calculae-A-n-0-dx-x-2-1-n-with-n-integr-natural-and-n-gt-0-

Question Number 67006 by mathmax by abdo last updated on 21/Aug/19

c a l c u l a e A_{n} = \int_{0}^{\infty} \frac{d x}{{(x^{2} + 1)}^{n}} w i t h n i n t e g r n a t u r a l a n d n > 0

Commented by mathmax by abdo last updated on 22/Aug/19

A_{n} = \int_{0}^{\infty} \frac{d x}{{(x^{2} + 1)}^{n}} c h a n g e m e n t x = t a n θ g i v e

A_{n} = \int_{0}^{\frac{π}{2}} \frac{(1 + {t a n}^{2} θ)}{{(1 + {t a n}^{2} θ)}^{n}} d θ = \int_{0}^{\frac{π}{2}} {({c o s}^{2} θ)}^{n - 1} d θ = \int_{0}^{\frac{π}{2}} {c o s}^{2 n - 2} θ d θ

(i t s a w a l l i s i n t e g r a l) w e h a v e A_{n + 1} = \int_{0}^{\frac{π}{2}} {c o s}^{2 n} θ d θ

= \int_{0}^{\frac{π}{2}} (1 - {s i n}^{2} θ) {c o s}^{2 n - 2} θ d θ = A_{n - 1} - \int_{0}^{\frac{π}{2}} {s i n}^{2} θ {c o s}^{2 n - 2} θ d θ

b y p a r t s u = s i n θ a n d v^{^{'}} = s i n θ {c o s}^{2 n - 2} θ \Rightarrow

\int_{0}^{\frac{π}{2}} {s i n}^{2} θ {c o s}^{2 n - 2} θ d θ = {[- \frac{1}{2 n - 1} s i n θ {c o s}^{2 n - 1} θ]}_{0}^{\frac{π}{2}} + \frac{1}{2 n - 1} \int_{0}^{\frac{π}{2}} c o s θ {c o s}^{2 n - 1} θ d θ

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

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

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

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

A_{n} = \frac{1.3.5 \dots \dots (2 n - 3)}{2^{n - 1} (n - 1)!} A_{1} = \frac{1.2.3.4.5 \dots . (2 n - 3) (2 n - 2)}{2^{n - 1} (n - 1)! 2.4 \dots . 2 (n - 1)} A_{1}

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