find-the-value-of-cosx-x-2-1-n-dx-n-fromN-and-n-1-

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 (\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) \dots . (p - k + 1) {(z + i)}^{p - k} \Rightarrow

{{(z + i)}^{- n}}^{(k)} = (- n) (- n - 1) \dots . (- n - k + 1) {(z + i)}^{- n - k}

= {(- 1)}^{k} n (n + 1) \dots (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) \dots (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) \dots . (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) \dots . (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) \dots (n + k - 1)}{2^{k}} \times C_{n}^{k}

= Re (\int \dots .) = \int_{- \infty}^{+ \infty} \frac{cosx}{{(x^{2} + 1)}^{n}} dx