Question Number 72012 by mathmax by abdo last updated on 23/Oct/19
$${find}\:{f}\left(\alpha\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left(\alpha{x}\right)}{\left({x}^{\mathrm{2}} +\mathrm{3}\right)^{\mathrm{2}} }{dx}\:\:{with}\:\alpha\:{real}. \\ $$
Commented by mathmax by abdo last updated on 07/Nov/19
$${we}\:{have}\:{f}\left(\alpha\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{{arctan}\left(\alpha{x}\right)}{\left({x}^{\mathrm{2}} \:+\mathrm{3}\right)^{\mathrm{2}} }{dx}\:\Rightarrow \\ $$$${f}^{'} \left(\alpha\right)\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{x}}{\left(\mathrm{1}+\alpha^{\mathrm{2}} {x}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} \:+\mathrm{3}\right)^{\mathrm{2}} }{dx}\:=_{\alpha{x}\:={t}} \:\:\int_{\mathrm{0}} ^{\infty} \:\frac{{t}}{\alpha\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(\frac{{t}^{\mathrm{2}} }{\alpha^{\mathrm{2}} }\:+\mathrm{3}\right)^{\mathrm{2}} }\frac{{dt}}{\alpha} \\ $$$$=\frac{\alpha^{\mathrm{4}} }{\alpha^{\mathrm{2}} }\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{tdt}}{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)\left({t}^{\mathrm{2}} \:+\mathrm{3}\alpha^{\mathrm{2}} \right)^{\mathrm{2}} }\:=\alpha^{\mathrm{2}\:} \:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{tdt}}{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)\left({t}^{\mathrm{2}} \:+\mathrm{3}\alpha^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$${let}\:{decompose}\:{F}\left({t}\right)\:=\frac{{t}}{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)\left({t}^{\mathrm{2}} \:+\mathrm{3}\alpha^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$${F}\left({t}\right)=\frac{{at}\:+{b}}{{t}^{\mathrm{2}\:} +\mathrm{1}}\:+\frac{{ct}+{d}}{{t}^{\mathrm{2}} \:+\mathrm{3}\alpha^{\mathrm{2}} }\:+\frac{{et}\:+{f}}{\left({t}^{\mathrm{2}} \:+\mathrm{3}\alpha^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$${F}\left(−{t}\right)={F}\left({t}\right)\:\Rightarrow\frac{−{at}+{b}}{{t}^{\mathrm{2}} \:+\mathrm{1}}+\frac{−{ct}+{d}}{{t}^{\mathrm{2}} \:+\mathrm{3}\alpha^{\mathrm{2}} }\:+\frac{−{et}+{f}}{\left({t}^{\mathrm{2}} \:+\mathrm{3}\alpha^{\mathrm{2}} \right)^{\mathrm{2}} }\:=\frac{−{at}−{b}}{{t}^{\mathrm{2}} \:+\mathrm{1}}+\frac{−{ct}−{d}}{{t}^{\mathrm{2}} \:+\mathrm{3}\alpha^{\mathrm{2}} } \\ $$$$+\frac{−{et}−{f}}{\left({t}^{\mathrm{2}} \:+\mathrm{3}\alpha^{\mathrm{2}} \right)^{\mathrm{2}} }\:\Rightarrow{b}={d}={f}=\mathrm{0}\:\Rightarrow \\ $$$${F}\left({t}\right)\:=\frac{{at}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:+\frac{{ct}}{{t}^{\mathrm{2}\:} +\mathrm{3}\alpha^{\mathrm{2}} }\:+\frac{{et}}{\left({t}^{\mathrm{2}} \:+\mathrm{3}\alpha^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$${lim}_{{t}\rightarrow+\infty} {tF}\left({t}\right)\:=\mathrm{0}\:={a}+{c}+{e}\:\Rightarrow{e}\:=−{a}−{c}\:\Rightarrow \\ $$$${F}\left({t}\right)=\frac{{at}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:+\frac{{ct}}{{t}^{\mathrm{2}} \:+\mathrm{3}\alpha^{\mathrm{2}} }−\frac{\left({a}+{c}\right){t}}{\left({t}^{\mathrm{2}} \:+\mathrm{3}\alpha^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$${F}\left(\mathrm{1}\right)\:=\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{1}+\mathrm{3}\alpha^{\mathrm{2}} \right)^{\mathrm{2}} }\:=\frac{{a}}{\mathrm{2}}\:+\frac{{c}}{\mathrm{1}+\mathrm{3}\alpha^{\mathrm{2}} }−\frac{{a}+{c}}{\left(\mathrm{1}+\mathrm{3}\alpha^{\mathrm{2}} \right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\mathrm{3}\alpha^{\mathrm{2}} \right)^{\mathrm{2}} {a}\:+\left(\mathrm{1}+\mathrm{3}\alpha^{\mathrm{2}} \right){c}−{a}−{c}\:\Rightarrow \\ $$$$\mathrm{1}\:=\left(\mathrm{1}+\mathrm{3}\alpha^{\mathrm{2}} \right)^{\mathrm{2}} {a}+\mathrm{2}\left(\mathrm{1}+\mathrm{3}\alpha^{\mathrm{2}} \right){c}−\mathrm{2}{a}−\mathrm{2}{c}\:\Rightarrow \\ $$$$\left\{\left(\mathrm{1}+\mathrm{3}\alpha^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2}\right\}{a}\:+\left\{\mathrm{2}\left(\mathrm{1}+\mathrm{3}\alpha^{\mathrm{2}} \right)−\mathrm{2}\right\}{c}\:=\mathrm{1} \\ $$$$….{be}\:{continued}… \\ $$