# calculate-0-arctan-x-2-1-x-2-dx-

Question Number 68409 by mathmax by abdo last updated on 10/Sep/19
$${calculate}\:\int_{\mathrm{0}} ^{+\infty} \:\:\frac{{arctan}\left({x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\$$
Commented by mathmax by abdo last updated on 11/Sep/19
$${let}\:{I}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{arctan}\left({x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:{changement}\:\:{x}=\frac{\mathrm{1}}{{t}}\:{give} \\$$$${I}\:=−\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left(\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right)}{\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}\left(−\frac{{dt}}{{t}^{\mathrm{2}} }\right)=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\frac{\pi}{\mathrm{2}}−{arctan}\left({t}^{\mathrm{2}} \right)}{{t}^{\mathrm{2}} \:+\mathrm{1}}{dt} \\$$$$=\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }−\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left({t}^{\mathrm{2}} \right)}{\mathrm{1}+{t}^{\mathrm{2}} }\:=\frac{\pi^{\mathrm{2}} }{\mathrm{4}}−{I}\:\Rightarrow\mathrm{2}{I}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\:\Rightarrow{I}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{8}} \\$$
Commented by mathmax by abdo last updated on 11/Sep/19
$${another}\:{way}\:\:\mathrm{2}{I}\:=\int_{−\infty} ^{+\infty} \:\frac{{arctan}\left({x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} \:+\mathrm{1}}{dx}\:\:{let}\:\:{W}\left({z}\right)\:=\frac{{arctan}\left({z}^{\mathrm{2}} \right)}{{z}^{\mathrm{2}} \:+\mathrm{1}} \\$$$${residus}\:{theorem}\:{give}\:\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi{Res}\left({W},{i}\right) \\$$$$=\mathrm{2}{i}\pi×\frac{\mid{arctan}\left({i}^{\mathrm{2}} \right)\mid}{\mathrm{2}{i}}\:=\pi\:×\mid−\frac{\pi}{\mathrm{4}}\mid=\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\:\:\Rightarrow{I}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\:\:\left(\:\:{x}\rightarrow\frac{{arctan}\left({x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }\right. \\$$$${is}\:{positive}\:{on}\left[\mathrm{0},+\infty\left[\right)\right. \\$$
Commented by mathmax by abdo last updated on 11/Sep/19
$${we}\:{must}\:{use}\:\:{a}\:{opposit}\:{contour}\:{in}\:{this}\:{case}\: \\$$
Answered by mind is power last updated on 10/Sep/19
$$=\int_{\mathrm{0}} ^{+\infty} \frac{{arctg}\left({x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}=\int_{\mathrm{0}} ^{+\infty} \frac{{arctg}\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)}{{x}^{\mathrm{2}} +\mathrm{1}}=\int_{\mathrm{0}} ^{+\infty} \frac{\frac{\pi}{\mathrm{2}}−{arctg}\left({x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} +\mathrm{1}} \\$$$$\Rightarrow\mathrm{2}\int_{\mathrm{0}} ^{+\infty} \frac{{arctg}\left({x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} +\mathrm{1}}{dx}=\int_{\mathrm{0}} ^{\infty} \frac{\pi}{\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{dx}=\frac{\pi^{\mathrm{2}} }{\mathrm{4}} \\$$$$\Rightarrow\int_{\mathrm{0}} ^{+\infty} \frac{{arctg}\left({x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}=\frac{\pi^{\mathrm{2}} }{\mathrm{8}} \\$$