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Question Number 70594 by mathmax by abdo last updated on 06/Oct/19

calculate by residus method the integral ∫_0 ^∞ (dx/((1+x^2 )^n )) with n integr and n≥1

$${calculate}\:{by}\:{residus}\:{method}\:{the}\:{integral}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{{n}} } \\ $$$${with}\:{n}\:{integr}\:{and}\:{n}\geqslant\mathrm{1} \\ $$

Commented by mathmax by abdo last updated on 23/Oct/19

let A_n =∫_0 ^∞ (dx/((x^2 +1)^n )) ⇒2A_n =∫_(−∞) ^(+∞) (dx/((x^2 +1)^n )) let W(z)=(1/((z^2 +1)^n )) ⇒W(z)=(1/((z−i)^n (z+i)^n )) so the poles of W are i and −i at ordr nresidus theorem give ∫_(−∞) ^(+∞) W(z)dz =2iπ Res(W,i) Res(W,i)=lim_(z→i) (1/((n−1)!)){(z−i)^n W(z)}^((n−1)) =lim_(z→i) (1/((n−1)!)){(z+i)^(−n) }^((n−1)) let first find{ (z+i)^(−n) }^((p)) we have {(z+i)^(−n) }^((1)) =−n(z+i)^(−n−1) =−n(z+i)^(−(n+1)) {(z+i)}^((2)) =(−1)^2 n(n+1)(z+i)^(−(n+2)) ...⇒{(z+i)^(−n) }^((p)) =(−1)^p n(n+1)...(n+p−1)(z+i)^(−(n+p)) p=n−1 ⇒{(z+i)^(−n) }^((n−1)) =(−1)^(n−1) n(n+1)...(2n−2)(z+i)^(−(2n−1)) Res(W,i) =lim_(z→i) (1/((n−1)!))(−1)^(n−1) n(n+1)....(2n−2)(z+i)^(−2n+1) =(((−1)^(n−1) n(n+1)(n+2)...(2n−2))/((n−1)!))×(1/((2i)^(2n−1) )) =(((−1)^(n−1) n(n+1)....(2n−2))/(2^(2n−1) (−1)^n ))×i =−i((n(n+1)(n+2)...(2n−2))/2^(2n−1) ) ⇒ ∫_(−∞) ^(+∞) W(z)dz =2iπ×−i((n(n+1)(n+2)....(2n−2))/2^(2n−1) ) =(π/4^(n−1) )×n(n+1)....(2n−2) =2A_n ⇒ A_n =((πn(n+1)(n+2)...(2n−2))/(2×4^(n−1) )) with n>1 and A_1 =(π/2)

