calculate ∫_1 ^∞ ((arctan((3/x)))/(2x^2 +1))dx


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Question Number 147205 by mathmax by abdo last updated on 18/Jul/21

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