tan-1-x-2-2-x-2-1-x-2-2-dx-

Question Number 142103 by rs4089 last updated on 26/May/21

\int_{- \infty}^{\infty} \frac{{t a n}^{- 1} (\sqrt{x^{2} + 2})}{(x^{2} + 1) \sqrt{x^{2} + 2}} d x

Answered by mathmax by abdo last updated on 26/May/21

f (a) = \int_{- \infty}^{+ \infty} \frac{\arctan (a \sqrt{x^{2} + 2})}{(x^{2} + 1) \sqrt{x^{2} + 2}} dx \Rightarrow

f^{^{'}} (a) = \int_{- \infty}^{+ \infty} \frac{dx}{(x^{2} + 1) (1 + a^{2} (x^{2} + 2))} = 2 \int_{0}^{\infty} \frac{dx}{(x^{2} + 1) (a^{2} x^{2} + {2 a}^{2} + 1)}

=_{ax = t} 2 \int_{0}^{\infty} \frac{dt}{a (\frac{t^{2}}{a^{2}} + 1) (t^{2} + {2 a}^{2} + 1)} = 2 a \int_{0}^{\infty} \frac{dt}{(t^{2} + a^{2}) (t^{2} + {2 a}^{2} + 1)}

= \frac{2 a}{a^{2} + 1} \int_{0}^{\infty} (\frac{1}{t^{2} + a^{2}} - \frac{1}{t^{2} + {2 a}^{2} + 1}) dt

= \frac{2 a}{a^{2} + 1} {\int_{0}^{\infty} \frac{dt}{t^{2} + a^{2}} - \int_{0}^{\infty} \frac{dt}{t^{2} + {2 a}^{2} + 1}} we have

{\int_{0}^{\infty} \frac{dt}{t^{2} + a^{2}} =_{t = ay} \int_{0}^{\infty} \frac{ady}{a^{2} (y^{2} + 1)} = \frac{1}{a} \arctan (\frac{t}{a})]}_{0}^{\infty} = \frac{π}{2 a}

\int_{0}^{\infty} \frac{dt}{t^{2} + {2 a}^{2} + 1} =_{t = \sqrt{{2 a}^{2} + 1} y} \int_{0}^{\infty} \frac{\sqrt{{2 a}^{2} + 1}}{({2 a}^{2} + 1) (y^{2} + 1)} dy

{= \frac{1}{\sqrt{{2 a}^{2} + 1}} \arctan (\frac{t}{\sqrt{{2 a}^{2} + 1}})]}_{0}^{\infty} = \frac{π}{2 \sqrt{{2 a}^{2} + 1}}

f^{^{'}} (a) = \frac{2 a}{a^{2} + 1} \frac{π}{2 a} - \frac{2 a}{a^{2} + 1} \frac{π}{2 \sqrt{{2 a}^{2} + 1}} = \frac{π}{a^{2} + 1} - \frac{a π}{(a^{2} + 1) \sqrt{{2 a}^{2} + 1}} \Rightarrow

f (a) = π \int_{0}^{a} \frac{dx}{x^{2} + 1} - π \int_{0}^{a} \frac{xdx}{(x^{2} + 1) \sqrt{{2 x}^{2} + 1}} + C

f (0) = 0 = C \Rightarrow f (a) = π \int_{0}^{a} \frac{dx}{x^{2} + 1} - π \int_{0}^{a} \frac{xdx}{(x^{2} + 1) \sqrt{{2 x}^{2} + 1}}

and \int_{- \infty}^{+ \infty} \frac{\arctan (\sqrt{x^{2} + 2})}{(x^{2} + 1) \sqrt{x^{2} + 2}} dx = f (1) = π \int_{0}^{1} \frac{dx}{x^{2} + 1} - π \int_{0}^{1} \frac{xdx}{(x^{2} + 1) \sqrt{{2 x}^{2} + 1}}

we have \int_{0}^{1} \frac{dx}{x^{2} + 1} = \frac{π}{4}

\int_{0}^{1} \frac{xdx}{(x^{2} + 1) \sqrt{{2 x}^{2} + 1}} =_{\sqrt{2} x = shy} \int_{0}^{argsh (\sqrt{2})} \frac{shy}{\sqrt{2} (\frac{{sh}^{2} y}{2} + 1) chy} \frac{chy}{\sqrt{2}} dy

= \frac{1}{2} \int_{0}^{\log (\sqrt{2} + \sqrt{3})} \frac{shy}{\frac{1}{2} {sh}^{2} y + 1} dy = \int_{0}^{\log (\sqrt{2} + \sqrt{3})} \frac{shy}{{sh}^{2} y + 2} dy

= \int_{0}^{\log (\sqrt{2} + \sqrt{3})} \frac{shy}{\frac{ch (2 y) - 1}{2} + 2} dy = 2 \int_{0}^{\log (\sqrt{2} + \sqrt{3})} \frac{shy}{ch (2 y) + 3} dy

