Question Number 40829 by math khazana by abdo last updated on 28/Jul/18
$${let}\:{f}\left({t}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left({tx}\right)}{{x}^{\mathrm{3}} +\mathrm{8}}{dx} \\ $$$$\left.\mathrm{1}\right){find}\:{a}\:{simple}\:{form}\:{of}\:{f}\left({t}\right) \\ $$$$\left.\mathrm{2}\right){calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{arctan}\left({x}\right)}{{x}^{\mathrm{3}} \:+\mathrm{8}}{dx}\:. \\ $$
Answered by maxmathsup by imad last updated on 29/Jul/18
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{f}^{'} \left({t}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{x}}{\left(\mathrm{1}+{t}^{\mathrm{2}} {x}^{\mathrm{2}} \right)\left({x}^{\mathrm{3}} +\mathrm{8}\right)}{dx}\:=_{{tx}\:=\alpha} \:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{\mathrm{1}}{\left(\mathrm{1}+\alpha^{\mathrm{2}} \right)\left(\frac{\alpha^{\mathrm{3}} }{{t}^{\mathrm{3}} }\:+\mathrm{8}\right)}\frac{\alpha}{{t}}\:\frac{{d}\alpha}{{t}} \\ $$$$=\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\infty} \:\:\:\:{t}^{\mathrm{3}} \:\:\frac{\alpha{d}\alpha}{\left(\mathrm{1}+\alpha^{\mathrm{2}} \right)\left(\alpha^{\mathrm{3}} \:+\mathrm{8}{t}^{\mathrm{3}} \right)}\:=\:{t}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{xdx}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)\left({x}^{\mathrm{3}} +\mathrm{8}{t}^{\mathrm{3}} \right)}\:{let}\:{decompose} \\ $$$${F}\left({x}\right)=\frac{{x}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)\left({x}^{\mathrm{3}} +\mathrm{8}{t}^{\mathrm{3}} \right)} \\ $$$${F}\left({x}\right)\:=\:\:\frac{{x}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)\left({x}+\mathrm{2}{t}\right)\left({x}^{\mathrm{2}} −\mathrm{2}{xt}\:+\mathrm{4}{t}^{\mathrm{2}} \right)}\:=\frac{{a}}{{x}+\mathrm{2}{t}}\:+\frac{{bx}+{c}}{{x}^{\mathrm{2}} \:+\mathrm{1}}\:+\frac{{dx}+{e}}{{x}^{\mathrm{2}} \:−\mathrm{2}{xt}\:+\mathrm{4}{t}^{\mathrm{2}} } \\ $$$${a}\:={lim}_{{x}\rightarrow−\mathrm{2}{t}} \:\:\:\:\left({x}+\mathrm{2}{t}\right){F}\left({x}\right)=\:\frac{−\mathrm{2}{t}}{\left(\mathrm{4}{t}^{\mathrm{2}} \:+\mathrm{1}\right)\left(\mathrm{12}{t}^{\mathrm{2}} \right)}\:=\:\frac{−\mathrm{1}}{\mathrm{6}{t}\left(\mathrm{4}{t}^{\mathrm{2}} \:+\mathrm{1}\right)} \\ $$$${lim}_{{x}\rightarrow+\infty} {xF}\left({x}\right)=\mathrm{0}\:={a}\:+{b}\:+{d}\:\Rightarrow{b}+{d}\:=−{a} \\ $$$${F}\left({o}\right)\:=\mathrm{0}\:=\frac{{a}}{\mathrm{2}{t}}\:+{c}\:+\frac{{e}}{\mathrm{4}{t}^{\mathrm{2}} }\:\Rightarrow\mathrm{2}{at}\:+\mathrm{4}{t}^{\mathrm{2}} {c}\:+{e}\:=\mathrm{0}\:….{be}\:{continued}… \\ $$