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Question Number 40829 by math khazana by abdo last updated on 28/Jul/18

let f(t) = ∫_0 ^∞ ((arctan(tx))/(x^3 +8))dx 1)find a simple form of f(t) 2)calculate ∫_0 ^∞ ((arctan(x))/(x^3 +8))dx .

$${let}\:{f}\left({t}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left({tx}\right)}{{x}^{\mathrm{3}} +\mathrm{8}}{dx} \\ $$$$\left.\mathrm{1}\right){find}\:{a}\:{simple}\:{form}\:{of}\:{f}\left({t}\right) \\ $$$$\left.\mathrm{2}\right){calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{arctan}\left({x}\right)}{{x}^{\mathrm{3}} \:+\mathrm{8}}{dx}\:. \\ $$

Answered by maxmathsup by imad last updated on 29/Jul/18

1) we have f^′ (t) = ∫_0 ^∞ (x/((1+t^2 x^2 )(x^3 +8)))dx =_(tx =α) ∫_0 ^∞ (1/((1+α^2 )((α^3 /t^3 ) +8)))(α/t) (dα/t) =(1/t^2 )∫_0 ^∞ t^3 ((αdα)/((1+α^2 )(α^3 +8t^3 ))) = t ∫_0 ^∞ ((xdx)/((x^2 +1)(x^3 +8t^3 ))) let decompose F(x)=(x/((x^2 +1)(x^3 +8t^3 ))) F(x) = (x/((x^2 +1)(x+2t)(x^2 −2xt +4t^2 ))) =(a/(x+2t)) +((bx+c)/(x^2 +1)) +((dx+e)/(x^2 −2xt +4t^2 )) a =lim_(x→−2t) (x+2t)F(x)= ((−2t)/((4t^2 +1)(12t^2 ))) = ((−1)/(6t(4t^2 +1))) lim_(x→+∞) xF(x)=0 =a +b +d ⇒b+d =−a F(o) =0 =(a/(2t)) +c +(e/(4t^2 )) ⇒2at +4t^2 c +e =0 ....be continued...

$$\left.\mathrm{1}\right)\:{we}\:{have}\:{f}^{'} \left({t}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{x}}{\left(\mathrm{1}+{t}^{\mathrm{2}} {x}^{\mathrm{2}} \right)\left({x}^{\mathrm{3}} +\mathrm{8}\right)}{dx}\:=_{{tx}\:=\alpha} \:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{\mathrm{1}}{\left(\mathrm{1}+\alpha^{\mathrm{2}} \right)\left(\frac{\alpha^{\mathrm{3}} }{{t}^{\mathrm{3}} }\:+\mathrm{8}\right)}\frac{\alpha}{{t}}\:\frac{{d}\alpha}{{t}} \\ $$$$=\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\infty} \:\:\:\:{t}^{\mathrm{3}} \:\:\frac{\alpha{d}\alpha}{\left(\mathrm{1}+\alpha^{\mathrm{2}} \right)\left(\alpha^{\mathrm{3}} \:+\mathrm{8}{t}^{\mathrm{3}} \right)}\:=\:{t}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{xdx}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)\left({x}^{\mathrm{3}} +\mathrm{8}{t}^{\mathrm{3}} \right)}\:{let}\:{decompose} \\ $$$${F}\left({x}\right)=\frac{{x}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)\left({x}^{\mathrm{3}} +\mathrm{8}{t}^{\mathrm{3}} \right)} \\ $$$${F}\left({x}\right)\:=\:\:\frac{{x}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)\left({x}+\mathrm{2}{t}\right)\left({x}^{\mathrm{2}} −\mathrm{2}{xt}\:+\mathrm{4}{t}^{\mathrm{2}} \right)}\:=\frac{{a}}{{x}+\mathrm{2}{t}}\:+\frac{{bx}+{c}}{{x}^{\mathrm{2}} \:+\mathrm{1}}\:+\frac{{dx}+{e}}{{x}^{\mathrm{2}} \:−\mathrm{2}{xt}\:+\mathrm{4}{t}^{\mathrm{2}} } \\ $$$${a}\:={lim}_{{x}\rightarrow−\mathrm{2}{t}} \:\:\:\:\left({x}+\mathrm{2}{t}\right){F}\left({x}\right)=\:\frac{−\mathrm{2}{t}}{\left(\mathrm{4}{t}^{\mathrm{2}} \:+\mathrm{1}\right)\left(\mathrm{12}{t}^{\mathrm{2}} \right)}\:=\:\frac{−\mathrm{1}}{\mathrm{6}{t}\left(\mathrm{4}{t}^{\mathrm{2}} \:+\mathrm{1}\right)} \\ $$$${lim}_{{x}\rightarrow+\infty} {xF}\left({x}\right)=\mathrm{0}\:={a}\:+{b}\:+{d}\:\Rightarrow{b}+{d}\:=−{a} \\ $$$${F}\left({o}\right)\:=\mathrm{0}\:=\frac{{a}}{\mathrm{2}{t}}\:+{c}\:+\frac{{e}}{\mathrm{4}{t}^{\mathrm{2}} }\:\Rightarrow\mathrm{2}{at}\:+\mathrm{4}{t}^{\mathrm{2}} {c}\:+{e}\:=\mathrm{0}\:....{be}\:{continued}... \\ $$

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