$${let}\:{A}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{{n}} }\:\Rightarrow\mathrm{2}{A}_{{n}} =\int_{−\infty} ^{+\infty} \:\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{{n}} }\:{let}\:{W}\left({z}\right)=\frac{\mathrm{1}}{\left({z}^{\mathrm{2}} +\mathrm{1}\right)^{{n}} } \\ $$$$\Rightarrow{W}\left({z}\right)=\frac{\mathrm{1}}{\left({z}−{i}\right)^{{n}} \left({z}+{i}\right)^{{n}} }\:{so}\:{the}\:{poles}\:{of}\:{W}\:{are}\:{i}\:{and}\:−{i} \\ $$$${at}\:{ordr}\:{nresidus}\:{theorem}\:{give}\:\int_{−\infty} ^{+\infty} {W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left({W},{i}\right) \\ $$$${Res}\left({W},{i}\right)={lim}_{{z}\rightarrow{i}} \:\:\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!}\left\{\left({z}−{i}\right)^{{n}} {W}\left({z}\right)\right\}^{\left({n}−\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!}\left\{\left({z}+{i}\right)^{−{n}} \right\}^{\left({n}−\mathrm{1}\right)} \:\:{let}\:{first}\:{find}\left\{\:\left({z}+{i}\right)^{−{n}} \right\}^{\left({p}\right)} \\ $$$${we}\:{have}\:\left\{\left({z}+{i}\right)^{−{n}} \right\}^{\left(\mathrm{1}\right)} =−{n}\left({z}+{i}\right)^{−{n}−\mathrm{1}} =−{n}\left({z}+{i}\right)^{−\left({n}+\mathrm{1}\right)} \\ $$$$\left\{\left({z}+{i}\right)\right\}^{\left(\mathrm{2}\right)} =\left(−\mathrm{1}\right)^{\mathrm{2}} {n}\left({n}+\mathrm{1}\right)\left({z}+{i}\right)^{−\left({n}+\mathrm{2}\right)} \\ $$$$...\Rightarrow\left\{\left({z}+{i}\right)^{−{n}} \right\}^{\left({p}\right)} =\left(−\mathrm{1}\right)^{{p}} {n}\left({n}+\mathrm{1}\right)...\left({n}+{p}−\mathrm{1}\right)\left({z}+{i}\right)^{−\left({n}+{p}\right)} \\ $$$${p}={n}−\mathrm{1}\:\Rightarrow\left\{\left({z}+{i}\right)^{−{n}} \right\}^{\left({n}−\mathrm{1}\right)} =\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {n}\left({n}+\mathrm{1}\right)...\left(\mathrm{2}{n}−\mathrm{2}\right)\left({z}+{i}\right)^{−\left(\mathrm{2}{n}−\mathrm{1}\right)} \\ $$$${Res}\left({W},{i}\right)\:={lim}_{{z}\rightarrow{i}} \:\:\:\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!}\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {n}\left({n}+\mathrm{1}\right)....\left(\mathrm{2}{n}−\mathrm{2}\right)\left({z}+{i}\right)^{−\mathrm{2}{n}+\mathrm{1}} \\ $$$$=\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)...\left(\mathrm{2}{n}−\mathrm{2}\right)}{\left({n}−\mathrm{1}\right)!}×\frac{\mathrm{1}}{\left(\mathrm{2}{i}\right)^{\mathrm{2}{n}−\mathrm{1}} } \\ $$$$=\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {n}\left({n}+\mathrm{1}\right)....\left(\mathrm{2}{n}−\mathrm{2}\right)}{\mathrm{2}^{\mathrm{2}{n}−\mathrm{1}} \:\left(−\mathrm{1}\right)^{{n}} }×{i}\:=−{i}\frac{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)...\left(\mathrm{2}{n}−\mathrm{2}\right)}{\mathrm{2}^{\mathrm{2}{n}−\mathrm{1}} }\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi×−{i}\frac{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)....\left(\mathrm{2}{n}−\mathrm{2}\right)}{\mathrm{2}^{\mathrm{2}{n}−\mathrm{1}} } \\ $$$$=\frac{\pi}{\mathrm{4}^{{n}−\mathrm{1}} }×{n}\left({n}+\mathrm{1}\right)....\left(\mathrm{2}{n}−\mathrm{2}\right)\:=\mathrm{2}{A}_{{n}} \:\Rightarrow \\ $$$${A}_{{n}} =\frac{\pi{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)...\left(\mathrm{2}{n}−\mathrm{2}\right)}{\mathrm{2}×\mathrm{4}^{{n}−\mathrm{1}} }\:\:{with}\:{n}>\mathrm{1}\:{and} \\ $$$${A}_{\mathrm{1}} =\frac{\pi}{\mathrm{2}} \\ $$

Answered by mind is power last updated on 06/Oct/19

f(z)=(1/((1+z^2 )^n )) lim zf(z)=0 ∫_0 ^∞ f(z)dz=(1/2)∫_(−∞) ^(+∞) (dz/((1+z^2 )^n ))=2iπRes(f,i) Res(f,i)=(1/((n−1)!))(d^(n−1) /dz^(n−1) )(z−i)^n .(1/((z−i)^n (z+i)^n ))∣_(z=i) =(1/((n−1)!)).(d^(n−1) /dz^(n−1) ).(1/((z+i)^n )) ((1/((z−i)^n )))^((k)) =(((−1)^k n...(n+k−1))/((z+i)^(n+k) )) ⇒(d^(n−1) /dz^(n−1) ).(1/((z+i)^n ))=(((−1)^n .n.....(2n−2))/((z+i)^(2n−1) )) Res(f,i)=(1/((n−1)!)).(((−1)^(n−1) ×n....(2n−2))/((2i)^(2n−1) ))=((2i.(−1)^(n−1) .n...(2n−2))/((n−1)!.(−4)^n .)) ∫_(−∞) ^(+∞) f(z)dz=((4π.n..(2n−2))/((n−1)!.4^n ))=((4π.(2n−2)!)/((n−1)!4^n )) ⇒∫_0 ^(+∞) (dx/((1+x^2 )^n ))=((2.4^(−n) π(2n−2)!)/((n−1)!))

