let-n-from-N-and-find-the-value-of-A-n-1-dt-t-n-t-1-

Question Number 38120 by maxmathsup by imad last updated on 22/Jun/18

l e t n f r o m N a n d

f i n d t h e v a l u e o f A_{n} = \int_{1}^{+ \infty} \frac{d t}{t^{n} \sqrt{t - 1}}

Commented by math khazana by abdo last updated on 26/Jun/18

l e t A_{n} = \int_{1}^{+ \infty} \frac{d t}{t^{n} \sqrt{t - 1}} c h a n g e m e n t \sqrt{t - 1} = x

g i v e A_{n} = \int_{0}^{+ \infty} \frac{2 x d x}{{(x^{2} + 1)}^{n} x} = \int_{- \infty}^{+ \infty} \frac{d x}{{(x^{2} + 1)}^{n}}

l e t φ (z) = \frac{1}{{(z^{2} + 1)}^{n}} h a v e i a n d - i f o r p o l e s

(w i t h m u l t i p l i c i t y n)

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

= {l i m}_{z \to i} \frac{1}{(n - 1)!} {{(z + i)}^{- n}}^{(n - 1)}

l e t d e t e m i n e {{(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) (n + 2) \dots (n + p - 1) {(z + i)}^{- (n + p)}

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

R e s (φ, i) = \frac{1}{(n - 1)!} {(- 1)}^{n - 1} n (n + 1) \dots (2 n - 2) \frac{1}{{(2 i)}^{2 n - 1}}

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

\int_{- \infty}^{+ \infty} φ (z) d z =

2 i π (- i) \frac{n (n + 1) (n + 2) \dots (2 n - 2)}{2^{2 n - 1}}

= 4 π \frac{n (n + 1) (n + 2) \dots . (2 n - 2)}{2^{2 n}} \Rightarrow

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

Commented by math khazana by abdo last updated on 26/Jun/18

\int_{- \infty}^{+ \infty} φ (z) d z = \frac{4 π}{(n - 1)!} \frac{n (n + 1) (n + 2) \dots (2 n - 2)}{2^{2 n}}

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