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Question Number 68409 by mathmax by abdo last updated on 10/Sep/19
calculate ∫_0 ^(+∞) ((arctan(x^2 ))/(1+x^2 ))dx
$${calculate}\:\int_{\mathrm{0}} ^{+\infty} \:\:\frac{{arctan}\left({x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$
Commented by mathmax by abdo last updated on 11/Sep/19
let I =∫_0 ^∞ ((arctan(x^2 ))/(1+x^2 ))dx changement x=(1/t) give I =−∫_0 ^∞ ((arctan((1/t^2 )))/(1+(1/t^2 )))(−(dt/t^2 ))=∫_0 ^∞ (((π/2)−arctan(t^2 ))/(t^2 +1))dt =(π/2)∫_0 ^∞ (dt/(1+t^2 ))−∫_0 ^∞ ((arctan(t^2 ))/(1+t^2 )) =(π^2 /4)−I ⇒2I =(π^2 /4) ⇒I =(π^2 /8)
$${let}\:{I}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{arctan}\left({x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:{changement}\:\:{x}=\frac{\mathrm{1}}{{t}}\:{give} \\ $$$${I}\:=−\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left(\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right)}{\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}\left(−\frac{{dt}}{{t}^{\mathrm{2}} }\right)=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\frac{\pi}{\mathrm{2}}−{arctan}\left({t}^{\mathrm{2}} \right)}{{t}^{\mathrm{2}} \:+\mathrm{1}}{dt} \\ $$$$=\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }−\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left({t}^{\mathrm{2}} \right)}{\mathrm{1}+{t}^{\mathrm{2}} }\:=\frac{\pi^{\mathrm{2}} }{\mathrm{4}}−{I}\:\Rightarrow\mathrm{2}{I}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\:\Rightarrow{I}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{8}} \\ $$
Commented by mathmax by abdo last updated on 11/Sep/19
another way 2I =∫_(−∞) ^(+∞) ((arctan(x^2 ))/(x^2 +1))dx let W(z) =((arctan(z^2 ))/(z^2 +1)) residus theorem give ∫_(−∞) ^(+∞) W(z)dz =2iπRes(W,i) =2iπ×((∣arctan(i^2 )∣)/(2i)) =π ×∣−(π/4)∣=(π^2 /4) ⇒I =(π^2 /8) ( x→((arctan(x^2 ))/(1+x^2 )) is positive on[0,+∞[)
$${another}\:{way}\:\:\mathrm{2}{I}\:=\int_{−\infty} ^{+\infty} \:\frac{{arctan}\left({x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} \:+\mathrm{1}}{dx}\:\:{let}\:\:{W}\left({z}\right)\:=\frac{{arctan}\left({z}^{\mathrm{2}} \right)}{{z}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${residus}\:{theorem}\:{give}\:\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi{Res}\left({W},{i}\right) \\ $$$$=\mathrm{2}{i}\pi×\frac{\mid{arctan}\left({i}^{\mathrm{2}} \right)\mid}{\mathrm{2}{i}}\:=\pi\:×\mid−\frac{\pi}{\mathrm{4}}\mid=\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\:\:\Rightarrow{I}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\:\:\left(\:\:{x}\rightarrow\frac{{arctan}\left({x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }\right. \\ $$$${is}\:{positive}\:{on}\left[\mathrm{0},+\infty\left[\right)\right. \\ $$
Commented by mathmax by abdo last updated on 11/Sep/19
we must use a opposit contour in this case
$${we}\:{must}\:{use}\:\:{a}\:{opposit}\:{contour}\:{in}\:{this}\:{case}\: \\ $$
Answered by mind is power last updated on 10/Sep/19
=∫_0 ^(+∞) ((arctg(x^2 ))/(1+x^2 ))dx=∫_0 ^(+∞) ((arctg((1/x^2 )))/(x^2 +1))=∫_0 ^(+∞) (((π/2)−arctg(x^2 ))/(x^2 +1)) ⇒2∫_0 ^(+∞) ((arctg(x^2 ))/(x^2 +1))dx=∫_0 ^∞ (π/(2(x^2 +1)))dx=(π^2 /4) ⇒∫_0 ^(+∞) ((arctg(x^2 ))/(1+x^2 ))dx=(π^2 /8)
$$=\int_{\mathrm{0}} ^{+\infty} \frac{{arctg}\left({x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}=\int_{\mathrm{0}} ^{+\infty} \frac{{arctg}\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)}{{x}^{\mathrm{2}} +\mathrm{1}}=\int_{\mathrm{0}} ^{+\infty} \frac{\frac{\pi}{\mathrm{2}}−{arctg}\left({x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\Rightarrow\mathrm{2}\int_{\mathrm{0}} ^{+\infty} \frac{{arctg}\left({x}^{\mathrm{2}} \right)}{{x}^{\mathrm{2}} +\mathrm{1}}{dx}=\int_{\mathrm{0}} ^{\infty} \frac{\pi}{\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{dx}=\frac{\pi^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{+\infty} \frac{{arctg}\left({x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}=\frac{\pi^{\mathrm{2}} }{\mathrm{8}} \\ $$

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