Question Number 57899 by maxmathsup by imad last updated on 13/Apr/19
$${let}\:{f}\left({x}\right)\:=\int_{\mathrm{0}} ^{+\infty} \:\:\frac{{dt}}{\left({t}^{\mathrm{2}} \:+{x}^{\mathrm{2}} \right)^{\mathrm{3}} }\:\:{with}\:{x}>\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{a}\:{explicit}\:{form}\:{off}\:\left({x}\right) \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dx}}{\left({t}^{\mathrm{2}} \:+\mathrm{3}\right)^{\mathrm{3}} }\:\:{and}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dt}}{\left({t}^{\mathrm{2}} \:+\mathrm{4}\right)^{\mathrm{3}} } \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:{A}\left(\theta\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dt}}{\left({t}^{\mathrm{2}} \:+{sin}^{\mathrm{2}} \theta\right)^{\mathrm{3}} }\:\:{with}\:\mathrm{0}<\theta<\pi. \\ $$
Commented by maxmathsup by imad last updated on 17/Apr/19
$$\left.\mathrm{1}\right)\:{we}\:{have}\:\:{f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{dt}}{\left({t}^{\mathrm{2}} \:+{x}^{\mathrm{2}} \right)^{\mathrm{3}} }\:\:{let}\:{consider}\:{the}\:{complex}\:{function} \\ $$$$\varphi\left({z}\right)\:=\frac{\mathrm{1}}{\left({z}^{\mathrm{2}} \:+{x}^{\mathrm{2}} \right)^{\mathrm{3}} }\:\:{poles}\:{of}\:\varphi? \\ $$$${we}\:{have}\:\varphi\left({z}\right)\:=\frac{\mathrm{1}}{\left({z}−{ix}\right)^{\mathrm{3}} \left({z}+{ix}\right)^{\mathrm{3}} }\:\:{so}\:\:{the}\:{poles}\:{of}\:{are}\:\overset{−} {+}{ix}\:\left({triples}\right)\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{ix}\right) \\ $$$${Res}\left(\varphi,{ix}\right)\:={lim}_{{z}\rightarrow{ix}} \:\:\:\frac{\mathrm{1}}{\left(\mathrm{3}−\mathrm{1}\right)!}\left\{\:\left({z}−{ix}\right)^{\mathrm{3}} \varphi\left({z}\right)\right\}^{\left(\mathrm{2}\right)} \\ $$$$={lim}_{{z}\rightarrow{ix}} \:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\left\{\:\:\frac{\mathrm{1}}{\left({z}+{ix}\right)^{\mathrm{3}} }\right\}^{\left(\mathrm{2}\right)} \:\:{we}\:{have}\: \\ $$$$\left\{\frac{\mathrm{1}}{\left({z}+{ix}\right)^{\mathrm{3}} }\right\}^{\left(\mathrm{1}\right)} \:=−\frac{\mathrm{3}\left({z}+{ix}\right)^{\mathrm{2}} }{\left({z}+{ix}\right)^{\mathrm{6}} }\:=−\frac{\mathrm{3}}{\left({z}+{ix}\right)^{\mathrm{4}} }\:\Rightarrow\left\{\frac{\mathrm{1}}{\left({z}+{ix}\right)^{\mathrm{3}} }\right\}^{\left(\mathrm{2}\right)} \\ $$$$=−\mathrm{3}\:\frac{−\mathrm{4}\left({z}+{ix}\right)^{\mathrm{3}} }{\left({z}+{ix}\right)^{\mathrm{8}} }\:=\frac{\mathrm{12}}{\left({z}+{ix}\right)^{\mathrm{5}} }\:\Rightarrow{Res}\left(\varphi,{ix}\right)\:={lim}_{{z}\rightarrow{ix}} \:\:\:\:\frac{\mathrm{6}}{\left({z}+{ix}\right)^{\mathrm{5}} } \\ $$$$=\frac{\mathrm{6}}{\left(\mathrm{2}{ix}\right)^{\mathrm{5}} }\:=\frac{\mathrm{6}}{\mathrm{2}^{\mathrm{5}} {x}^{\mathrm{5}} {i}}\:=\frac{\mathrm{6}}{\mathrm{32}{ix}^{\mathrm{5}} }\:=\frac{\mathrm{3}}{\mathrm{16}{ix}^{\mathrm{5}} }\:\Rightarrow\:\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\frac{\mathrm{3}}{\mathrm{16}\:{x}^{\mathrm{5}} {i}}\:=\frac{\mathrm{3}\pi}{\mathrm{8}\:{x}^{\mathrm{5}} }\:\Rightarrow \\ $$$${f}\left({x}\right)\:=\frac{\mathrm{3}\pi}{\mathrm{16}\:{x}^{\mathrm{5}} }\:\:\:\left(\:\:{with}\:{x}>\mathrm{0}\right) \\ $$$$\left.\mathrm{2}\right)\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dt}}{\left({t}^{\mathrm{2}} \:+\mathrm{3}\right)^{\mathrm{3}} }\:={f}\left(\sqrt{\mathrm{3}}\right)\:=\:\frac{\mathrm{3}\pi}{\mathrm{16}\:\left(\sqrt{\mathrm{3}}\right)^{\mathrm{5}} }\:=\frac{\mathrm{3}\pi}{\mathrm{16}\:\left(\sqrt{\mathrm{3}}\right)^{\mathrm{4}} \sqrt{\mathrm{3}}}\:=\frac{\mathrm{3}\pi}{\mathrm{9}×\mathrm{16}\:\sqrt{\mathrm{3}}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{dt}}{\left({t}^{\mathrm{2}} \:+\mathrm{4}\right)^{\mathrm{3}} }\:={f}\left(\mathrm{2}\right)\:=\frac{\mathrm{3}\pi}{\mathrm{16}\:\left(\mathrm{2}\right)^{\mathrm{5}} }\:=\frac{\mathrm{3}\pi}{\mathrm{16}×\mathrm{32}} \\ $$
Commented by maxmathsup by imad last updated on 17/Apr/19
$$\left.\mathrm{3}\right)\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dt}}{\left({t}^{\mathrm{2}} \:+{sin}^{\mathrm{2}} \theta\right)^{\mathrm{3}} }\:={f}\left({sin}\theta\right)\:=\frac{\mathrm{3}\pi}{\mathrm{16}\:{sin}^{\mathrm{5}} \theta} \\ $$