calculate-x-n-1-x-2-n-dx-with-n-natral-integr-

Question Number 39711 by math khazana by abdo last updated on 10/Jul/18

c a l c u l a t e \int_{- \infty}^{+ \infty} \frac{x^{n}}{{(1 + x^{2})}^{n}} d x w i t h n n a t r a l i n t e g r

Commented by maxmathsup by imad last updated on 11/Jul/18

l e t A_{n} = \int_{- \infty}^{+ \infty} \frac{x^{n} d x}{{(1 + x^{2})}^{n}} a n d l e t φ (z) = \frac{z^{n}}{{(1 + z^{2})}^{n}}

φ (z) = \frac{z^{n}}{{(z - i)}^{n} {(z + i)}^{n}}

\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, i) b u t

R e s (φ, i) = {l i m}_{z \to i} \frac{1}{(n - 1)!} {{(z - i)}^{n} φ (z)}^{(n - 1)}

= {l i m}_{z \to i} \frac{1}{(n - 1)!} {x^{n} {(z + i)}^{- n}}^{(n - 1)} a n d

{x^{n} {(z + i)}^{- n}}^{(n - 1)} = \sum_{k = 0}^{n - 1} C_{n - 1}^{k} {{(z + i)}^{- n}}^{(k)} {(x^{n})}^{(n - 1 - k)} b u t

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

{(x^{n})}^{(p)} = n (n - 1) \dots (n - p + 1) x^{n - p} \Rightarrow {(x^{n})}^{(n - 1 - k)} = n (n - 1) \dots (k + 2) x^{k + 1} \Rightarrow

R e s (φ, i) = \frac{1}{(n - 1)!} \sum_{k = 0}^{n - 1} C_{n - 1}^{k} {(- 1)}^{k} n (n + 1) \dots (n + k - 1) {(2 i)}^{- n - k} (n - 1) \dots (k + 2) i^{k + 1}

A_{n} = 2 i π R e s (φ, i)