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Question Number 77886 by mathmax by abdo last updated on 11/Jan/20

calculate ∫_0 ^∞   ((arctan(x^2  +x^(−2) ))/(x^2  +a^2 ))dx  with a>0  2) find the value of ∫_0 ^∞   ((arctan(x^2  +x^(−2) ))/(x^2  +1))dx

$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left({x}^{\mathrm{2}} \:+{x}^{−\mathrm{2}} \right)}{{x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} }{dx}\:\:{with}\:{a}>\mathrm{0} \\ $$ $$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left({x}^{\mathrm{2}} \:+{x}^{−\mathrm{2}} \right)}{{x}^{\mathrm{2}} \:+\mathrm{1}}{dx} \\ $$

Commented bymathmax by abdo last updated on 12/Jan/20

1) let A =∫_0 ^∞  ((arctan(x^2  +(1/x^2 )))/(x^2  +a^2 ))dx  the convergence of A is asdured  because ∣arctan(x^2  +x^(−2) )∣<(π/2)  for x≠0  we have  2A = ∫_(−∞) ^(+∞)  ((arctan(x^2 +x^(−2) ))/(x^2  +a^2 ))dx let W(z) =((arctan(z^2  +z^(−2) ))/(z^2  +a^2 ))  ∫_(−∞) ^(+∞)  W(z)dz =2iπRes(W,ia) =2iπ((∣arctan(−a^2 −(1/a^2 ))∣)/(2ia))  =(π/a)arctan(a^2  +(1/a^2 ))⇒A =(π/(2a))arctan(a^2  +(1/a^2 ))  2) a=1 ⇒∫_0 ^∞   ((arctan(x^2  +x^(−2) ))/(x^2  +1))dx =(π/2)arctan(2).

$$\left.\mathrm{1}\right)\:{let}\:{A}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{arctan}\left({x}^{\mathrm{2}} \:+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)}{{x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} }{dx}\:\:{the}\:{convergence}\:{of}\:{A}\:{is}\:{asdured} \\ $$ $${because}\:\mid{arctan}\left({x}^{\mathrm{2}} \:+{x}^{−\mathrm{2}} \right)\mid<\frac{\pi}{\mathrm{2}}\:\:{for}\:{x}\neq\mathrm{0} \\ $$ $${we}\:{have}\:\:\mathrm{2}{A}\:=\:\int_{−\infty} ^{+\infty} \:\frac{{arctan}\left({x}^{\mathrm{2}} +{x}^{−\mathrm{2}} \right)}{{x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} }{dx}\:{let}\:{W}\left({z}\right)\:=\frac{{arctan}\left({z}^{\mathrm{2}} \:+{z}^{−\mathrm{2}} \right)}{{z}^{\mathrm{2}} \:+{a}^{\mathrm{2}} } \\ $$ $$\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi{Res}\left({W},{ia}\right)\:=\mathrm{2}{i}\pi\frac{\mid{arctan}\left(−{a}^{\mathrm{2}} −\frac{\mathrm{1}}{{a}^{\mathrm{2}} }\right)\mid}{\mathrm{2}{ia}} \\ $$ $$=\frac{\pi}{{a}}{arctan}\left({a}^{\mathrm{2}} \:+\frac{\mathrm{1}}{{a}^{\mathrm{2}} }\right)\Rightarrow{A}\:=\frac{\pi}{\mathrm{2}{a}}{arctan}\left({a}^{\mathrm{2}} \:+\frac{\mathrm{1}}{{a}^{\mathrm{2}} }\right) \\ $$ $$\left.\mathrm{2}\right)\:{a}=\mathrm{1}\:\Rightarrow\int_{\mathrm{0}} ^{\infty} \:\:\frac{{arctan}\left({x}^{\mathrm{2}} \:+{x}^{−\mathrm{2}} \right)}{{x}^{\mathrm{2}} \:+\mathrm{1}}{dx}\:=\frac{\pi}{\mathrm{2}}{arctan}\left(\mathrm{2}\right). \\ $$

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