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Question Number 31296 by abdo imad last updated on 05/Mar/18

find ∫_0 ^(+∞) (dx/((1+x^2 )^n )) with n integr and n≥1 .

$${find}\:\:\int_{\mathrm{0}} ^{+\infty} \:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{{n}} }\:\:{with}\:{n}\:{integr}\:{and}\:{n}\geqslant\mathrm{1}\:. \\ $$

Commented by NECx last updated on 07/Mar/18

thats the greatest lie of the century.

$${thats}\:{the}\:{greatest}\:{lie}\:{of}\:{the}\:{century}. \\ $$

Commented by abdo imad last updated on 06/Mar/18

let put I_n =∫_0 ^∞ (dx/((1+x^2 )^n )) ⇒ I_n =(1/2) ∫_(−∞) ^(+∞) (dx/((1+x^2 )^n )) let put ϕ(z)= (1/((1+z^2 )^n )) we have ϕ(z)= (1/((z−i)^n (z+i)^n )) so thepoles of f are i and −i (poles with ordre n) ∫_(−∞) ^(+∞) ϕ(z)dz=2iπ Res(ϕ,i) but Res(ϕ,i)=lim_(z→i) (1/((n−1)!))( (z−i)^n f(z))^((n−1)) =lim_(z→i) (1/((n−1)!))( (1/((z+i)^n )))^((n−1)) first let find ((z+i)^p )^((k)) with k ≤p (z+i)^p )^((1)) =p(z+i)^(p−1) ,((z+i)^p )^((2)) =p(p−1)(z+i)^(p−2) ⇒ ((z+i)^p )^((k)) =p(p−1)...(p−k+1)(z+i)^(p−k) and for p=−n and k=n−1 weget ((z+i)^(−n) )^((n−1)) =(−n)(−n−1)....(−n −n+1+1)(z+i)^(−n−n+1) =(−1)^(n−1) n(n+1)(n+2)....(2n−2)(z+i)^(−2n+1) Res(ϕ,i)= (1/((n−1)!)) (−1)^(n−1) n(n+1)(n+2)...(2n−2)(2i)^(−2n+1) =(((−1)^(n−1) )/(2^(2n−1) (n−1)! i^(2n−1) )) n(n+1)(n+2)....(2n−2) = ((−i)/2^(2n−1) ) (1/((n−1)!))n(n+1)(n+2)...(2n −2)⇒ I_n =(1/2)2iπ ((−i)/2^(2n−1) )(1/((n−1)!))n(n+1)(n+2)...(2n−2) I_n = (π/2^(2n−1) ) n(n+1)(n+2)....(2n−2)=((π(2n−2)!)/(((n−1)!)^2 2^(2n−1) )) .

