Question Number 67465 by ~ À ® @ 237 ~ last updated on 27/Aug/19
$$ \\ $$$$ \\ $$$$\:\:{let}\:{consider}\:{a}\:{function}\:{g}\:{defined}\:{by}\:\:\:{g}\left({a}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{dx}}{\sqrt{\left(\mathrm{1}−{x}\right)\left(\mathrm{1}+{ax}\right)}}\:\: \\ $$$${Give}\:{the}\:{defined}\:{Domain}\:{of}\:{g}\:\:{and}\:{simplify}\:{g}. \\ $$
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)))](Q67491.png)
$$ \\ $$$$ \\ $$$${D}_{{g}} =\left\{\:{a}\in\mathbb{R}/\:{g}\left({a}\right)\in\mathbb{R}\right\} \\ $$$${we}\:{know}\:{that}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dx}}{\sqrt{\mathrm{1}−{u}^{\mathrm{2}} }\:}=\:\frac{\pi}{\mathrm{2}}\:\left(\:{so}\:\:{it}\:{converges}\right).\:{Then}\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{1}−{u}^{\mathrm{2}} }\:}{\sqrt{\left(\mathrm{1}−{u}\right)\left(\mathrm{1}+{au}\right)}}\:=\sqrt{\frac{\mathrm{2}}{\mathrm{1}+{a}}}\:\: \\ $$$${by}\:{convergence}\:{theorem}\:\:{g}\left({a}\right)\:{exist}\:{if}\:\:\sqrt{\frac{\mathrm{2}}{\mathrm{1}+{a}}\:\:}\:\in\:\mathbb{R}\:\:\Leftrightarrow\:\:{a}>−\mathrm{1} \\ $$$$\left.{So}\:{D}_{{g}} =\:\right]−\mathrm{1};+\infty\left[\right. \\ $$$${If}\:\:{a}=\mathrm{0}\:\:\:{g}\left(\mathrm{0}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\:{dx}}{\sqrt{\mathrm{1}−{x}}\:}=\left[−\mathrm{2}\sqrt{\mathrm{1}−{x}}\:\right]_{\mathrm{0}} ^{\mathrm{1}} =\mathrm{2} \\ $$$$\left.{Now}\:{if}\:\:{a}\neq\mathrm{0}\:\:{and}\:{a}\in\right]−\mathrm{1};+\infty\left[\right. \\ $$$${g}\left({a}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\frac{{du}}{\sqrt{{u}\left(\mathrm{1}+{a}−{au}\right)}}\:\:\:\left(\:{with}\:\:{u}=\mathrm{1}−{x}\:\right) \\ $$$$\:\:\:\:\:\:\:=\:\frac{\mathrm{1}}{\sqrt{\mathrm{1}+{a}}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{u}^{\frac{−\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}−\frac{{au}}{\mathrm{1}+{a}}\right)^{\frac{−\mathrm{1}}{\mathrm{2}}} {du}=\frac{\mathrm{1}}{\sqrt{\mathrm{1}+{a}}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{u}^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} \left(\mathrm{1}−{u}\right)^{\frac{\mathrm{3}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} \left(\mathrm{1}−\frac{{a}}{\mathrm{1}+{a}}{u}\right)^{\frac{−\mathrm{1}}{\mathrm{2}}} {du} \\ $$$$\:\:\:\:\:\:\:=\:\frac{\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)}{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\mathrm{1}\right)\sqrt{\mathrm{1}+{a}}\:}\:.\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{3}}{\mathrm{2}},\:\frac{{a}}{\mathrm{1}+{a}}\right)\:\:\:\:\:\:{because}\:\:\mid\frac{{a}}{\mathrm{1}+{a}}\mid<\mathrm{1} \\ $$$${using}\:{the}\:{formula}\:\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{3}}{\mathrm{2}},{z}^{\mathrm{2}} \right)=\frac{{arcsinz}}{{z}}\:\: \\ $$$$\left.{if}\:\:\:{a}\in\right]−\mathrm{1};\mathrm{0}\left[\:\:\:\:\:,\:\:\frac{{a}}{\mathrm{1}+{a}}<\mathrm{0}\:\:\:\:\:{let}\:\:{take}\:{z}={i}\sqrt{\mid\frac{{a}}{\mathrm{1}+{a}}\mid}\:\:\right. \\ $$$${g}\left({a}\right)=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{1}+{a}}\:}\:.