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

Question Number 36944 by maxmathsup by imad last updated on 07/Jun/18

f i n d φ (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.com last updated on 07/Jun/18

c h a n g e m e n t x = a c h (t) g i v e

φ (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} c h (2 t)}

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

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

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

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

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

l e t F (x) = \frac{1}{a^{2} x^{2} + 2 (2 + a^{2}) x + a^{2}}

Δ^{^{'}} = {(2 + a^{2})}^{2} - a^{4} = 4 + 4 a^{2} + a^{4} - a^{4}

= 4 a^{2} + 4

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

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

F (x) = \frac{1}{x_{1} - x_{2}} {\frac{1}{x - x_{1}} - \frac{1}{x - x_{2}}} \Rightarrow

\int_{1}^{+ \infty} F (x) d x = \frac{a^{2}}{4 \sqrt{1 + a^{2}}} \int_{1}^{+ \infty} (\frac{1}{x - x_{1}} - \frac{1}{x - x_{2}}) d x

= \frac{a^{2}}{4 \sqrt{1 + a^{2}}} {[l n ∣ \frac{x - x_{1}}{x - x_{2}} ∣]}_{1}^{+ \infty}

= \frac{a^{2}}{4 \sqrt{1 + a^{2}}} l n ∣ \frac{1 - x_{2}}{1 - x_{1}} ∣ = φ (a) .

Answered by tanmay.chaudhury50@gmail.com last updated on 07/Jun/18

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

d t = - t^{2} d x

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

\int_{\frac{1}{a}}^{0} \frac{- d t}{t^{2}} \times \frac{1}{1 + \frac{1}{t^{2}}} \times \frac{1}{\sqrt{\frac{1}{t^{2}} - a^{2}}}

\int_{\frac{1}{a}}^{0} \frac{- d t}{t^{2}} \times \frac{t^{2}}{1 + t^{2}} \times \frac{t}{\sqrt{1 - a^{2} t^{2}}}

\int_{0}^{\frac{1}{a}} \frac{t d t}{(1 + t^{2}) \times a \sqrt{\frac{1}{a^{2}} - t^{2}}}

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

k^{2} = \frac{1}{a^{2}} - t^{2}

2 k d k = - 2 t d t

= \frac{1}{a} \int_{\frac{1}{a}}^{0} \frac{- k d k}{(1 + \frac{1}{a^{2}} - k^{2}) k}

= \frac{1}{a} \int_{0}^{\frac{1}{a}} \frac{d k}{{(\sqrt{1 + \frac{1}{a^{2}}})}^{2} - k^{2}}

n o w u s e 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 \times \sqrt{1 + \frac{1}{a^{2}}}} {l n ∣ \frac{\sqrt{1 + \frac{1}{a^{2}}} + k}{\sqrt{1 + \frac{1}{a^{2}}} - k} ∣}_{0}^{\frac{1}{a}}

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

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