calculate-1-arctan-3-x-2x-2-1-dx-

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 = \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

\dots . be continued \dots