find-f-a-a-dx-1-x-2-x-2-a-2-with-a-gt-0-

Question Number 50423 by Abdo msup. last updated on 16/Dec/18

f i n d f (a) = \int_{a}^{+ \infty} \frac{d x}{(1 + x^{2}) \sqrt{x^{2} - a^{2}}} w i t h a > 0

Commented by Abdo msup. last updated on 17/Dec/18

c h a n g e m e n t x = a c h (t) g i v e t = a r g c h (\frac{x}{a})

f (a) = \int_{0}^{+ \infty} \frac{a s h (t) d t}{(1 + a^{2} {c h}^{2} (t)) a s h (t)}

= \int_{0}^{\infty} \frac{d t}{1 + a^{2} \frac{1 + c h (2 t)}{2}} = \int_{0}^{\infty} \frac{2 d t}{2 + a^{2} + a^{2} \frac{e^{2 t} + e^{- 2 t}}{2}}

= \int_{0}^{\infty} \frac{4 d t}{4 + 2 a^{2} + a^{2} e^{2 t} + a^{2} e^{- 2 t}}

=_{e^{2 t} = u} \int_{1}^{\infty} \frac{4}{4 + 2 a^{2} + a^{2} u + a^{2} u^{- 1}} \frac{d u}{2 u}

= 2 \int_{1}^{\infty} \frac{d u}{(4 + 2 a^{2}) u + a^{2} u^{2} + a^{2}}

= \int_{1}^{\infty} \frac{2 d u}{a^{2} u^{2} + (4 + 2 a^{2}) u + a^{2}}

Δ^{^{'}} = {(2 + a^{2})}^{2} - a^{4} = 4 + 4 a^{2} > 0 \Rightarrow

u_{1} = \frac{- 2 - a^{2} + 2 \sqrt{1 + a^{2}}}{a^{2}}

u_{2} = \frac{- 2 - a^{2} - 2 \sqrt{1 + a^{2}}}{a^{2}}

F (u) = \frac{2}{(u - u_{1}) (u - u_{2})} = \frac{2}{u_{1} - u_{2}} (\frac{1}{u - u_{1}} - \frac{1}{u - u_{2}})

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

\int_{1}^{\infty} F (u) d u = \frac{a^{2}}{2 \sqrt{1 + a^{2}}} {[l n ∣ \frac{u - u_{1}}{u - u_{2}} ∣]}_{1}^{+ \infty}

= \frac{a^{2}}{2 \sqrt{1 + a^{2}}} l n ∣ \frac{1 - u_{2}}{1 - u_{1}} ∣

= \frac{a^{2}}{2 \sqrt{1 + a^{2}}} l n ∣ \frac{1 - \frac{- 2 - a^{2} - 2 \sqrt{1 + a^{2}}}{a^{2}}}{1 - \frac{- 2 - a^{2} + 2 \sqrt{1 + a^{2}}}{a^{2}}} ∣

= \frac{a^{2}}{2 \sqrt{1 + a^{2}}} l n ∣ \frac{2 a^{2} + 2 + 2 \sqrt{1 + a^{2}}}{2 a^{2} + 2 - 2 \sqrt{1 + a^{2}}} ∣

= \frac{a^{2}}{2 \sqrt{1 + a^{2}}} l n ∣ \frac{a^{2} + 1 + \sqrt{1 + a^{2}}}{a^{2} + 1 - \sqrt{1 + a^{2}}} ∣ = f (a)

Answered by tanmay.chaudhury50@gmail.com last updated on 17/Dec/18

\int \frac{d x}{(1 + x^{2}) \sqrt{x^{2} - a^{2}}}

t = \frac{1}{x} x = \frac{1}{t} d x = \frac{- 1}{t^{2}} d t

\int \frac{- d t}{t^{2} (1 + \frac{1}{t^{2}}) \sqrt{\frac{1}{t^{2}} - a^{2}}}

\int \frac{- t d t}{(t^{2} + 1) \sqrt{1 - a^{2} t^{2}}}

= \frac{- 1}{a} \int \frac{t d t}{(t^{2} + 1) \sqrt{\frac{1}{a^{2}} - t^{2}}}

k = t^{2} d k = 2 t d t

= \frac{- 1}{2 a} \int \frac{d k}{(k + 1) \sqrt{\frac{1}{a^{2}} - k}}

p^{2} = \frac{1}{a^{2}} - k 2 p d p = - d k

= \frac{- 1}{2 a} \int \frac{- 2 p d p}{(\frac{1}{a^{2}} - p^{2} + 1) p}

= \frac{1}{a} \int \frac{d p}{(\frac{1}{a^{2}} + 1) - p^{2}} f o r m u l a [\int \frac{d x}{a^{2} - x^{2}} = \frac{1}{2 a} l n (\frac{a + x}{a - x})]

= \frac{1}{a} \times \frac{1}{2 \sqrt{(\frac{1}{a^{2}} + 1)}} \times l n (\frac{\sqrt{\frac{1}{a^{2}} + 1} + p}{\sqrt{\frac{1}{a^{2}} + 1} - p})

= \frac{1}{2 \sqrt{1 + a^{2}}} l n (\frac{\sqrt{\frac{1}{a^{2}} + 1} + \sqrt{\frac{1}{a^{2}} - k}}{\sqrt{\frac{1}{a^{2}} + 1} - \sqrt{\frac{1}{a^{2}} - k}})

= \frac{1}{2 \sqrt{1 + a^{2}}} l n (\frac{\sqrt{1 + a^{2}} + \sqrt{1 - a^{2} k}}{\sqrt{1 + a^{2}} - \sqrt{1 - a^{2} k}})

= \frac{1}{2 \sqrt{1 + a^{2}}} l n (\frac{\sqrt{1 + a^{2}} + \sqrt{1 - a^{2} t^{2}}}{\sqrt{1 + a^{2}} - \sqrt{1 - a^{2} t^{2}}})

= \frac{1}{2 \sqrt{1 + a^{2}}} ∣ l n (\frac{\sqrt{1 + a^{2}} + \sqrt{1 - \frac{a^{2}}{x^{2}}}}{\sqrt{1 + a^{2}} - \sqrt{1 - \frac{a^{2}}{x^{2}}}}) ∣_{a}^{\infty}

= \frac{1}{2 \sqrt{1 + a^{2}}} [{l n (\frac{\sqrt{1 + a^{2}} + \sqrt{1 - 0}}{\sqrt{1 + a^{2}} - \sqrt{1 - 0}})} - l n (\frac{\sqrt{1 + a^{2}} + \sqrt{0}}{\sqrt{1 + a^{2}} - \sqrt{0}})}]

= \frac{1}{\sqrt{1 + a^{2}}} l n (\frac{\sqrt{1 + a^{2}} + 1}{\sqrt{1 + a^{2}} - 1})

Commented by Abdo msup. last updated on 17/Dec/18

t h a n k y o u s i r t a n m a y .