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Question Number 44476 by abdo.msup.com last updated on 29/Sep/18

l e t f (x) = \int_{0}^{\infty} \frac{d t}{t^{2} + 2 x t - 1}

1) f i n d a e x p l i c i t f o r m o f f (x)

2) c s l v u l s t e \int_{0}^{\infty} \frac{d t}{t^{2} + t - 1}

3) c a l c u l a t e A (θ) = \int_{0}^{\infty} \frac{d t}{t^{2} + 2 t a n (θ) t - 1}

4) c a l c u l a t e g (x) = \int_{0}^{\infty} \frac{t d t}{{(t^{2} + 2 x t - 1)}^{2}}

5) f i n d t h e v a l u e o f \int_{0}^{\infty} \frac{t d t}{{(t^{2} + 4 t - 1)}^{2}}

Commented by maxmathsup by imad last updated on 01/Oct/18

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

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

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

= \frac{1}{2 \sqrt{1 + x^{2}}} {[l n ∣ \frac{t + x - \sqrt{1 + x^{2}}}{t + x + \sqrt{1 + x^{2}}}]}_{0}^{+ \infty}

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

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

Commented by maxmathsup by imad last updated on 01/Oct/18

2) \int_{0}^{\infty} \frac{d t}{t^{2} + t - 1} = f (\frac{1}{2}) = \frac{1}{2 \sqrt{1 + \frac{1}{4}}} l n ∣ \frac{\frac{1}{2} + \sqrt{1 + \frac{1}{4}}}{\frac{1}{2} - \sqrt{1 + \frac{1}{4}}} ∣

= \frac{1}{\sqrt{5}} l n (\frac{1 + \sqrt{5}}{\sqrt{5} - 1}) .

Commented by maxmathsup by imad last updated on 01/Oct/18

3) w e h a v e \int_{0}^{\infty} \frac{d t}{t^{2} - 2 t a n θ t - 1} = f (t a n θ)

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

= \frac{∣ c o s θ ∣}{2} l n ∣ \frac{t a n θ + \frac{1}{∣ c o s θ ∣}}{t a n θ - \frac{1}{∣ c o s θ ∣}} ∣ = \frac{∣ c o s θ ∣}{2} l n ∣ \frac{ξ (θ) s i n θ + 1}{ξ (θ) - 1} ∣ w i t h ξ (θ) = \frac{c o s θ}{∣ c o s θ ∣} = \bar{+} 1

Commented by maxmathsup by imad last updated on 01/Oct/18

4) w e h a v e f (x) = \int_{0}^{\infty} \frac{d t}{t^{2} + 2 x t - 1} \Rightarrow f^{^{'}} (x) = - \int_{0}^{\infty} \frac{2 t}{{(t^{2} + 2 x t - 1)}^{2}} d t \Rightarrow

\int_{0}^{\infty} \frac{t}{{(t^{2} + 2 x t - 1)}^{2}} d t = - \frac{1}{2} f^{^{'}} (x)

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

f^{^{'}} (x) = \frac{- x}{2} {(1 + x^{2})}^{- \frac{3}{2}} {l n ∣ x + \sqrt{1 + x^{2} ∣} - l n ∣ x - \sqrt{1 + x^{2}} ∣}

+ \frac{1}{2 \sqrt{1 + x^{2}}} {\frac{1 + \frac{x}{\sqrt{1 + x^{2}}}}{x + \sqrt{1 + x^{2}}} - \frac{1 - \frac{x}{\sqrt{1 + x^{2}}}}{x - \sqrt{1 + x^{2}}}}

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

+ \frac{1}{2 \sqrt{1 + x^{2}}} {\frac{1}{\sqrt{1 + x^{2}}} + \frac{1}{\sqrt{1 + x^{2}}}} \Rightarrow g (x) = \frac{x}{2 (1 + x^{2}) \sqrt{1 + x^{2}}} l n ∣ \frac{x + \sqrt{1 + x^{2}}}{x - \sqrt{1 + x^{2}}} ∣

+ \frac{1}{1 + x^{2}}

Commented by maxmathsup by imad last updated on 01/Oct/18

5) \int_{0}^{\infty} \frac{t d t}{{(t^{2} + 4 t - 1)}^{2}} = g (2)

= \frac{1}{10 \sqrt{5}} l n ∣ \frac{2 + \sqrt{5}}{2 - \sqrt{5}} ∣ - \frac{1}{10} .

Commented by maxmathsup by imad last updated on 01/Oct/18

g (x) = - \frac{1}{2} f^{^{'}} (x) \Rightarrow g (x) = \frac{x}{4 (1 + x^{2}) \sqrt{1 + x^{2}}} l n ∣ \frac{x + \sqrt{1 + x^{2}}}{x - \sqrt{1 + x^{2}}} ∣ - \frac{1}{2 (1 + x^{2})}