Question Number 193734 by Mastermind last updated on 18/Jun/23
$$\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{x}^{\mathrm{2}} \left(\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{x}\right)\right)^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{2}}\mathrm{dx} \\ $$$$ \\ $$$$\mathrm{Help}! \\ $$
Answered by witcher3 last updated on 18/Jun/23
![I didnt Converge x>1 ((x^2 (tan^− (x))^2 )/(x^2 −x+2))≥(π^2 /(16)).(x^2 /(x^2 −x+2))≥(π^2 /(16)).((2x−1)/(x^2 −x+2)),x^2 ≥2x−1 I≥(π^2 /(16))[ln(x^2 −x+2)]_0 ^a ,∀a≥1 ⇒I≥lim_(a→∞) (π^2 /(16))ln[((a^2 −a+2)/2)]=∞](https://www.tinkutara.com/question/Q193737.png)
$$\mathrm{I}\:\mathrm{didnt}\:\mathrm{Converge} \\ $$$$\mathrm{x}>\mathrm{1}\: \\ $$$$\frac{\mathrm{x}^{\mathrm{2}} \left(\mathrm{tan}^{−} \left(\mathrm{x}\right)\right)^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{2}}\geqslant\frac{\pi^{\mathrm{2}} }{\mathrm{16}}.\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{2}}\geqslant\frac{\pi^{\mathrm{2}} }{\mathrm{16}}.\frac{\mathrm{2x}−\mathrm{1}}{\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{2}},\mathrm{x}^{\mathrm{2}} \geqslant\mathrm{2x}−\mathrm{1} \\ $$$$\mathrm{I}\geqslant\frac{\pi^{\mathrm{2}} }{\mathrm{16}}\left[\mathrm{ln}\left(\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{2}\right)\right]_{\mathrm{0}} ^{\mathrm{a}} ,\forall\mathrm{a}\geqslant\mathrm{1} \\ $$$$\Rightarrow\mathrm{I}\geqslant\underset{\mathrm{a}\rightarrow\infty} {\mathrm{lim}}\frac{\pi^{\mathrm{2}} }{\mathrm{16}}\mathrm{ln}\left[\frac{\mathrm{a}^{\mathrm{2}} −\mathrm{a}+\mathrm{2}}{\mathrm{2}}\right]=\infty \\ $$$$ \\ $$
Commented by Mastermind last updated on 20/Jun/23
$$\mathrm{Thank}\:\mathrm{you} \\ $$