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Question Number 39023 by maxmathsup by imad last updated on 01/Jul/18

$$letg\left(x\right)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{arctan\left(x(1+t^2)\right)}{1+t^2}dtwithx>0$$ $$findasimpleformofg\left(x\right).$$

Commented bymath khazana by abdo last updated on 08/Jul/18

$$wehave{g}^{{}^{\prime }}\left(x\right)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{1}{1+x^2{(1+t^2)}^2}dt$$ $$={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dt}{(x(1+t^2)-i)(x(1+t^2)+i)}$$ $$={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dt}{({xt}^2+x-i)({xt}^2+x+i)}$$ $$let\phi \left(z\right)=\frac{1}{({xz}^2+x-i)({xz}^2+x+i)}$$ $$\phi \left(z\right)=\frac{1}{x^2(z^2+1-\frac{i}{x})(z^2+1+\frac{i}{x})}$$ $$=\frac{1}{x^2(z-i\sqrt{1-\frac{i}{x}})(z+i\sqrt{1-\frac{i}{x}})(z-i\sqrt{1+\frac{i}{x}})(z+i\sqrt{1+\frac{i}{x}})}$$ $$wehave\mid 1-\frac{i}{x}\mid =\sqrt{1+\frac{1}{x^2}}=\frac{\sqrt{x^2+1}}{x}$$ $$1-\frac{i}{x}=\frac{\sqrt{x^2+1}}{x}(\frac{x}{\sqrt{x^2+1}}-\frac{i}{x}\frac{x}{\sqrt{x^2+1}})$$ $$=r{e}^{i\theta }\Rightarrow r=\frac{\sqrt{x^2+1}}{x}andcos\theta =\frac{x}{\sqrt{x^2+1}}$$ $$sin\theta =-\frac{1}{\sqrt{x^2+1}}\Rightarrow tan\theta =-\frac{1}{x}\Rightarrow$$ $$\theta =-arctan\left(x\right)\Rightarrow 1-\frac{i}{x}=r\left(x\right)e^{-iarctan\left(\frac{1}{x}\right)}$$ $$\sqrt{1-\frac{i}{x}}=\sqrt{r\left(x\right)}{e}^{-\frac{i}{2}arctan\left(\frac{1}{x}\right)}and$$ $$\sqrt{1+\frac{i}{x}}=\sqrt{r\left(x\right)}{e}^{\frac{i}{2}arctan\left(\frac{1}{x}\right)}$$ $$i\sqrt{1-\frac{i}{x}}=\sqrt{r\left(x\right)}{e}^{i\frac{\pi }{2}-\frac{i}{2}arctan\left(\frac{1}{x}\right)}$$ $$=\sqrt{r\left(x\right)}=\sqrt{r\left(x\right)}{e}^{i(\frac{\pi }{2}-\frac{1}{2}(\frac{\pi }{2}-arctanx))}$$ $$=\sqrt{r\left(x\right)}{e}^{i(\frac{\pi }{4}+\frac{1}{2}arctan\left(x\right))}also$$ $$i\sqrt{1+\frac{i}{x}}=\sqrt{r\left(x\right)}{e}^{i\frac{\pi }{2}+\frac{i}{2}arctan\left(\frac{1}{x}\right)}$$ $$=\sqrt{r\left(x\right)}{e}^{i(\frac{\pi }{2}+\frac{1}{2}(\frac{\pi }{2}-arctanx))}$$ $$=\sqrt{r\left(x\right)}{e}^{i(\frac{3\pi }{4}-\frac{1}{2}actan\left(x\right))}....becontinued...$$ $$$$

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