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.


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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
$D_g ={ a∈R/ g(a)∈R} we know that ∫_0 ^1 (dx/((√(1−u^2 )) ))= (π/2) ( so it converges). Then lim_(x→1) (((√(1−u^2 )) )/(√((1−u)(1+au)))) =(√(2/(1+a))) by convergence theorem g(a) exist if (√((2/(1+a)) )) ∈ R ⇔ a>−1 So D_g = ]−1;+∞[ If a=0 g(0)=∫_0 ^1 (( dx)/((√(1−x)) ))=[−2(√(1−x)) ]_0 ^1 =2 Now if a≠0 and a∈]−1;+∞[ g(a)=∫_0 ^1 (du/(√(u(1+a−au)))) ( with u=1−x ) = (1/(√(1+a))) ∫_0 ^1 u^((−1)/2) (1−((au)/(1+a)))^((−1)/2) du=(1/(√(1+a))) ∫_0 ^1 u^((1/2)−1) (1−u)^((3/2)−(1/2)−1) (1−(a/(1+a))u)^((−1)/2) du = ((Γ((3/2)))/(Γ((1/2))Γ(1)(√(1+a)) )) . _2 F_1 ((1/2),(1/2),(3/2), (a/(1+a))) because ∣(a/(1+a))∣<1 using the formula _2 F_1 ((1/2),(1/2),(3/2),z^2 )=((arcsinz)/z) if a∈]−1;0[ , (a/(1+a))<0 let take z=i(√(∣(a/(1+a))∣)) g(a)=(1/(2(√(1+a)) )) . (1/(i(√(∣(a/(1+a))∣)) )) arcsin(i(√(∣(a/(1+a))∣)) )=_ ((arcsin(i(√((∣a∣)/(1+a))) ))/(2i(√(∣a∣)) )) =((argsh((√(((∣a∣)/(1+a)) )) ))/(2(√(∣a∣)) )) because sin(x)=((e^(ix) −e^(−ix) )/(2i))= ((sh(ix))/i) ⇒ sin( arcsin(ix))=((sh(iarcsin(ix)))/i) ⇒ iargshx=arcsin(ix) if a>0 then (a/(1+a))>0 we take z=(√(a/(1+a))) g(a)=((arcsin((√((a/(1+a)) )) ))/(2(√a)))$
$D_{g} = {a \in R / g (a) \in R}$ $w e k n o w t h a t \int_{0}^{1} \frac{d x}{\sqrt{1 - u^{2}}} = \frac{π}{2} (s o i t c o n v e r g e s) . T h e n \lim_{x \to 1} \frac{\sqrt{1 - u^{2}}}{\sqrt{(1 - u) (1 + a u)}} = \sqrt{\frac{2}{1 + a}}$ $b y c o n v e r g e n c e t h e o r e m g (a) e x i s t i f \sqrt{\frac{2}{1 + a}} \in R \Leftrightarrow a > - 1$ $S o D_{g} =] - 1; + \infty [$ $I f a = 0 g (0) = \int_{0}^{1} \frac{d x}{\sqrt{1 - x}} = {[- 2 \sqrt{1 - x}]}_{0}^{1} = 2$ $N o w i f a \neq 0 a n d a \in] - 1; + \infty [$ $g (a) = \int_{0}^{1} \frac{d u}{\sqrt{u (1 + a - a u)}} (w i t h u = 1 - x)$ $= \frac{1}{\sqrt{1 + a}} \int_{0}^{1} u^{\frac{- 1}{2}} {(1 - \frac{a u}{1 + a})}^{\frac{- 1}{2}} d u = \frac{1}{\sqrt{1 + a}} \int_{0}^{1} u^{\frac{1}{2} - 1} {(1 - u)}^{\frac{3}{2} - \frac{1}{2} - 1} {(1 - \frac{a}{1 + a} u)}^{\frac{- 1}{2}} d u$ $= \frac{Γ (\frac{3}{2})}{Γ (\frac{1}{2}) Γ (1) \sqrt{1 + a}} ._{2} F_{1} (\frac{1}{2}, \frac{1}{2}, \frac{3}{2}, \frac{a}{1 + a}) b e c a u s e ∣ \frac{a}{1 + a} ∣< 1$ $u s i n g t h e f o r m u l a_{2} F_{1} (\frac{1}{2}, \frac{1}{2}, \frac{3}{2}, z^{2}) = \frac{a r c s i n z}{z}$ $i f a \in] - 1; 0 [, \frac{a}{1 + a} < 0 l e t t a k e z = i \sqrt{∣ \frac{a}{1 + a} ∣}$ $g (a) = \frac{1}{2 \sqrt{1 + a}} . \frac{1}{i \sqrt{∣ \frac{a}{1 + a} ∣}} a r c s i n (i \sqrt{∣ \frac{a}{1 + a} ∣}) =_{} \frac{a r c s i n (i \sqrt{\frac{∣ a ∣}{1 + a}})}{2 i \sqrt{∣ a ∣}} = \frac{a r g s h (\sqrt{\frac{∣ a ∣}{1 + a}})}{2 \sqrt{∣ a ∣}} b e c a u s e s i n (x) = \frac{e^{i x} - e^{- i x}}{2 i} = \frac{s h (i x)}{i} \Rightarrow s i n (a r c s i n (i x)) = \frac{s h (i a r c s i n (i x))}{i} \Rightarrow i a r g s h x = a r c s i n (i x)$ $i f a > 0 t h e n \frac{a}{1 + a} > 0 w e t a k e z = \sqrt{\frac{a}{1 + a}}$ $g (a) = \frac{a r c s i n (\sqrt{\frac{a}{1 + a}})}{2 \sqrt{a}}$
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

$2) w e h a v e g (a) = \int_{0}^{1} \frac{d x}{\sqrt{(1 - x) (1 + a x)}}$ $c h a n g r m e n t \sqrt{1 - x} = t g i v e 1 - x = t^{2} \Rightarrow x = 1 - t^{2}$ $a > - 1 \Rightarrow a x > - x \Rightarrow 1 + a x > 1 - x > 0$ $g (a) = - \int_{0}^{1} \frac{(- 2 t) d t}{t \sqrt{1 + a (1 - t^{2})}} = 2 \int_{0}^{1} \frac{d t}{\sqrt{- {a t}^{2} + a + 1}}$ $- {a t}^{2} + a + 1 = 0 \Rightarrow {a t}^{2} - (a + 1) = 0$ $Δ^{^{'}} = 0 + a (a + 1) = a^{2} + a = a (a + 1)$ $c a s e - 1 < a < 0 \Rightarrow Δ^{^{'}} < 0 \Rightarrow n o r e a l r o o t s$ $a n d - {a t}^{2} + a + 1 = - a (t^{2} - \frac{a + 1}{a}) a n d - \frac{a + 1}{a} > 0 \Rightarrow$ $w e d o t h e c h a n g e m e n t t = \sqrt{- \frac{a + 1}{a}} u \Rightarrow$ $g (a) = 2 \int_{0}^{\sqrt{- \frac{a}{a + 1}}} \frac{1}{\sqrt{- a} \sqrt{- \frac{a + 1}{a}} (1 + u^{2})} \sqrt{- \frac{a + 1}{a}} d u$ $= \frac{2}{\sqrt{- a}} \int_{0}^{\sqrt{- \frac{a}{a + 1}}} \frac{d u}{1 + u^{2}} = \frac{2}{\sqrt{- a}} a r c t a n (\sqrt{- \frac{a}{a + 1}})$ $. . . . b e c o n t i n u e d . . .$
Commented by mathmax by abdo last updated on 28/Aug/19
$case 2 if a>0 ⇒Δ^′ >0 ⇒t_1 =((√(a^2 +a))/a) and t_2 =−((√(a^2 +a))/a) g(a) = 2∫_0 ^1 (dt/(√(−a(t−t_1 )(t−t_2 )))) =(2/(√a)) ∫_0 ^1 (dt/(√((t_1 −t)(t−t_2 )))) we use the changement (√(t_1 −t))=u ⇒t_1 −t =u^2 ⇒t =t_1 −u^2 g(a) =(2/(√a)) ∫_(√t_1 ) ^(√(t_1 −1)) (((−2u)du)/(u(√(t_1 −u^2 −t_2 )))) =((−4)/(√a)) ∫_(√t_1 ) ^(√(t_1 −1)) (du/(√(t_1 −t_2 −u^2 ))) =((−4)/(√a)) ∫_(√t_1 ) ^(√(t_1 −1)) (du/(√(((√(t_1 −t_2 )))^2 −u^2 ))) =_(u=(√(t_1 −t_2 ))z) ((−4)/(√a)) ∫_((√t_1 )/(√(t_1 −t_2 ))) ^((√(t_1 −1))/(√(t_1 −t_2 ))) (((√(t_1 −t_2 )) dz)/((√(t_1 −t_2 ))((√(1−z^2 ))))) =((−4)/(√a))[arcsin(z)]_(√(t_1 /(t_1 −t_2 ))) ^(√((t_1 −1)/(t_1 −t_2 ))) =((−4)/(√a)){ arcsin((√((t_1 −1)/(t_1 −t_2 ))))−arcsin((√(t_1 /(t_1 −t_2 ))))}$
$c a s e 2 i f a > 0 \Rightarrow Δ^{^{'}} > 0 \Rightarrow t_{1} = \frac{\sqrt{a^{2} + a}}{a} a n d t_{2} = - \frac{\sqrt{a^{2} + a}}{a}$ $g (a) = 2 \int_{0}^{1} \frac{d t}{\sqrt{- a (t - t_{1}) (t - t_{2})}} = \frac{2}{\sqrt{a}} \int_{0}^{1} \frac{d t}{\sqrt{(t_{1} - t) (t - t_{2})}}$ $w e u s e t h e c h a n g e m e n t \sqrt{t_{1} - t} = u \Rightarrow t_{1} - t = u^{2} \Rightarrow t = t_{1} - u^{2}$ $g (a) = \frac{2}{\sqrt{a}} \int_{\sqrt{t_{1}}}^{\sqrt{t_{1} - 1}} \frac{(- 2 u) d u}{u \sqrt{t_{1} - u^{2} - t_{2}}} = \frac{- 4}{\sqrt{a}} \int_{\sqrt{t_{1}}}^{\sqrt{t_{1} - 1}} \frac{d u}{\sqrt{t_{1} - t_{2} - u^{2}}}$ $= \frac{- 4}{\sqrt{a}} \int_{\sqrt{t_{1}}}^{\sqrt{t_{1} - 1}} \frac{d u}{\sqrt{{(\sqrt{t_{1} - t_{2}})}^{2} - u^{2}}}$ $=_{u = \sqrt{t_{1} - t_{2}} z} \frac{- 4}{\sqrt{a}} \int_{\frac{\sqrt{t_{1}}}{\sqrt{t_{1} - t_{2}}}}^{\frac{\sqrt{t_{1} - 1}}{\sqrt{t_{1} - t_{2}}}} \frac{\sqrt{t_{1} - t_{2}} d z}{\sqrt{t_{1} - t_{2}} (\sqrt{1 - z^{2}})}$ $= \frac{- 4}{\sqrt{a}} {[a r c s i n (z)]}_{\sqrt{\frac{t_{1}}{t_{1} - t_{2}}}}^{\sqrt{\frac{t_{1} - 1}{t_{1} - t_{2}}}} = \frac{- 4}{\sqrt{a}} {a r c s i n (\sqrt{\frac{t_{1} - 1}{t_{1} - t_{2}}}) - a r c s i n (\sqrt{\frac{t_{1}}{t_{1} - t_{2}}})}$
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