Question Number 70594 by mathmax by abdo last updated on 06/Oct/19
$${calculate}\:{by}\:{residus}\:{method}\:{the}\:{integral}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{{n}} } \\ $$$${with}\:{n}\:{integr}\:{and}\:{n}\geqslant\mathrm{1} \\ $$
Commented by mathmax by abdo last updated on 23/Oct/19
$${let}\:{A}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{{n}} }\:\Rightarrow\mathrm{2}{A}_{{n}} =\int_{−\infty} ^{+\infty} \:\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{{n}} }\:{let}\:{W}\left({z}\right)=\frac{\mathrm{1}}{\left({z}^{\mathrm{2}} +\mathrm{1}\right)^{{n}} } \\ $$$$\Rightarrow{W}\left({z}\right)=\frac{\mathrm{1}}{\left({z}−{i}\right)^{{n}} \left({z}+{i}\right)^{{n}} }\:{so}\:{the}\:{poles}\:{of}\:{W}\:{are}\:{i}\:{and}\:−{i} \\ $$$${at}\:{ordr}\:{nresidus}\:{theorem}\:{give}\:\int_{−\infty} ^{+\infty} {W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left({W},{i}\right) \\ $$$${Res}\left({W},{i}\right)={lim}_{{z}\rightarrow{i}} \:\:\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!}\left\{\left({z}−{i}\right)^{{n}} {W}\left({z}\right)\right\}^{\left({n}−\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!}\left\{\left({z}+{i}\right)^{−{n}} \right\}^{\left({n}−\mathrm{1}\right)} \:\:{let}\:{first}\:{find}\left\{\:\left({z}+{i}\right)^{−{n}} \right\}^{\left({p}\right)} \\ $$$${we}\:{have}\:\left\{\left({z}+{i}\right)^{−{n}} \right\}^{\left(\mathrm{1}\right)} =−{n}\left({z}+{i}\right)^{−{n}−\mathrm{1}} =−{n}\left({z}+{i}\right)^{−\left({n}+\mathrm{1}\right)} \\ $$$$\left\{\left({z}+{i}\right)\right\}^{\left(\mathrm{2}\right)} =\left(−\mathrm{1}\right)^{\mathrm{2}} {n}\left({n}+\mathrm{1}\right)\left({z}+{i}\right)^{−\left({n}+\mathrm{2}\right)} \\ $$$$…\Rightarrow\left\{\left({z}+{i}\right)^{−{n}} \right\}^{\left({p}\right)} =\left(−\mathrm{1}\right)^{{p}} {n}\left({n}+\mathrm{1}\right)…\left({n}+{p}−\mathrm{1}\right)\left({z}+{i}\right)^{−\left({n}+{p}\right)} \\ $$$${p}={n}−\mathrm{1}\:\Rightarrow\left\{\left({z}+{i}\right)^{−{n}} \right\}^{\left({n}−\mathrm{1}\right)} =\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {n}\left({n}+\mathrm{1}\right)…\left(\mathrm{2}{n}−\mathrm{2}\right)\left({z}+{i}\right)^{−\left(\mathrm{2}{n}−\mathrm{1}\right)} \\ $$$${Res}\left({W},{i}\right)\:={lim}_{{z}\rightarrow{i}} \:\:\:\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!}\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {n}\left({n}+\mathrm{1}\right)….\left(\mathrm{2}{n}−\mathrm{2}\right)\left({z}+{i}\right)^{−\mathrm{2}{n}+\mathrm{1}} \\ $$$$=\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)…\left(\mathrm{2}{n}−\mathrm{2}\right)}{\left({n}−\mathrm{1}\right)!}×\frac{\mathrm{1}}{\left(\mathrm{2}{i}\right)^{\mathrm{2}{n}−\mathrm{1}} } \\ $$$$=\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {n}\left({n}+\mathrm{1}\right)….