calculate-A-n-0-dx-x-2-1-n-n-integr-and-n-1-

Question Number 128952 by mathmax by abdo last updated on 11/Jan/21

calculate A_{n} = \int_{0}^{\infty} \frac{dx}{{(x^{2} + 1)}^{n}}, n integr and n ⩾ 1

Answered by Dwaipayan Shikari last updated on 11/Jan/21

\int_{0}^{\infty} \frac{d x}{{(x^{2} + 1)}^{n}}

= \frac{1}{2} \int_{0}^{\infty} \frac{t^{\frac{1}{2} - 1}}{{(t + 1)}^{(n - \frac{1}{2}) + \frac{1}{2}}} d t = β (\frac{1}{2}, n - \frac{1}{2}) = \frac{Γ (n - \frac{1}{2}) Γ (\frac{1}{2})}{2 Γ (n)}

Answered by mathmax by abdo last updated on 11/Jan/21

A_{n} = \int_{0}^{\infty} \frac{x^{2} + 1}{{(x^{2} + 1)}^{n + 1}} dx = \int_{0}^{\infty} \frac{x^{2}}{{(x^{2} + 1)}^{n + 1}} dx + A_{n + 1} we have

\int_{0}^{\infty} \frac{x^{2}}{{(x^{2} + 1)}^{n + 1}} dx = \int_{0}^{\infty} x (x {(x^{2} + 1)}^{- n - 1}) dx by parts u = x and

v^{^{'}} = x {(x^{2} + 1)}^{- n - 1} \Rightarrow v = - \frac{1}{2 n} {(x^{2} + 1)}^{- n}

\Rightarrow \int_{0}^{\infty} \frac{x^{2} dx}{{(x^{2} + 1)}^{n + 1}} = {[- \frac{x}{2 n} {(x^{2} + 1)}^{- n}]}_{0}^{\infty} + \frac{1}{2 n} \int_{0}^{\infty} \frac{dx}{{(x^{2} + 1)}^{n}}

= \frac{1}{2 n} A_{n} \Rightarrow A_{n} = \frac{1}{2 n} A_{n} + A_{n + 1} \Rightarrow A_{n + 1} = (1 - \frac{1}{2 n}) A_{n} \Rightarrow

A_{n + 1} = \frac{2 n - 1}{2 n} A_{n} \Rightarrow \frac{A_{n + 1}}{A_{n}} = \frac{2 n - 1}{2 n} \Rightarrow \prod_{k = 1}^{n - 1} \frac{A_{k + 1}}{A_{k}} = \prod_{k = 1}^{n - 1} \frac{2 k - 1}{2 k}

\Rightarrow \frac{A_{2}}{A_{1}} . \frac{A_{3}}{A_{2}} \dots . . \frac{A_{n}}{A_{n - 1}} = \frac{\prod_{k = 1}^{n - 1} (2 k - 1)}{2^{n - 1} (n - 1)!} \Rightarrow

A_{n} = A_{1} \times \frac{1.3 \dots . (2 n - 3)}{2^{n - 1} (n - 1)!} = A_{1} \times \frac{1.2.3 \dots . . (2 n - 4) (2 n - 3) (2 n - 2)}{{(2^{n - 1})}^{2} (n - 1)!^{2}}

= \frac{(2 n - 2)!}{2^{2 n - 2} {((n - 1)!)}^{2}} A_{1} = \frac{(2 n - 2)!}{2^{2 n - 2} (n - 1)!^{2}} . \frac{π}{2} (n > 1) \Rightarrow

A_{n} = π \times \frac{(2 n - 2)!}{2^{2 n - 1} {(n - 1)!^{2}}}

Answered by mathmax by abdo last updated on 11/Jan/21

residus method A_{n} = \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{dx}{{(x^{2} + 1)}^{n}} let φ (z) = \frac{1}{{(z^{2} + 1)}^{n}} \Rightarrow

φ (z) = \frac{1}{{(z - i)}^{n} {(z + i)}^{n}} residus give \int_{R} φ (z) dz = 2 i π Res (φ, i)

Res (φ, i) = \lim_{z \to i} \frac{1}{(n - 1)!} {{(z - i)}^{n} φ (z)}^{(n - 1)} = \lim_{z \to i} \frac{1}{(n - 1)!} {{(z + i)}^{- n}}^{(n - 1)}

{that lead to find {(z + i)}^{- n}}}^{(p)}

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

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

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

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

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

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

Res (φ, i) = \lim_{z \to i} \frac{1}{(n - 1)!} \times \frac{{(- 1)}^{n - 1} n (n + 1) \dots . (2 n - 2)}{{(z + i)}^{2 n - 1}}

= \frac{{(- 1)}^{n - 1} n (n + 1) \dots . (2 n - 2)}{(n - 1)! {(2 i)}^{2 n - 1}} = i \frac{{(- 1)}^{n - 1} n (n + 1) \dots . (2 n - 2)}{(n - 1)! 2^{2 n - 1} {(- 1)}^{n}}

= - i \frac{n (n + 1) \dots . (2 n - 2)}{2^{2 n - 1} (n - 1)!} \Rightarrow \int_{R} φ (z) dz = 2 i π \times (- i) \times \frac{n (n + 1) \dots . (2 n - 2)}{2^{2 n - 1} (n - 1)!}

= π \times \frac{n (n + 1) (n + 2) \dots (2 n - 2)}{2^{2 n - 2} (n - 1)!} = {2 A}_{n} \Rightarrow

A_{n} = \frac{π}{2} \times \frac{n (n + 1) (n + 2) \dots . (2 n - 2)}{2^{2 n - 2} (n - 1)!}