$${f}\left({z}\right)=\frac{\mathrm{1}}{\left(\mathrm{1}+{z}^{\mathrm{2}} \right)^{{n}} } \\ $$$${lim}\:{zf}\left({z}\right)=\mathrm{0} \\ $$$$\int_{\mathrm{0}} ^{\infty} {f}\left({z}\right){dz}=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{+\infty} \frac{{dz}}{\left(\mathrm{1}+{z}^{\mathrm{2}} \right)^{{n}} }=\mathrm{2}{i}\pi{Res}\left({f},{i}\right) \\ $$$${Res}\left({f},{i}\right)=\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!}\frac{{d}^{{n}−\mathrm{1}} }{{dz}^{{n}−\mathrm{1}} }\left({z}−{i}\right)^{{n}} .\frac{\mathrm{1}}{\left({z}−{i}\right)^{{n}} \left({z}+{i}\right)^{{n}} }\mid_{{z}={i}} \\ $$$$=\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!}.\frac{{d}^{{n}−\mathrm{1}} }{{dz}^{{n}−\mathrm{1}} }.\frac{\mathrm{1}}{\left({z}+{i}\right)^{{n}} } \\ $$$$\left(\frac{\mathrm{1}}{\left({z}−{i}\right)^{{n}} }\right)^{\left({k}\right)} =\frac{\left(−\mathrm{1}\right)^{{k}} {n}...\left({n}+{k}−\mathrm{1}\right)}{\left({z}+{i}\right)^{{n}+{k}} } \\ $$$$\Rightarrow\frac{{d}^{{n}−\mathrm{1}} }{{dz}^{{n}−\mathrm{1}} }.\frac{\mathrm{1}}{\left({z}+{i}\right)^{{n}} }=\frac{\left(−\mathrm{1}\right)^{{n}} .{n}.....\left(\mathrm{2}{n}−\mathrm{2}\right)}{\left({z}+{i}\right)^{\mathrm{2}{n}−\mathrm{1}} } \\ $$$${Res}\left({f},{i}\right)=\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!}.\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} ×{n}....\left(\mathrm{2}{n}−\mathrm{2}\right)}{\left(\mathrm{2}{i}\right)^{\mathrm{2}{n}−\mathrm{1}} }=\frac{\mathrm{2}{i}.\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} .{n}...\left(\mathrm{2}{n}−\mathrm{2}\right)}{\left({n}−\mathrm{1}\right)!.\left(−\mathrm{4}\right)^{{n}} .} \\ $$$$\int_{−\infty} ^{+\infty} {f}\left({z}\right){dz}=\frac{\mathrm{4}\pi.{n}..\left(\mathrm{2}{n}−\mathrm{2}\right)}{\left({n}−\mathrm{1}\right)!.\mathrm{4}^{{n}} }=\frac{\mathrm{4}\pi.\left(\mathrm{2}{n}−\mathrm{2}\right)!}{\left({n}−\mathrm{1}\right)!\mathrm{4}^{{n}} } \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{+\infty} \frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{{n}} }=\frac{\mathrm{2}.\mathrm{4}^{−{n}} \pi\left(\mathrm{2}{n}−\mathrm{2}\right)!}{\left({n}−\mathrm{1}\right)!} \\ $$

Commented by mathmax by abdo last updated on 06/Oct/19

thank you sir.

$${thank}\:{you}\:{sir}. \\ $$

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