$${let}\:{put}\:{I}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{{n}} }\:\Rightarrow\:{I}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{{n}} }\: \\ $$$${let}\:{put}\:\varphi\left({z}\right)=\:\frac{\mathrm{1}}{\left(\mathrm{1}+{z}^{\mathrm{2}} \right)^{{n}} }\:{we}\:{have}\:\varphi\left({z}\right)=\:\:\frac{\mathrm{1}}{\left({z}−{i}\right)^{{n}} \left({z}+{i}\right)^{{n}} }\:{so} \\ $$$${thepoles}\:{of}\:{f}\:{are}\:{i}\:{and}\:−{i}\:\left({poles}\:{with}\:{ordre}\:{n}\right) \\ $$$$\int_{−\infty} ^{+\infty} \varphi\left({z}\right){dz}=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{i}\right)\:\:{but} \\ $$$${Res}\left(\varphi,{i}\right)={lim}_{{z}\rightarrow{i}} \:\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!}\left(\:\left({z}−{i}\right)^{{n}} {f}\left({z}\right)\right)^{\left({n}−\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!}\left(\:\:\:\frac{\mathrm{1}}{\left({z}+{i}\right)^{{n}} }\right)^{\left({n}−\mathrm{1}\right)} \:{first}\:{let}\:{find} \\ $$$$\left(\left({z}+{i}\right)^{{p}} \right)^{\left({k}\right)} \:{with}\:{k}\:\leqslant{p} \\ $$$$\left.\left({z}+{i}\right)^{{p}} \right)^{\left(\mathrm{1}\right)} ={p}\left({z}+{i}\right)^{{p}−\mathrm{1}} \:\:\:,\left(\left({z}+{i}\right)^{{p}} \right)^{\left(\mathrm{2}\right)} ={p}\left({p}−\mathrm{1}\right)\left({z}+{i}\right)^{{p}−\mathrm{2}} \Rightarrow \\ $$$$\left(\left({z}+{i}\right)^{{p}} \right)^{\left({k}\right)} ={p}\left({p}−\mathrm{1}\right)...\left({p}−{k}+\mathrm{1}\right)\left({z}+{i}\right)^{{p}−{k}} \:{and} \\ $$$${for}\:{p}=−{n}\:{and}\:{k}={n}−\mathrm{1}\:{weget} \\ $$$$\left(\left({z}+{i}\right)^{−{n}} \right)^{\left({n}−\mathrm{1}\right)} =\left(−{n}\right)\left(−{n}−\mathrm{1}\right)....\left(−{n}\:−{n}+\mathrm{1}+\mathrm{1}\right)\left({z}+{i}\right)^{−{n}−{n}+\mathrm{1}} \\ $$$$=\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)....\left(\mathrm{2}{n}−\mathrm{2}\right)\left({z}+{i}\right)^{−\mathrm{2}{n}+\mathrm{1}} \\ $$$${Res}\left(\varphi,{i}\right)=\:\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!}\:\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)...\left(\mathrm{2}{n}−\mathrm{2}\right)\left(\mathrm{2}{i}\right)^{−\mathrm{2}{n}+\mathrm{1}} \\ $$$$=\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{\mathrm{2}^{\mathrm{2}{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!\:{i}^{\mathrm{2}{n}−\mathrm{1}} }\:{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)....\left(\mathrm{2}{n}−\mathrm{2}\right) \\ $$$$=\:\frac{−{i}}{\mathrm{2}^{\mathrm{2}{n}−\mathrm{1}} }\:\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!}{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)...\left(\mathrm{2}{n}\:−\mathrm{2}\right)\Rightarrow \\ $$$${I}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}\mathrm{2}{i}\pi\:\frac{−{i}}{\mathrm{2}^{\mathrm{2}{n}−\mathrm{1}} }\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!}{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)...\left(\mathrm{2}{n}−\mathrm{2}\right) \\ $$$${I}_{{n}} =\:\frac{\pi}{\mathrm{2}^{\mathrm{2}{n}−\mathrm{1}} }\:{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)....\left(\mathrm{2}{n}−\mathrm{2}\right)=\frac{\pi\left(\mathrm{2}{n}−\mathrm{2}\right)!}{\left(\left({n}−\mathrm{1}\right)!\right)^{\mathrm{2}} \:\mathrm{2}^{\mathrm{2}{n}−\mathrm{1}} }\:. \\ $$

Commented by NECx last updated on 07/Mar/18

wow..... Prof Abdo Imad,I′m so amased by the questions you solve.Thanks for you help on the platform.

$${wow}.....\:{Prof}\:{Abdo}\:{Imad},{I}'{m}\:{so} \\ $$$${amased}\:{by}\:{the}\:{questions}\:{you} \\ $$$${solve}.{Thanks}\:{for}\:{you}\:{help}\:{on}\:{the} \\ $$$${platform}. \\ $$

Commented by NECx last updated on 07/Mar/18

If I may ask,which aspect of integration is this ?

$${If}\:{I}\:{may}\:{ask},{which}\:{aspect}\:{of} \\ $$$${integration}\:{is}\:{this}\:\:? \\ $$

Commented by NECx last updated on 07/Mar/18

this is because I always notice that you always apply complex values to your integrals.

$${this}\:{is}\:{because}\:{I}\:{always}\:{notice} \\ $$$${that}\:{you}\:{always}\:{apply}\:{complex} \\ $$$${values}\:{to}\:{your}\:{integrals}. \\ $$

Commented by prof Abdo imad last updated on 07/Mar/18

its only integration by residus theorem....

$${its}\:{only}\:{integration}\:{by}\:{residus}\:{theorem}.... \\ $$

Commented by prof Abdo imad last updated on 07/Mar/18

I am given all my time to integration because its contain all comsepts of analysis...

$${I}\:{am}\:{given}\:{all}\:{my}\:{time}\:{to}\:{integration}\:{because} \\ $$$${its}\:{contain}\:{all}\:{comsepts}\:{of}\:{analysis}... \\ $$

Commented by NECx last updated on 07/Mar/18

wow.... i′m most grateful

$${wow}....\:{i}'{m}\:{most}\:{grateful} \\ $$

Commented by NECx last updated on 07/Mar/18

I′ve heard of the theorem : cauchy residue theorem.It has to do with comlex analysis.Though its not in Class XI or XII syllabus.u

$${I}'{ve}\:{heard}\:{of}\:{the}\:{theorem}\::\:{cauchy} \\ $$$${residue}\:{theorem}.{It}\:{has}\:{to}\:{do}\:{with} \\ $$$${comlex}\:{analysis}.{Though}\:{its}\:{not} \\ $$$${in}\:{Class}\:{XI}\:{or}\:{XII}\:{syllabus}.{u} \\ $$

Commented by rahul 19 last updated on 07/Mar/18

so u r also a prof ?

$${so}\:{u}\:{r}\:{also}\:\:{a}\:{prof}\:? \\ $$

Commented by rahul 19 last updated on 07/Mar/18

then?

$${then}? \\ $$

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