n fromN and n ⩾ 1)$
	Answered by mathmax by abdo last updated on 05/Jan/22
	$A_n =Re(∫_(−∞) ^(+∞) (e^(ix) /((x^2 +1)^n ))dx) let Ψ(z)=(e^(iz) /((z^2 +1)^n )) ⇒ Ψ(z)=(e^(iz) /((z−i)^n (z+i)^n )) the poles of Ψ are i and −i(ordre=n) ∫_R Ψ(z)dz =2iπ Res(Ψ,i) Res(Ψ,i)=lim_(z→i) (1/((n−1)!)){(z−i)^n Ψ(z)}^()n−1)) =(1/((n−1)!))lim_(z→i) {(z+i)^(−n) e^(iz) }^((n−1)) =(1/((n−1)!))lim_(z→i) Σ_(k=0) ^(n−1) C_(n−1) ^k {(z+i)^(−n) }^((k)) (e^(iz) )^((n−1−k)) we have (e^(iz) )^((n−1−k)) =i^(n−1−k) e^(iz) let find (z+i)^p }^((k)) (z+i)^p }^((1)) =p(z+i)^(p−1) (z+i)^p }^((2)) =p(p−1)(z+i)^(p−2) ⇒by recurrence (z+i)^p }^((k)) =p(p−1)....(p−k+1)(z+i)^(p−k) ⇒ {(z+i)^(−n) }^((k)) =(−n)(−n−1)....(−n−k+1)(z+i)^(−n−k) =(−1)^k n(n+1)...(n+k−1)(z+i)^(−n−k) ⇒ Res(Ψ,i)=(1/((n−1)!)) e^(−1) Σ_(k=0) ^(n−1) C_(n−1) ^k i^(n−1−k) (−1)^k n(n+1)...(n+k−1)(2i)^(−n−k) =(1/((n−1)!))e^(−1) Σ_(k=0) ^(n−1) C_(n−1) ^k ((i^(n−1−k) (−1)^k n(n+1)....(n+k−1)i^(−n−k) )/(2^(n+k) )) =(e^(−1) /(2^n (n−1)!))Σ_(k=0) ^(n−1) C_(n−1) ^k (−i)(−1)^(2k) ((n(n+1)....(n+k−1))/2^k ) ⇒ ∫_R Ψ(z)dz =((2πe^(−1) )/(2^n (n−1)!))Σ_(k=0) ^(n−1) ((n(n+1)...(n+k−1))/2^k )×C_n ^k =Re(∫....)=∫_(−∞) ^(+∞) ((cosx)/((x^2 +1)^n ))dx$
	$A_{n} = Re (\int_{- \infty}^{+ \infty} \frac{e^{ix}}{{(x^{2} + 1)}^{n}} dx) let Ψ (z) = \frac{e^{iz}}{{(z^{2} + 1)}^{n}} \Rightarrow$ $Ψ (z) = \frac{e^{iz}}{{(z - i)}^{n} {(z + i)}^{n}} the poles of Ψ are i and - i (ordre = n)$ $\int_{R} Ψ (z) dz = 2 i π Res (Ψ, i)$ $Res (Ψ, i) = \lim_{z \to i} \frac{1}{(n - 1)!} {{(z - i)}^{n} Ψ (z)}^{) n - 1)}$ $= \frac{1}{(n - 1)!} \lim_{z \to i} {{(z + i)}^{- n} e^{iz}}^{(n - 1)}$ $= \frac{1}{(n - 1)!} \lim_{z \to i} \sum_{k = 0}^{n - 1} C_{n - 1}^{k} {{(z + i)}^{- n}}^{(k)} {(e^{iz})}^{(n - 1 - k)}$ $we have {(e^{iz})}^{(n - 1 - k)} = i^{n - 1 - k} e^{iz}$ ${let find {(z + i)}^{p}}}^{(k)}$ ${(z + i)}^{p}}^{(1)} = p {(z + i)}^{p - 1}$ ${(z + i)}^{p}}^{(2)} = p (p - 1) {(z + i)}^{p - 2} \Rightarrow by recurrence$ ${(z + i)}^{p}}^{(k)} = p (p - 1) . . . . (p - k + 1) {(z + i)}^{p - k} \Rightarrow$ ${{(z + i)}^{- n}}^{(k)} = (- n) (- n - 1) . . . . (- n - k + 1) {(z + i)}^{- n - k}$ $= {(- 1)}^{k} n (n + 1) . . . (n + k - 1) {(z + i)}^{- n - k} \Rightarrow$ $Res (Ψ, i) = \frac{1}{(n - 1)!} e^{- 1} \sum_{k = 0}^{n - 1} C_{n - 1}^{k} i^{n - 1 - k} {(- 1)}^{k} n (n + 1) . . . (n + k - 1) {(2 i)}^{- n - k}$ $= \frac{1}{(n - 1)!} e^{- 1} \sum_{k = 0}^{n - 1} C_{n - 1}^{k} \frac{i^{n - 1 - k} {(- 1)}^{k} n (n + 1) . . . . (n + k - 1) i^{- n - k}}{2^{n + k}}$ $= \frac{e^{- 1}}{2^{n} (n - 1)!} \sum_{k = 0}^{n - 1} C_{n - 1}^{k} (- i) {(- 1)}^{2 k} \frac{n (n + 1) . . . . (n + k - 1)}{2^{k}} \Rightarrow$ $\int_{R} Ψ (z) dz = \frac{2 π e^{- 1}}{2^{n} (n - 1)!} \sum_{k = 0}^{n - 1} \frac{n (n + 1) . . . (n + k - 1)}{2^{k}} \times C_{n}^{k}$ $= Re (\int . . . .) = \int_{- \infty}^{+ \infty} \frac{cosx}{{(x^{2} + 1)}^{n}} dx$
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