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

Question Number 122323 by mathmax by abdo last updated on 15/Nov/20

calculate A_{n} = \int_{0}^{\infty} \frac{dx}{(x^{2} + 1) (x^{2} + 2) \dots . (x^{2} + n)}

wth n integr natural and n ⩾ 1

Commented by Dwaipayan Shikari last updated on 16/Nov/20

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

= \frac{π}{2} (1 - \frac{1}{\sqrt{2}})

\int_{0}^{\infty} \frac{1}{(x^{2} + 1) (x^{2} + 2) (x^{2} + 3)} = \frac{1}{2} \int_{0}^{\infty} \frac{1}{(x^{2} + 1)} - \frac{1}{(x^{2} + 3)} - \int_{0}^{\infty} \frac{1}{(x^{2} + 2)} - \frac{1}{(x^{2} + 3)}

= \frac{1}{2} (\frac{π}{2} - \frac{π}{2 \sqrt{3}}) - (\frac{π}{2 \sqrt{2}} - \frac{π}{2 \sqrt{3}})

= \frac{π}{4} - \frac{π}{2 \sqrt{2}} + \frac{π}{4 \sqrt{3}}

\int_{0}^{\infty} \frac{1}{(x^{2} + 1) (x^{2} + 2) (x^{2} + 3) (x^{2} + 4)} d x

= \frac{1}{3} \int_{0}^{\infty} \frac{1}{(x^{2} + 1) (x^{2} + 2) (x^{2} + 3)} - \frac{1}{3} \int_{0}^{\infty} \frac{1}{(x^{2} + 2) (x^{2} + 3) (x^{2} + 4)}

= \frac{π}{12} - \frac{π}{6 \sqrt{2}} + \frac{π}{6 \sqrt{3}} - \frac{1}{6} \int_{0}^{\infty} \frac{1}{x^{2} + 2} - \frac{1}{x^{2} + 4} + \frac{1}{3} \int_{0}^{\infty} \frac{1}{x^{2} + 3} - \frac{1}{x^{2} + 4}

= \frac{π}{12} - \frac{π}{6 \sqrt{2}} + \frac{π}{6 \sqrt{3}} - \frac{1}{6} (\frac{π}{2 \sqrt{2}} - \frac{π}{2 \sqrt{4}}) + \frac{1}{3} (\frac{π}{2 \sqrt{3}} - \frac{π}{2 \sqrt{4}})

= \frac{π}{12} - \frac{π}{4 \sqrt{2}} + \frac{π}{3 \sqrt{3}}

\dots .

Answered by mathmax by abdo last updated on 16/Nov/20

let decompose F (u) = \frac{1}{(u + 1) (u + 2) \dots . (u + n)} = \frac{1}{\prod_{k = 1}^{n} (u + k)}

F (u) = \sum_{k = 1}^{n} \frac{a_{k}}{u + k} with a_{k} = \lim_{u \to - k} (u + k) F (u)

we have F (u) = \frac{1}{(u + 1) \dots . (u + k - 1) (u + k) (u + k + 1) \dots (u + n)}

(u + k) F (u) = \frac{1}{(u + 1) (u + 2) \dots (u + k - 1) (u + k + 1) \dots . (u + n)} \Rightarrow

a_{k} = \frac{1}{(- k + 1) (- k + 2) \dots . (- 1) (- k + k + 1) (- k + k + 2) \dots (- k + n)}

= \frac{1}{{(- 1)}^{k - 1} (k - 1)!} \times \frac{1}{(n - k)!} = \frac{{(- 1)}^{k - 1}}{(k - 1)! (n - k)!} \Rightarrow

F (u) = \sum_{k = 1}^{n} \frac{{(- 1)}^{k - 1}}{(k - 1)! (n - k)! (u + k)}

{2 A}_{n} = \int_{- \infty}^{+ \infty} F (x^{2}) dx = \sum_{k = 1}^{n} \frac{{(- 1)}^{k - 1}}{(k - 1)! (n - k)!} \int_{- \infty}^{+ \infty} \frac{dx}{x^{2} + k}

\int_{- \infty}^{+ \infty} \frac{dx}{x^{2} + k} = \int_{- \infty}^{+ \infty} \frac{dx}{(x - i \sqrt{k}) (x + i \sqrt{k})} = 2 i π Res (f, i \sqrt{k})

= 2 i π \times \frac{1}{2 i \sqrt{k}} = \frac{π}{\sqrt{k}} \Rightarrow A_{n} = \frac{π}{2} \sum_{k = 1}^{n} \frac{{(- 1)}^{k - 1}}{(k - 1)! (n - k)! \sqrt{k}}