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Question Number 74343 by mathmax by abdo last updated on 22/Nov/19

$$calculatef\left(\alpha \right)={\int }_0^{\mathrm{\infty }}\frac{arctan\left(\alpha x^2\right)}{x^2+9}dxwith\alpha real.$$

Commented by mathmax by abdo last updated on 24/Nov/19

$$case1\alpha >0wehave2f\left(\alpha \right)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{arctan\left(\alpha x^2\right)}{x^2+9}dx$$$${=}_{x=3t}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{arctan\left(9\alpha t^2\right)}{9(t^2+1)}\left(3\right)dt=\frac{1}{3}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{arctan\left(9\alpha t^2\right)}{t^2+1}dt$$$$x\to \frac{arctan\left(9\alpha t^2\right)}{t^2+1}ispositive\Rightarrow {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}(...)dt⩾0let$$$$W\left(z\right)=\frac{arctan\left(9\alpha z^2\right)}{z^2+1}\Rightarrow W\left(z\right)=\frac{arctan\left(9\alpha z^2\right)}{(z-i)(z+i)}and$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}W\left(z\right)dz=2i\pi Res(W,i)=2i\pi \frac{\mid arctan\left(9\alpha (-1)\right)\mid }{2i}$$$$=\pi arctan\left(9\alpha \right)\left(butthisresulteedaproof\right)\Rightarrow$$$$f\left(\alpha \right)=\frac{\pi }{6}arctan\left(9\alpha \right)$$$$case2\alpha <0let{\alpha }^{{}^{\prime }}=-\alpha >0\Rightarrow f\left(\alpha \right)={\int }_0^{\mathrm{\infty }}\frac{arctan(-{\alpha }^{{}^{\prime }}{x}^2)}{x^2+9}dx$$$$=-{\int }_0^{\mathrm{\infty }}\frac{arctan\left({\alpha }^{{}^{\prime }}{x}^2\right)}{x^2+9}=-\frac{\pi }{6}arctan\left(9{\alpha }^{{}^{\prime }}\right)=\frac{\pi }{6}arctan\left(9\alpha \right)$$

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