calculateA-n-0-dx-x-2-n-x-2-2n-with-n-integr-natural-1-

Question Number 109214 by mathmax by abdo last updated on 22/Aug/20

{calculateA}_{n} = \int_{0}^{\infty} \frac{dx}{(x^{2} + n) (x^{2} + 2 n)} with n integr natural ⩾ 1

Answered by mathmax by abdo last updated on 22/Aug/20

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

φ (z) = \frac{1}{(z - i \sqrt{n}) (z + i \sqrt{n}) (z - i \sqrt{2 n}) (z + i \sqrt{2 n})} residus theorem \Rightarrow

\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π {Res (φ, i \sqrt{n}) + Res (φ, - i \sqrt{n})}

Res (φ, i \sqrt{n}) = \frac{1}{2 i \sqrt{n} (- n + 2 n)} = \frac{1}{2 in \sqrt{n}}

Res (φ, i \sqrt{2 n}) = \frac{1}{2 i \sqrt{2 n} (- 2 n + n)} = - \frac{1}{2 in \sqrt{2 n}} \Rightarrow

\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π {\frac{1}{2 in \sqrt{n}} - \frac{1}{2 in \sqrt{2 n}}} = \frac{π}{n \sqrt{n}} - \frac{π}{n \sqrt{2} \sqrt{n}}

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