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

$$letf\left(t\right)={\int }_0^{\mathrm{\infty }}\frac{arctan\left(tx\right)}{x^3+8}dx$$$$1)findasimpleformoff\left(t\right)$$$$2)calculate{\int }_0^{\mathrm{\infty }}\frac{arctan\left(x\right)}{x^3+8}dx.$$

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

$$1)wehave{f}^{{}^{\prime }}\left(t\right)={\int }_0^{\mathrm{\infty }}\frac{x}{(1+t^2{x}^2)(x^3+8)}dx{=}_{tx=\alpha }{\int }_0^{\mathrm{\infty }}\frac{1}{(1+{\alpha }^2)(\frac{{\alpha }^3}{t^3}+8)}\frac{\alpha }{t}\frac{d\alpha }{t}$$$$=\frac{1}{t^2}{\int }_0^{\mathrm{\infty }}{t}^3\frac{\alpha d\alpha }{(1+{\alpha }^2)({\alpha }^3+8t^3)}=t{\int }_0^{\mathrm{\infty }}\frac{xdx}{(x^2+1)(x^3+8t^3)}letdecompose$$$$F\left(x\right)=\frac{x}{(x^2+1)(x^3+8t^3)}$$$$F\left(x\right)=\frac{x}{(x^2+1)(x+2t)(x^2-2xt+4t^2)}=\frac{a}{x+2t}+\frac{bx+c}{x^2+1}+\frac{dx+e}{x^2-2xt+4t^2}$$$$a={lim}_{x\to -2t}(x+2t)F\left(x\right)=\frac{-2t}{(4t^2+1)\left(12t^2\right)}=\frac{-1}{6t(4t^2+1)}$$$${lim}_{x\to +\mathrm{\infty }}xF\left(x\right)=0=a+b+d\Rightarrow b+d=-a$$$$F\left(o\right)=0=\frac{a}{2t}+c+\frac{e}{4t^2}\Rightarrow 2at+4t^{2}c+e=0....becontinued...$$

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