\left(\mathrm{2}{n}−\mathrm{2}\right)}{\mathrm{2}^{\mathrm{2}{n}−\mathrm{1}} \:\left(−\mathrm{1}\right)^{{n}} }×{i}\:=−{i}\frac{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)…\left(\mathrm{2}{n}−\mathrm{2}\right)}{\mathrm{2}^{\mathrm{2}{n}−\mathrm{1}} }\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi×−{i}\frac{{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)….\left(\mathrm{2}{n}−\mathrm{2}\right)}{\mathrm{2}^{\mathrm{2}{n}−\mathrm{1}} } \\ $$$$=\frac{\pi}{\mathrm{4}^{{n}−\mathrm{1}} }×{n}\left({n}+\mathrm{1}\right)….\left(\mathrm{2}{n}−\mathrm{2}\right)\:=\mathrm{2}{A}_{{n}} \:\Rightarrow \\ $$$${A}_{{n}} =\frac{\pi{n}\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)…\left(\mathrm{2}{n}−\mathrm{2}\right)}{\mathrm{2}×\mathrm{4}^{{n}−\mathrm{1}} }\:\:{with}\:{n}>\mathrm{1}\:{and} \\ $$$${A}_{\mathrm{1}} =\frac{\pi}{\mathrm{2}} \\ $$
Answered by mind is power last updated on 06/Oct/19
$${f}\left({z}\right)=\frac{\mathrm{1}}{\left(\mathrm{1}+{z}^{\mathrm{2}} \right)^{{n}} } \\ $$$${lim}\:{zf}\left({z}\right)=\mathrm{0} \\ $$$$\int_{\mathrm{0}} ^{\infty} {f}\left({z}\right){dz}=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{+\infty} \frac{{dz}}{\left(\mathrm{1}+{z}^{\mathrm{2}} \right)^{{n}} }=\mathrm{2}{i}\pi{Res}\left({f},{i}\right) \\ $$$${Res}\left({f},{i}\right)=\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!}\frac{{d}^{{n}−\mathrm{1}} }{{dz}^{{n}−\mathrm{1}} }\left({z}−{i}\right)^{{n}} .\frac{\mathrm{1}}{\left({z}−{i}\right)^{{n}} \left({z}+{i}\right)^{{n}} }\mid_{{z}={i}} \\ $$$$=\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!}.\frac{{d}^{{n}−\mathrm{1}} }{{dz}^{{n}−\mathrm{1}} }.\frac{\mathrm{1}}{\left({z}+{i}\right)^{{n}} } \\ $$$$\left(\frac{\mathrm{1}}{\left({z}−{i}\right)^{{n}} }\right)^{\left({k}\right)} =\frac{\left(−\mathrm{1}\right)^{{k}} {n}…\left({n}+{k}−\mathrm{1}\right)}{\left({z}+{i}\right)^{{n}+{k}} } \\ $$$$\Rightarrow\frac{{d}^{{n}−\mathrm{1}} }{{dz}^{{n}−\mathrm{1}} }.\frac{\mathrm{1}}{\left({z}+{i}\right)^{{n}} }=\frac{\left(−\mathrm{1}\right)^{{n}} .{n}…..\left(\mathrm{2}{n}−\mathrm{2}\right)}{\left({z}+{i}\right)^{\mathrm{2}{n}−\mathrm{1}} } \\ $$$${Res}\left({f},{i}\right)=\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!}.\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} ×{n}….\left(\mathrm{2}{n}−\mathrm{2}\right)}{\left(\mathrm{2}{i}\right)^{\mathrm{2}{n}−\mathrm{1}} }=\frac{\mathrm{2}{i}.\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} .{n}…\left(\mathrm{2}{n}−\mathrm{2}\right)}{\left({n}−\mathrm{1}\right)!.\left(−\mathrm{4}\right)^{{n}} .} \\ $$$$\int_{−\infty} ^{+\infty} {f}\left({z}\right){dz}=\frac{\mathrm{4}\pi.{n}..\left(\mathrm{2}{n}−\mathrm{2}\right)}{\left({n}−\mathrm{1}\right)!.\mathrm{4}^{{n}} }=\frac{\mathrm{4}\pi.\left(\mathrm{2}{n}−\mathrm{2}\right)!}{\left({n}−\mathrm{1}\right)!\mathrm{4}^{{n}} } \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{+\infty} \frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{{n}} }=\frac{\mathrm{2}.\mathrm{4}^{−{n}} \pi\left(\mathrm{2}{n}−\mathrm{2}\right)!}{\left({n}−\mathrm{1}\right)!} \\ $$
Commented by mathmax by abdo last updated on 06/Oct/19
$${thank}\:{you}\:{sir}. \\ $$