Menu Close

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

Tinku Tara 0 Comments
Pin




Question Number 38120 by maxmathsup by imad last updated on 22/Jun/18
let n from N and find the value of A_n = ∫_1 ^(+∞) (dt/(t^n (√(t−1))))
$${let}\:\:{n}\:{from}\:{N}\:{and} \\ $$$${find}\:{the}\:{value}\:{of}\:\:{A}_{{n}} =\:\int_{\mathrm{1}} ^{+\infty} \:\:\:\frac{{dt}}{{t}^{{n}} \sqrt{{t}−\mathrm{1}}} \\ $$
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) ) .
$${let}\:{A}_{{n}} =\:\int_{\mathrm{1}} ^{+\infty} \:\:\:\frac{{dt}}{{t}^{{n}} \sqrt{{t}−\mathrm{1}}}\:{changement}\:\sqrt{{t}−\mathrm{1}}={x}\: \\ $$$${give}\:{A}_{{n}} =\:\int_{\mathrm{0}} ^{+\infty} \:\:\:\:\frac{\mathrm{2}{xdx}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{{n}} {x}}\:=\int_{−\infty} ^{+\infty} \:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{{n}} } \\ $$$${let}\:\varphi\left({z}\right)\:=\:\frac{\mathrm{1}}{\left({z}^{\mathrm{2}} +\mathrm{1}\right)^{{n}} }\:\:{have}\:{i}\:{and}\:−{i}\:{for}\:{poles} \\ $$$$\left({with}\:{multiplicity}\:{n}\right) \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{i}\right) \\ $$$${Res}\left(\varphi,{i}\right)={lim}_{{z}\rightarrow{i}} \:\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!}\left\{\:\left({z}−{i}\right)^{{n}} \varphi\left({z}\right)\right\}^{\left({n}−\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!}\:\left\{\:\:\frac{\mathrm{1}}{\left({z}+{i}\right)^{{n}} }\right\}^{\left({n}−\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!}\:\left\{\:\left({z}+{i}\right)^{−{n}} \right\}^{\left({n}−\mathrm{1}\right)} \\ $$$${let}\:{detemine}\:\left\{\left({z}+{i}\right)^{−{n}} \right\}^{\left({p}\right)} \\ $$$$\left\{\left({z}+{i}\right)^{−{n}} \right\}^{\left(\mathrm{1}\right)} =−{n}\left({z}+{i}\right)^{−{n}−\mathrm{1}} \\ $$$$\left\{\left({z}+{i}\right)^{−{n}} \right\}^{\left(\mathrm{2}\right)} =\left(−\mathrm{1}\right)^{\mathrm{2}} {n}\left({n}+\mathrm{1}\right)\left({z}+{i}\right)^{−\left({n}+\mathrm{2}\right)} \\ $$$$\left\{\left({z}+{i}\right)^{−{n}} \right\}^{\left({p}\right)} =\left(−\mathrm{1}\right)^{{p}} {n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)…\left({n}+{p}−\mathrm{1}\right)\left({z}+{i}\right)^{−\left({n}+{p}\right)} \\ $$$$\left\{\left({z}+{i}\right)^{−{n}} \right\}^{\left({n}−\mathrm{1}\right)} =\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {n}\left({n}+\mathrm{1}\right)….\left(\mathrm{2}{n}−\mathrm{2}\right)\left({z}+{i}\right)^{−\left(\mathrm{2}{n}−\mathrm{1}\right)} \\ $$$${Res}\left(\varphi,{i}\right)=\:\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!}\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {n}\left({n}+\mathrm{1}\right)…\left(\mathrm{2}{n}−\mathrm{2}\right)\:\frac{\mathrm{1}}{\left(\mathrm{2}{i}\right)^{\mathrm{2}{n}−\mathrm{1}} } \\ $$$$=\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!}\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {n}\left({n}+\mathrm{1}\right)…\left(\mathrm{2}{n}−\mathrm{2}\right).\frac{{i}}{\mathrm{2}^{\mathrm{2}{n}−\mathrm{1}} .\left(−\mathrm{1}\right)^{{n}} } \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}= \\ $$$$\mathrm{2}{i}\pi\:\left(−{i}\right)\frac{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)…\left(\mathrm{2}{n}−\mathrm{2}\right)}{\mathrm{2}^{\mathrm{2}{n}−\mathrm{1}} } \\ $$$$=\mathrm{4}\pi\:\frac{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)….\left(\mathrm{2}{n}−\mathrm{2}\right)}{\mathrm{2}^{\mathrm{2}{n}} }\:\Rightarrow \\ $$$${A}_{{n}} =\:\frac{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)….\left(\mathrm{2}{n}−\mathrm{2}\right)}{\mathrm{2}^{\mathrm{2}{n}−\mathrm{2}} }\:. \\ $$$$ \\ $$
Commented by math khazana by abdo last updated on 26/Jun/18
∫_(−∞) ^(+∞) ϕ(z)dz= ((4π)/((n−1)!)) ((n(n+1)(n+2)...(2n−2))/2^(2n) ) A_n = (π/((n−1)!)) ((n(n+1)(n+2)....(2n−2))/2^(2n−2) ) .
$$\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}=\:\frac{\mathrm{4}\pi}{\left({n}−\mathrm{1}\right)!}\:\frac{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)…\left(\mathrm{2}{n}−\mathrm{2}\right)}{\mathrm{2}^{\mathrm{2}{n}} } \\ $$$${A}_{{n}} =\:\frac{\pi}{\left({n}−\mathrm{1}\right)!}\:\frac{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)….\left(\mathrm{2}{n}−\mathrm{2}\right)}{\mathrm{2}^{\mathrm{2}{n}−\mathrm{2}} }\:. \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *