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Question Number 147205 by mathmax by abdo last updated on 18/Jul/21
calculate ∫_1 ^∞  ((arctan((3/x)))/(2x^2  +1))dx
$$\mathrm{calculate}\:\int_{\mathrm{1}} ^{\infty} \:\frac{\mathrm{arctan}\left(\frac{\mathrm{3}}{\mathrm{x}}\right)}{\mathrm{2x}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{dx} \\ $$
Answered by mathmax by abdo last updated on 20/Jul/21
Ψ=∫_1 ^∞  ((arctan((3/x)))/(2x^2  +1))dx ⇒Ψ=∫_1 ^∞  (((π/2)−arcan((x/3)))/(2x^2  +1))dx  =(π/2)∫_1 ^∞  (dx/(2x^2  +1))−∫_1 ^∞  ((arctan((x/3)))/(2x^2  +1))dx =(π/2)I−J  I=(1/2)∫_1 ^∞  (dx/(x^2  +(1/2))) =_(x=(1/( (√2)))y)   (1/2)∫_(√2) ^∞  (dy/( (√2)×(1/2)(1+y^2 )))  =(1/( (√2)))[arctany]_(√2) ^∞  =(1/( (√2)))((π/2) −arctan((√2)))  we consider f(a)=∫_1 ^∞  ((arctan(ax))/(2x^2  +1))dx     (o<a<1)  f^′ (a)=∫_1 ^∞  (x/((1+a^2 x^2 )(2x^2  +1)))dx  =_(ax=y)     ∫_a ^∞  (dy/(a(1+y^2 )(2(y^2 /a^2 ) +1)))=∫_a ^∞   ((a^2 dy)/(a(y^2 +1)(2y^2  +a^2 )))  =a∫_a ^∞  (dy/((y^2  +1)(2y^2  +a^2 )))=2a∫_a ^∞  (dy/((2y^2 +2)(2y^2  +a^2 )))  =((2a)/(a^2 −2)) ∫_a ^∞ ((1/(2y^2 +2))−(1/(2y^2 +a^2 )))dy  =(a/(a^2 −2))∫_a ^∞  (dy/(y^2  +1))−(a/(a^2 −2))∫_a ^∞  (dy/(y^2  +(a^2 /2)))(→y=(a/( (√2)))z)  =(a/(a^2 −2))((π/2)−arctana)−(a/(a^2 −2)).(a/( (√2)))∫_(√2) ^∞  (dz/((a^2 /2)(1+z^2 )))  =(a/(a^2 −2))((π/2)−arctana)−((√2)/(a^2 −2))((π/2)−arctan((√2))) ⇒  f(a)=∫ {(a/(a^2 −2))((π/2)−arctana)−((√2)/(a^2 −2))((π/2)−arctan(√2))}da  ....be continued...
$$\Psi=\int_{\mathrm{1}} ^{\infty} \:\frac{\mathrm{arctan}\left(\frac{\mathrm{3}}{\mathrm{x}}\right)}{\mathrm{2x}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{dx}\:\Rightarrow\Psi=\int_{\mathrm{1}} ^{\infty} \:\frac{\frac{\pi}{\mathrm{2}}−\mathrm{arcan}\left(\frac{\mathrm{x}}{\mathrm{3}}\right)}{\mathrm{2x}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{dx} \\ $$$$=\frac{\pi}{\mathrm{2}}\int_{\mathrm{1}} ^{\infty} \:\frac{\mathrm{dx}}{\mathrm{2x}^{\mathrm{2}} \:+\mathrm{1}}−\int_{\mathrm{1}} ^{\infty} \:\frac{\mathrm{arctan}\left(\frac{\mathrm{x}}{\mathrm{3}}\right)}{\mathrm{2x}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{dx}\:=\frac{\pi}{\mathrm{2}}\mathrm{I}−\mathrm{J} \\ $$$$\mathrm{I}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{1}} ^{\infty} \:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{2}}}\:=_{\mathrm{x}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mathrm{y}} \:\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\sqrt{\mathrm{2}}} ^{\infty} \:\frac{\mathrm{dy}}{\:\sqrt{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\mathrm{y}^{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left[\mathrm{arctany}\right]_{\sqrt{\mathrm{2}}} ^{\infty} \:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left(\frac{\pi}{\mathrm{2}}\:−\mathrm{arctan}\left(\sqrt{\mathrm{2}}\right)\right) \\ $$$$\mathrm{we}\:\mathrm{consider}\:\mathrm{f}\left(\mathrm{a}\right)=\int_{\mathrm{1}} ^{\infty} \:\frac{\mathrm{arctan}\left(\mathrm{ax}\right)}{\mathrm{2x}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{dx}\:\:\:\:\:\left(\mathrm{o}<\mathrm{a}<\mathrm{1}\right) \\ $$$$\mathrm{f}^{'} \left(\mathrm{a}\right)=\int_{\mathrm{1}} ^{\infty} \:\frac{\mathrm{x}}{\left(\mathrm{1}+\mathrm{a}^{\mathrm{2}} \mathrm{x}^{\mathrm{2}} \right)\left(\mathrm{2x}^{\mathrm{2}} \:+\mathrm{1}\right)}\mathrm{dx} \\ $$$$=_{\mathrm{ax}=\mathrm{y}} \:\:\:\:\int_{\mathrm{a}} ^{\infty} \:\frac{\mathrm{dy}}{\mathrm{a}\left(\mathrm{1}+\mathrm{y}^{\mathrm{2}} \right)\left(\mathrm{2}\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} }\:+\mathrm{1}\right)}=\int_{\mathrm{a}} ^{\infty} \:\:\frac{\mathrm{a}^{\mathrm{2}} \mathrm{dy}}{\mathrm{a}\left(\mathrm{y}^{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{2y}^{\mathrm{2}} \:+\mathrm{a}^{\mathrm{2}} \right)} \\ $$$$=\mathrm{a}\int_{\mathrm{a}} ^{\infty} \:\frac{\mathrm{dy}}{\left(\mathrm{y}^{\mathrm{2}} \:+\mathrm{1}\right)\left(\mathrm{2y}^{\mathrm{2}} \:+\mathrm{a}^{\mathrm{2}} \right)}=\mathrm{2a}\int_{\mathrm{a}} ^{\infty} \:\frac{\mathrm{dy}}{\left(\mathrm{2y}^{\mathrm{2}} +\mathrm{2}\right)\left(\mathrm{2y}^{\mathrm{2}} \:+\mathrm{a}^{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{2a}}{\mathrm{a}^{\mathrm{2}} −\mathrm{2}}\:\int_{\mathrm{a}} ^{\infty} \left(\frac{\mathrm{1}}{\mathrm{2y}^{\mathrm{2}} +\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2y}^{\mathrm{2}} +\mathrm{a}^{\mathrm{2}} }\right)\mathrm{dy} \\ $$$$=\frac{\mathrm{a}}{\mathrm{a}^{\mathrm{2}} −\mathrm{2}}\int_{\mathrm{a}} ^{\infty} \:\frac{\mathrm{dy}}{\mathrm{y}^{\mathrm{2}} \:+\mathrm{1}}−\frac{\mathrm{a}}{\mathrm{a}^{\mathrm{2}} −\mathrm{2}}\int_{\mathrm{a}} ^{\infty} \:\frac{\mathrm{dy}}{\mathrm{y}^{\mathrm{2}} \:+\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{2}}}\left(\rightarrow\mathrm{y}=\frac{\mathrm{a}}{\:\sqrt{\mathrm{2}}}\mathrm{z}\right) \\ $$$$=\frac{\mathrm{a}}{\mathrm{a}^{\mathrm{2}} −\mathrm{2}}\left(\frac{\pi}{\mathrm{2}}−\mathrm{arctana}\right)−\frac{\mathrm{a}}{\mathrm{a}^{\mathrm{2}} −\mathrm{2}}.\frac{\mathrm{a}}{\:\sqrt{\mathrm{2}}}\int_{\sqrt{\mathrm{2}}} ^{\infty} \:\frac{\mathrm{dz}}{\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{2}}\left(\mathrm{1}+\mathrm{z}^{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{a}}{\mathrm{a}^{\mathrm{2}} −\mathrm{2}}\left(\frac{\pi}{\mathrm{2}}−\mathrm{arctana}\right)−\frac{\sqrt{\mathrm{2}}}{\mathrm{a}^{\mathrm{2}} −\mathrm{2}}\left(\frac{\pi}{\mathrm{2}}−\mathrm{arctan}\left(\sqrt{\mathrm{2}}\right)\right)\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{a}\right)=\int\:\left\{\frac{\mathrm{a}}{\mathrm{a}^{\mathrm{2}} −\mathrm{2}}\left(\frac{\pi}{\mathrm{2}}−\mathrm{arctana}\right)−\frac{\sqrt{\mathrm{2}}}{\mathrm{a}^{\mathrm{2}} −\mathrm{2}}\left(\frac{\pi}{\mathrm{2}}−\mathrm{arctan}\sqrt{\mathrm{2}}\right)\right\}\mathrm{da} \\ $$$$….\mathrm{be}\:\mathrm{continued}… \\ $$$$ \\ $$

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