Calculate-for-n-N-0-dt-t-2-1-n-Notice-1-1-t-2-t-2-

Question Number 169797 by mathocean1 last updated on 09/May/22

C a l c u l a t e f o r n \in N^{*} : \int_{0}^{+ \infty} \frac{d t}{{(t^{2} + 1)}^{n}}

(N o t i c e : 1 = (1 + t^{2}) - t^{2})

Answered by floor(10²Eta[1]) last updated on 09/May/22

t = tg θ \Rightarrow dt = \sec^{2} θ d θ

\int_{0}^{π / 2} \frac{\sec^{2} d θ}{\sec^{2 n} θ} = \int_{0}^{π / 2} \cos^{2 n - 2} θ d θ, use reduction formula .

Answered by Mathspace last updated on 09/May/22

u_{n} = \int_{0}^{\infty} \frac{d t}{{(t^{2} + 1)}^{n}} \Rightarrow

w e h a v e B (x, y) = \int_{0}^{\infty} \frac{t^{x - 1}}{{(1 + t)}^{x + y}} d t

s o \int_{0}^{\infty} \frac{d t}{{(1 + t^{2})}^{n}} =_{t = \sqrt{x}} \frac{1}{2} \int_{0}^{\infty} \frac{x^{- \frac{1}{2}}}{{(1 + x)}^{n}} d x

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

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

= \frac{1}{2} B (\frac{1}{2}, n - \frac{1}{2})

= \frac{1}{2} \times \frac{Γ (\frac{1}{2}) . Γ (n - \frac{1}{2})}{Γ (n)}

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

= \frac{\sqrt{π}}{2} \times \frac{(n - \frac{3}{2})!}{(n - 1)!} (n > 0)

Answered by Mathspace last updated on 09/May/22

r e s i d u s m e t h o d

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

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

\int_{R} φ 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)!} {{(z + i)}^{- n}}^{(n - 1)}

{w e h a v e {(z + i)}^{p}}}^{(1)} = p {(z + i)}^{p - 1}

Missing or unrecognized delimiter for \left

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

\dots . .