Question Number 66459 by mathmax by abdo last updated on 15/Aug/19
$$\left.\mathrm{1}\right)\:{calculate}\:{by}\:{residus}\:{method}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{3}} } \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}+{x}^{\mathrm{4}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{3}} }{dx} \\ $$
Commented by mathmax by abdo last updated on 17/Aug/19
$$\left.\mathrm{1}\right){let}\:{A}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{3}} }\:\Rightarrow\mathrm{2}{A}\:=\int_{−\infty} ^{+\infty} \:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$${let}\:{W}\left({z}\right)=\frac{\mathrm{1}}{\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{3}} }\:\Rightarrow{W}\left({z}\right)=\frac{\mathrm{1}}{\left({z}−{i}\right)^{\mathrm{3}} \left({z}+{i}\right)^{\mathrm{3}} }\:{so}\:{the}\:{poles}\:{of}\:{W}\:{are} \\ $$$$\overset{−} {+}{i}\:\:\left({triples}\right)\:\:{residus}\:{theoreme}\:{give}\:\int_{−\infty} ^{+\infty} {W}\left({z}\right){dz}=\mathrm{2}{i}\pi{Res}\left({W},{i}\right) \\ $$$${Res}\left({W},{i}\right)={lim}_{{z}\rightarrow{i}} \frac{\mathrm{1}}{\left(\mathrm{3}−\mathrm{1}\right)!}\left\{\left({z}−{i}\right)^{\mathrm{3}} {W}\left({z}\right)\right\}^{\left(\mathrm{2}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\:\frac{\mathrm{1}}{\mathrm{2}}\left\{\left({z}+{i}\right)^{−\mathrm{3}} \right\}^{\left(\mathrm{2}\right)} \:={lim}_{{z}\rightarrow{i}} \:\frac{\mathrm{1}}{\mathrm{2}}\left\{−\mathrm{3}\left({z}+{i}\right)^{−\mathrm{4}} \right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}} \:\frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{12}\left({z}+{i}\right)^{−\mathrm{5}} \right\}\:=\mathrm{6}\left(\mathrm{2}{i}\right)^{−\mathrm{5}} \:=\frac{\mathrm{6}}{\left(\mathrm{2}{i}\right)^{\mathrm{5}} }\:=\frac{\mathrm{6}}{\mathrm{2}^{\mathrm{5}} {i}}\:=\frac{\mathrm{3}}{\mathrm{16}{i}}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi×\frac{\mathrm{3}}{\mathrm{16}{i}}\:=\frac{\mathrm{3}\pi}{\mathrm{8}}\:\Rightarrow{A}\:=\frac{\mathrm{3}\pi}{\mathrm{16}} \\ $$$$\left.\mathrm{2}\right){we}\:{have}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{3}} }\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{3}} }\:+\int_{\mathrm{1}} ^{+\infty} \:\:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{3}} }\:\:{and} \\ $$$$\int_{\mathrm{1}} ^{+\infty} \:\:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{3}} }\:=_{{x}=\frac{\mathrm{1}}{{t}}} \:\:\:\:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{−}{\left(\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right)^{\mathrm{3}} }\frac{{dt}}{{t}^{\mathrm{2}} }\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{\mathrm{6}} }{{t}^{\mathrm{2}} \left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{3}} }{dt} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{{t}^{\mathrm{4}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{3}} }{dt}\:\Rightarrow\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{3}} }\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{3}} }\:+\int_{\mathrm{0}} ^{\infty} \:\:\frac{{x}^{\mathrm{4}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{3}} }{dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}+{x}^{\mathrm{4}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{3}} }{dx}\:\:\:\Rightarrow\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}+{x}^{\mathrm{4}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{3}} }=\frac{\mathrm{3}\pi}{\mathrm{16}}\:. \\ $$