let-n-from-N-and-A-n-cos-ax-x-2-x-1-n-dx-and-B-n-sin-ax-x-2-x-1-n-dx-find-the-value-of-A-n-and-B-n-

Question Number 38727 by maxmathsup by imad last updated on 28/Jun/18

l e t n f r o m N a n d A_{n} = \int_{- \infty}^{+ \infty} \frac{c o s (a x)}{{(x^{2} + x + 1)}^{n}} d x a n d

B_{n} = \int_{- \infty}^{+ \infty} \frac{s i n (a x)}{{(x^{2} + x + 1)}^{n}} d x

f i n d t h e v a l u e o f A_{n} a n d B_{n} .

Commented by math khazana by abdo last updated on 01/Jul/18

w e h a v e A_{n} + {i B}_{n} = \int_{- \infty}^{+ \infty} \frac{e^{i a x}}{{(x^{2} + x + 1)}^{n}} d x = W_{n}

W_{n} = \int_{- \infty}^{+ \infty} \frac{e^{i a x}}{{({(x + \frac{1}{2})}^{2} + \frac{3}{4})}^{n}} d x c h a n g e m e n t

x + \frac{1}{2} = \frac{\sqrt{3}}{2} t g i v e

W_{n} = {(\frac{4}{3})}^{n} \int_{- \infty}^{+ \infty} \frac{e^{i a (\frac{\sqrt{3}}{2} t - \frac{1}{2})}}{{(t^{2} + 1)}^{n}} \frac{\sqrt{3}}{2} d t

= \frac{\sqrt{3}}{2} {(\frac{4}{3})}^{n} e^{- \frac{i a}{2}} \int_{- \infty}^{+ \infty} \frac{e^{i \frac{\sqrt{3}}{2} a t}}{{(t^{2} + 1)}^{n}} d t l e t λ = \frac{\sqrt{3}}{2} a

a n d f i n d \int_{- \infty}^{+ \infty} \frac{e^{i λ t}}{{(t^{2} + 1)}^{n}} = I_{λ}

l e t φ (z) = \frac{e^{i λ z}}{{(z^{2} + 1)}^{n}}

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

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

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)!} {e^{i λ z} {(z + i)}^{- n}}^{(n - 1)} b u t

{e^{i λ z} {(z + i)}^{(- n)}}^{(p)} = \sum_{k = 0}^{p} C_{p}^{k} {{(z + i)}^{(- n)}}^{(k)} {(e^{i λ z})}^{(p - k)}

\dots b e c o n t i n u e d \dots

Commented by math khazana by abdo last updated on 02/Jul/18

l e t f i n d {({(z + i)}^{- n})}^{(k)}

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

{{(z + i)}^{- n}}^{(2)} = {(- 1)}^{2} n (n + 1) {(z + i)}^{- n - 2} \Rightarrow

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

a l s o w e h a v e {(e^{i λ z})}^{(1)} = i λ e^{i λ z}

{(e^{i λ z})}^{(2)} = {(i λ)}^{2} e^{i λ z} \Rightarrow {(e^{i λ z})}^{(p - k)} = {(i λ)}^{p - k} e^{i λ z} \Rightarrow

{e^{i λ z} {(z + i)}^{- n}}^{(p)} = \sum_{k = 0}^{p} C_{p}^{k} {(- 1)}^{k} n (n + 1) \dots (n + k - 1) {(z + i)}^{- n - k} {(i λ)}^{p - k} e^{i λ z}

p = n - 1 \Rightarrow

{e^{i λ z} {(z + i)}^{- n}}^{(n - 1)} = \sum_{k = 0}^{n - 1} C_{n - 1}^{k} {(- 1)}^{k} n (n + 1) \dots (2 n - 1) {(z + i)}^{- 2 n + 1} {(i λ)}^{n - 1 - k} e^{i λ z}

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