Question Number 143080 by Mathspace last updated on 09/Jun/21
$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left({x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$
Answered by qaz last updated on 10/Jun/21
$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{tan}^{−\mathrm{1}} \mathrm{x}^{\mathrm{2}} }{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{tan}^{−\mathrm{1}} \mathrm{x}^{\mathrm{2}} }{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}+\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{tan}^{−\mathrm{1}} \mathrm{x}^{\mathrm{2}} }{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{tan}^{−\mathrm{1}} \mathrm{x}^{\mathrm{2}} }{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$$$=\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{dx}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} } \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{8}} \\ $$
Commented by mathmax by abdo last updated on 10/Jun/21
$$\mathrm{thanks}\:\mathrm{sir}. \\ $$
Answered by mathmax by abdo last updated on 10/Jun/21
![Φ=∫_0 ^∞ ((arctan(x^2 ))/(1+x^2 ))dx ⇒Φ=∫_0 ^1 ((arctan(x^2 ))/(1+x^2 ))dx +∫_1 ^∞ ((arctan(x^2 ))/(1+x^2 ))(→x=(1/t)) =∫_0 ^1 ((arctan(x^2 ))/(1+x^2 ))dx∫_0 ^1 (((π/2)−arctan(t^2 ))/(1+(1/t^2 )))((dt/t^2 )) =(π/2)∫_0 ^1 (dt/(1+t^2 ))=(π/2)[arctant]_0 ^1 =(π/2).(π/4) ⇒Φ=(π^2 /8)](https://www.tinkutara.com/question/Q143145.png)
$$\Phi=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{arctan}\left(\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\:\Rightarrow\Phi=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{arctan}\left(\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\:+\int_{\mathrm{1}} ^{\infty} \:\frac{\mathrm{arctan}\left(\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\left(\rightarrow\mathrm{x}=\frac{\mathrm{1}}{\mathrm{t}}\right) \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{arctan}\left(\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\frac{\pi}{\mathrm{2}}−\mathrm{arctan}\left(\mathrm{t}^{\mathrm{2}} \right)}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{2}} }}\left(\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} }\right) \\ $$$$=\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{dt}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }=\frac{\pi}{\mathrm{2}}\left[\mathrm{arctant}\right]_{\mathrm{0}} ^{\mathrm{1}} =\frac{\pi}{\mathrm{2}}.\frac{\pi}{\mathrm{4}}\:\Rightarrow\Phi=\frac{\pi^{\mathrm{2}} }{\mathrm{8}} \\ $$