let n from N and find the value of A_n = ∫_1 ^(+∞) (dt/(t^n (√(t−1))))


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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
$let A_n = ∫_1 ^(+∞) (dt/(t^n (√(t−1)))) changement (√(t−1))=x give A_n = ∫_0 ^(+∞) ((2xdx)/((x^2 +1)^n x)) =∫_(−∞) ^(+∞) (dx/((x^2 +1)^n )) let ϕ(z) = (1/((z^2 +1)^n )) have i and −i for poles (with multiplicity n) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i) Res(ϕ,i)=lim_(z→i) (1/((n−1)!)){ (z−i)^n ϕ(z)}^((n−1)) =lim_(z→i) (1/((n−1)!)) { (1/((z+i)^n ))}^((n−1)) =lim_(z→i) (1/((n−1)!)) { (z+i)^(−n) }^((n−1)) let detemine {(z+i)^(−n) }^((p)) {(z+i)^(−n) }^((1)) =−n(z+i)^(−n−1) {(z+i)^(−n) }^((2)) =(−1)^2 n(n+1)(z+i)^(−(n+2)) {(z+i)^(−n) }^((p)) =(−1)^p n(n+1)(n+2)...(n+p−1)(z+i)^(−(n+p)) {(z+i)^(−n) }^((n−1)) =(−1)^(n−1) n(n+1)....(2n−2)(z+i)^(−(2n−1)) Res(ϕ,i)= (1/((n−1)!))(−1)^(n−1) n(n+1)...(2n−2) (1/((2i)^(2n−1) )) =(1/((n−1)!))(−1)^(n−1) n(n+1)...(2n−2).(i/(2^(2n−1) .(−1)^n )) ∫_(−∞) ^(+∞) ϕ(z)dz= 2iπ (−i)((n(n+1)(n+2)...(2n−2))/2^(2n−1) ) =4π ((n(n+1)(n+2)....(2n−2))/2^(2n) ) ⇒ A_n = ((n(n+1)(n+2)....(2n−2))/2^(2n−2) ) .$
$l e t A_{n} = \int_{1}^{+ \infty} \frac{d t}{t^{n} \sqrt{t - 1}} c h a n g e m e n t \sqrt{t - 1} = x$ $g i v e A_{n} = \int_{0}^{+ \infty} \frac{2 x d x}{{(x^{2} + 1)}^{n} x} = \int_{- \infty}^{+ \infty} \frac{d x}{{(x^{2} + 1)}^{n}}$ $l e t φ (z) = \frac{1}{{(z^{2} + 1)}^{n}} h a v e i a n d - i f o r p o l e s$ $(w i t h m u l t i p l i c i t y n)$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, i)$ $R e s (φ, i) = {l i m}_{z \to i} \frac{1}{(n - 1)!} {{(z - i)}^{n} φ (z)}^{(n - 1)}$ $= {l i m}_{z \to i} \frac{1}{(n - 1)!} {\frac{1}{{(z + i)}^{n}}}^{(n - 1)}$ $= {l i m}_{z \to i} \frac{1}{(n - 1)!} {{(z + i)}^{- n}}^{(n - 1)}$ $l e t d e t e m i n e {{(z + i)}^{- n}}^{(p)}$ ${{(z + i)}^{- n}}^{(1)} = - n {(z + i)}^{- n - 1}$ ${{(z + i)}^{- n}}^{(2)} = {(- 1)}^{2} n (n + 1) {(z + i)}^{- (n + 2)}$ ${{(z + i)}^{- n}}^{(p)} = {(- 1)}^{p} n (n + 1) (n + 2) . . . (n + p - 1) {(z + i)}^{- (n + p)}$ ${{(z + i)}^{- n}}^{(n - 1)} = {(- 1)}^{n - 1} n (n + 1) . . . . (2 n - 2) {(z + i)}^{- (2 n - 1)}$ $R e s (φ, i) = \frac{1}{(n - 1)!} {(- 1)}^{n - 1} n (n + 1) . . . (2 n - 2) \frac{1}{{(2 i)}^{2 n - 1}}$ $= \frac{1}{(n - 1)!} {(- 1)}^{n - 1} n (n + 1) . . . (2 n - 2) . \frac{i}{2^{2 n - 1} . {(- 1)}^{n}}$ $\int_{- \infty}^{+ \infty} φ (z) d z =$ $2 i π (- i) \frac{n (n + 1) (n + 2) . . . (2 n - 2)}{2^{2 n - 1}}$ $= 4 π \frac{n (n + 1) (n + 2) . . . . (2 n - 2)}{2^{2 n}} \Rightarrow$ $A_{n} = \frac{n (n + 1) (n + 2) . . . . (2 n - 2)}{2^{2 n - 2}} .$
Commented by math khazana by abdo last updated on 26/Jun/18

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