= 2 \int_{0}^{\log (\sqrt{2} + \sqrt{3})} \frac{\frac{e^{y} - e^{- y}}{2}}{\frac{e^{2 y} - e^{- 2 y}}{2} + 3} dy = 2 \int_{0}^{\log (\sqrt{2} + \sqrt{3})} \frac{e^{y} - e^{- y}}{e^{2 y} - e^{- 2 y} + 6} dy

= 2 \int_{0}^{\log (\sqrt{2} + \sqrt{3})} \frac{e^{3 y} - e^{y}}{e^{4 y} - 1 + {6 e}^{2 y}} dy

=_{e^{y} = u} 2 \int_{1}^{\sqrt{2} + \sqrt{3}} \frac{u^{3} - u}{u^{4} + {6 u}^{2} - 1} \frac{du}{u}

= 2 \int_{1}^{\sqrt{2} + \sqrt{3}} \frac{u^{2} - 1}{u^{4} + {6 u}^{2} - 1} du

u^{4} + {6 u}^{2} - 1 = 0 \Rightarrow z^{2} + 6 z - 1 = 0 (z = u^{2})

Δ^{^{'}} = 9 + 1 = 10 \Rightarrow u_{1} = - 3 + \sqrt{10} and u_{2} = - 3 - \sqrt{10}

u^{4} + {6 u}^{2} - 1 = (u^{2} - u_{1}) (u^{2} - u_{2}) \Rightarrow

Ψ (u) = \frac{u^{2} - 1}{u^{4} + {6 u}^{2} - 1} = (u^{2} - 1) (\frac{1}{u^{2} - u_{1}} - \frac{1}{u^{2} - u_{2}}) \times \frac{1}{u_{1} - u_{2}}

= \frac{1}{2 \sqrt{10}} (\frac{u^{2} - 1}{u^{2} - u_{1}} - \frac{u^{2} - 1}{u^{2} - u_{2}}) = \frac{1}{2 \sqrt{10}} (\frac{u^{2} - u_{1} + u_{1} - 1}{u^{2} - u_{1}} - \frac{u^{2} - u_{2} + u_{2} - 1}{u^{2} - u_{2}})

= \frac{1}{2 \sqrt{10}} (\frac{u_{1} - 1}{u^{2} - u_{1}} - \frac{u_{2} - 1}{u^{2} - u_{2}}) \Rightarrow

\int_{0}^{1} \frac{xdx}{(x^{2} + 1) \sqrt{{2 x}^{2} + 1}} = \frac{u_{1} - 1}{\sqrt{10}} \int_{1}^{\sqrt{2} + \sqrt{3}} \frac{du}{u^{2} - u_{1}} - \frac{u_{2} - 1}{\sqrt{10}} \int_{1}^{\sqrt{2} + \sqrt{3}} \frac{du}{u^{2} - u_{2}}

now its eazy to find those integrals \dots .

Answered by mindispower last updated on 26/May/21

\int_{- \infty}^{\infty} \frac{{t a n}^{-} (t \sqrt{x^{2} + 2})}{\sqrt{x^{2} + 2} (1 + x^{2})} d x = f (t)

f^{'} (t) = \int_{- \infty}^{\infty} \frac{1}{(1 + x^{2}) (1 + 2 t^{2} + t^{2} x^{2})}

= 2 i π . (\frac{1}{2 i (1 + t^{2})} + \frac{1}{1 - \frac{1 + 2 t^{2}}{t^{2}}} . \frac{1}{2 i t \sqrt{1 + 2 t^{2}}})

= \frac{π}{1 + t^{2}} + \frac{π t}{- (1 + t^{2}) \sqrt{1 + 2 t^{2}}} = f^{'} (t)

f (0) = 0 \Leftrightarrow \int_{- \infty}^{\infty} \frac{a r c t a n (\sqrt{x^{2} + 2})}{(1 + x^{2}) \sqrt{x^{2} + 2}} d x = f (1) = \int_{0}^{1} f^{'} (t) d t

= \int_{0}^{1} \frac{π}{1 + t^{2}} - \frac{π}{2} \int_{0}^{1} \frac{d x}{(1 + x) \sqrt{1 + 2 x}} d x

= \frac{π^{2}}{4} - \frac{π}{2} \int_{1}^{3} \frac{u d u}{\frac{u^{2} + 1}{2} u} = \frac{π^{2}}{4} - π \int_{1}^{\sqrt{3}} \frac{d u}{(u^{2} + 1)}

= \frac{π^{2}}{4} - π [_{1}^{\sqrt{3}} a r c t a n (u)]

= \frac{π^{2}}{4} - π . \frac{π}{12} = \frac{π^{2}}{6}

\int_{- \infty}^{\infty} \frac{\tan^{- 1} (\sqrt{2 x + 1})}{(1 + x^{2}) \sqrt{2 x + 1}} d x = \frac{π^{2}}{6}