let-consider-a-function-g-defined-by-g-a-0-1-dx-1-x-1-ax-Give-the-defined-Domain-of-g-and-simplify-g-

Question Number 67465 by ~ À ® @ 237 ~ last updated on 27/Aug/19

l e t c o n s i d e r a f u n c t i o n g d e f i n e d b y g (a) = \int_{0}^{1} \frac{d x}{\sqrt{(1 - x) (1 + a x)}}

G i v e t h e d e f i n e d D o m a i n o f g a n d s i m p l i f y g .

Commented by ~ À ® @ 237 ~ last updated on 28/Aug/19

D_{g} = {a \in R / g (a) \in R}

w e k n o w t h a t \int_{0}^{1} \frac{d x}{\sqrt{1 - u^{2}}} = \frac{π}{2} (s o i t c o n v e r g e s) . T h e n \lim_{x \to 1} \frac{\sqrt{1 - u^{2}}}{\sqrt{(1 - u) (1 + a u)}} = \sqrt{\frac{2}{1 + a}}

b y c o n v e r g e n c e t h e o r e m g (a) e x i s t i f \sqrt{\frac{2}{1 + a}} \in R \Leftrightarrow a > - 1

S o D_{g} =] - 1; + \infty [

I f a = 0 g (0) = \int_{0}^{1} \frac{d x}{\sqrt{1 - x}} = {[- 2 \sqrt{1 - x}]}_{0}^{1} = 2

N o w i f a \neq 0 a n d a \in] - 1; + \infty [

g (a) = \int_{0}^{1} \frac{d u}{\sqrt{u (1 + a - a u)}} (w i t h u = 1 - x)

= \frac{1}{\sqrt{1 + a}} \int_{0}^{1} u^{\frac{- 1}{2}} {(1 - \frac{a u}{1 + a})}^{\frac{- 1}{2}} d u = \frac{1}{\sqrt{1 + a}} \int_{0}^{1} u^{\frac{1}{2} - 1} {(1 - u)}^{\frac{3}{2} - \frac{1}{2} - 1} {(1 - \frac{a}{1 + a} u)}^{\frac{- 1}{2}} d u

= \frac{Γ (\frac{3}{2})}{Γ (\frac{1}{2}) Γ (1) \sqrt{1 + a}} ._{2} F_{1} (\frac{1}{2}, \frac{1}{2}, \frac{3}{2}, \frac{a}{1 + a}) b e c a u s e ∣ \frac{a}{1 + a} ∣< 1

u s i n g t h e f o r m u l a_{2} F_{1} (\frac{1}{2}, \frac{1}{2}, \frac{3}{2}, z^{2}) = \frac{a r c s i n z}{z}

i f a \in] - 1; 0 [, \frac{a}{1 + a} < 0 l e t t a k e z = i \sqrt{∣ \frac{a}{1 + a} ∣}

g (a) = \frac{1}{2 \sqrt{1 + a}} . \frac{1}{i \sqrt{∣ \frac{a}{1 + a} ∣}} a r c s i n (i \sqrt{∣ \frac{a}{1 + a} ∣}) =_{} \frac{a r c s i n (i \sqrt{\frac{∣ a ∣}{1 + a}})}{2 i \sqrt{∣ a ∣}} = \frac{a r g s h (\sqrt{\frac{∣ a ∣}{1 + a}})}{2 \sqrt{∣ a ∣}} b e c a u s e s i n (x) = \frac{e^{i x} - e^{- i x}}{2 i} = \frac{s h (i x)}{i} \Rightarrow s i n (a r c s i n (i x)) = \frac{s h (i a r c s i n (i x))}{i} \Rightarrow i a r g s h x = a r c s i n (i x)

i f a > 0 t h e n \frac{a}{1 + a} > 0 w e t a k e z = \sqrt{\frac{a}{1 + a}}

g (a) = \frac{a r c s i n (\sqrt{\frac{a}{1 + a}})}{2 \sqrt{a}}

Commented by prof Abdo imad last updated on 28/Aug/19

a t V (1) g (a) \sim \int_{0}^{1} \frac{d x}{\sqrt{(1 - x) (1 + a)}}

g (a) e x i s t \Leftrightarrow 1 + a > 0 \Rightarrow a > - 1 \Rightarrow

D_{g} =] - 1, + \infty [.

Commented by prof Abdo imad last updated on 28/Aug/19

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

c h a n g r m e n t \sqrt{1 - x} = t g i v e 1 - x = t^{2} \Rightarrow x = 1 - t^{2}

a > - 1 \Rightarrow a x > - x \Rightarrow 1 + a x > 1 - x > 0

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

- {a t}^{2} + a + 1 = 0 \Rightarrow {a t}^{2} - (a + 1) = 0

Δ^{^{'}} = 0 + a (a + 1) = a^{2} + a = a (a + 1)

c a s e - 1 < a < 0 \Rightarrow Δ^{^{'}} < 0 \Rightarrow n o r e a l r o o t s

a n d - {a t}^{2} + a + 1 = - a (t^{2} - \frac{a + 1}{a}) a n d - \frac{a + 1}{a} > 0 \Rightarrow

w e d o t h e c h a n g e m e n t t = \sqrt{- \frac{a + 1}{a}} u \Rightarrow

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

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

\dots . b e c o n t i n u e d \dots

Commented by mathmax by abdo last updated on 28/Aug/19

c a s e 2 i f a > 0 \Rightarrow Δ^{^{'}} > 0 \Rightarrow t_{1} = \frac{\sqrt{a^{2} + a}}{a} a n d t_{2} = - \frac{\sqrt{a^{2} + a}}{a}

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

w e u s e t h e c h a n g e m e n t \sqrt{t_{1} - t} = u \Rightarrow t_{1} - t = u^{2} \Rightarrow t = t_{1} - u^{2}

g (a) = \frac{2}{\sqrt{a}} \int_{\sqrt{t_{1}}}^{\sqrt{t_{1} - 1}} \frac{(- 2 u) d u}{u \sqrt{t_{1} - u^{2} - t_{2}}} = \frac{- 4}{\sqrt{a}} \int_{\sqrt{t_{1}}}^{\sqrt{t_{1} - 1}} \frac{d u}{\sqrt{t_{1} - t_{2} - u^{2}}}

= \frac{- 4}{\sqrt{a}} \int_{\sqrt{t_{1}}}^{\sqrt{t_{1} - 1}} \frac{d u}{\sqrt{{(\sqrt{t_{1} - t_{2}})}^{2} - u^{2}}}

=_{u = \sqrt{t_{1} - t_{2}} z} \frac{- 4}{\sqrt{a}} \int_{\frac{\sqrt{t_{1}}}{\sqrt{t_{1} - t_{2}}}}^{\frac{\sqrt{t_{1} - 1}}{\sqrt{t_{1} - t_{2}}}} \frac{\sqrt{t_{1} - t_{2}} d z}{\sqrt{t_{1} - t_{2}} (\sqrt{1 - z^{2}})}

= \frac{- 4}{\sqrt{a}} {[a r c s i n (z)]}_{\sqrt{\frac{t_{1}}{t_{1} - t_{2}}}}^{\sqrt{\frac{t_{1} - 1}{t_{1} - t_{2}}}} = \frac{- 4}{\sqrt{a}} {a r c s i n (\sqrt{\frac{t_{1} - 1}{t_{1} - t_{2}}}) - a r c s i n (\sqrt{\frac{t_{1}}{t_{1} - t_{2}}})}