Question Number 35678 by abdo imad last updated on 21/May/18
$${let}\:{f}\left({t}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−{tx}^{\mathrm{2}} } \:{arctan}\left({x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} }{dx}\:{with}\:{t}>\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:{study}\:{the}\:{existencte}\:{of}\:{f}\left({t}\right) \\ $$$$\left.\mathrm{2}\right){calculate}\:{f}^{'} \left({t}\right) \\ $$$$\left.\mathrm{3}\right){find}\:{a}\:{simple}\:{form}\:{of}\:{f}\left({t}\right). \\ $$
Commented by prof Abdo imad last updated on 23/May/18
$$\left.\mathrm{1}\right)\:{f}\left({t}\right)\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{e}^{−{tx}^{\mathrm{2}} } \:{arctan}\left({x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} }{dx}\:+\int_{\mathrm{1}} ^{+\infty} \:\frac{{e}^{−{tx}^{\mathrm{2}} } \:{arctan}\left({x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} }{dx} \\ $$$$=\:{I}\:+{J}\:{but}\:{we}\:{have}\:\:\frac{{e}^{−{tx}^{\mathrm{2}} } \:{arctan}\left({x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} }\sim\:{e}^{−{tx}^{\mathrm{2}} } \:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \:\frac{{e}^{−{tx}^{\mathrm{2}} } \:{arctan}\left({x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} }\:=\:\mathrm{1}\:{so}\:{I}\:{converges} \\ $$$${also}\:{we}\:{have}\:{lim}_{{x}\rightarrow+\infty} \:\:{x}^{\mathrm{2}} \:\:\frac{{e}^{−{tx}^{\mathrm{2}} } {arctan}\left({x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} }\:=\mathrm{0} \\ $$$${the}\:{comvergence}\:{of}\:{J}\:{is}\:{assured}\:{so}\:{f}\left({t}\right)?{exists} \\ $$$${for}\:{t}>\mathrm{0} \\ $$$$\left.\mathrm{2}\right)\:{f}^{'} \left({t}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\partial}{\partial{t}}\left\{\:\:\frac{{e}^{−{tx}^{\mathrm{2}} } \:{arctan}\left({x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} }\right\}{dx} \\ $$$$=−\int_{\mathrm{0}} ^{\infty} \:\:\:{e}^{−{tx}^{\mathrm{2}} } \:{arctan}\left({x}^{\mathrm{2}} \right){dx}\:. \\ $$