\:\frac{\mathrm{1}}{{i}\sqrt{\mid\frac{{a}}{\mathrm{1}+{a}}\mid}\:}\:{arcsin}\left({i}\sqrt{\mid\frac{{a}}{\mathrm{1}+{a}}\mid}\:\right)=_{\:} \frac{{arcsin}\left({i}\sqrt{\frac{\mid{a}\mid}{\mathrm{1}+{a}}}\:\right)}{\mathrm{2}{i}\sqrt{\mid{a}\mid}\:}\:\:=\frac{{argsh}\left(\sqrt{\frac{\mid{a}\mid}{\mathrm{1}+{a}}\:}\:\right)}{\mathrm{2}\sqrt{\mid{a}\mid}\:}\:\:\:\:\:{because}\:\:{sin}\left({x}\right)=\frac{{e}^{{ix}} −{e}^{−{ix}} }{\mathrm{2}{i}}=\:\frac{{sh}\left({ix}\right)}{{i}}\:\Rightarrow\:{sin}\left(\:{arcsin}\left({ix}\right)\right)=\frac{{sh}\left({iarcsin}\left({ix}\right)\right)}{{i}}\:\Rightarrow\:{iargshx}={arcsin}\left({ix}\right) \\ $$$${if}\:\:{a}>\mathrm{0}\:\:\:{then}\:\:\frac{{a}}{\mathrm{1}+{a}}>\mathrm{0}\:\:\:{we}\:\:{take}\:\:{z}=\sqrt{\frac{{a}}{\mathrm{1}+{a}}}\: \\ $$$${g}\left({a}\right)=\frac{{arcsin}\left(\sqrt{\frac{{a}}{\mathrm{1}+{a}}\:\:}\:\right)}{\mathrm{2}\sqrt{{a}}} \\ $$$$ \\ $$$$ \\ $$
Commented by prof Abdo imad last updated on 28/Aug/19
![at V(1) g(a) ∼ ∫_0 ^1 (dx/(√((1−x)(1+a)))) g(a) exist ⇔ 1+a>0 ⇒a>−1 ⇒ D_g =]−1,+∞[ .](Q67494.png)
$${at}\:{V}\left(\mathrm{1}\right)\:\:{g}\left({a}\right)\:\sim\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dx}}{\sqrt{\left(\mathrm{1}−{x}\right)\left(\mathrm{1}+{a}\right)}} \\ $$$${g}\left({a}\right)\:{exist}\:\Leftrightarrow\:\mathrm{1}+{a}>\mathrm{0}\:\Rightarrow{a}>−\mathrm{1}\:\Rightarrow \\ $$$$\left.{D}_{{g}} =\right]−\mathrm{1},+\infty\left[\:.\right. \\ $$
Commented by prof Abdo imad last updated on 28/Aug/19
$$\left.\mathrm{2}\right)\:{we}\:{have}\:{g}\left({a}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dx}}{\sqrt{\left(\mathrm{1}−{x}\right)\left(\mathrm{1}+{ax}\right)}} \\ $$$${changrment}\:\sqrt{\mathrm{1}−{x}}={t}\:{give}\:\mathrm{1}−{x}\:={t}^{\mathrm{2}} \:\Rightarrow{x}=\mathrm{1}−{t}^{\mathrm{2}} \\ $$$${a}>−\mathrm{1}\:\Rightarrow{ax}>−{x}\:\Rightarrow\mathrm{1}+{ax}>\mathrm{1}−{x}>\mathrm{0} \\ $$$${g}\left({a}\right)\:=−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\frac{\left(−\mathrm{2}\boldsymbol{{t}}\right)\boldsymbol{{dt}}}{\boldsymbol{{t}}\sqrt{\mathrm{1}+\boldsymbol{{a}}\left(\mathrm{1}−\boldsymbol{{t}}^{\mathrm{2}} \right)}}\:\:=\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dt}}{\sqrt{−{at}^{\mathrm{2}} \:+{a}+\mathrm{1}}} \\ $$$$−{at}^{\mathrm{2}} \:+{a}+\mathrm{1}\:=\mathrm{0}\:\Rightarrow{at}^{\mathrm{2}} −\left({a}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\Delta^{'} \:=\mathrm{0}+{a}\left({a}+\mathrm{1}\right)\:={a}^{\mathrm{2}} \:+{a}\:={a}\left({a}+\mathrm{1}\right) \\ $$$${case}\:−\mathrm{1}<\:{a}<\mathrm{0}\:\Rightarrow\Delta^{'} <\mathrm{0}\:\Rightarrow{no}\:{real}\:{roots} \\ $$$${and}\:\:−{at}^{\mathrm{2}} \:+{a}+\mathrm{1}=−{a}\left({t}^{\mathrm{2}} −\frac{{a}+\mathrm{1}}{{a}}\right)\:{and}\:−\frac{{a}+\mathrm{1}}{{a}}>\mathrm{0}\:\Rightarrow \\ $$$${we}\:{do}\:{the}\:{changement}\:\:{t}\:=\sqrt{−\frac{{a}+\mathrm{1}}{{a}}}{u}\:\Rightarrow \\ $$$${g}\left({a}\right)\:=\:\mathrm{2}\:\int_{\mathrm{0}} ^{\sqrt{−\frac{{a}}{{a}+\mathrm{1}}}} \:\:\:\:\:\:\:\frac{\mathrm{1}}{\sqrt{−{a}}\sqrt{−\frac{{a}+\mathrm{1}}{{a}}}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}\sqrt{−\frac{{a}+\mathrm{1}}{{a}}}{du} \\ $$$$=\frac{\mathrm{2}}{\sqrt{−{a}}}\:\int_{\mathrm{0}} ^{\sqrt{−\frac{{a}}{{a}+\mathrm{1}}}} \:\:\:\:\:\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} }\:=\frac{\mathrm{2}}{\sqrt{−{a}}}\:{arctan}\left(\sqrt{−\frac{{a}}{{a}+\mathrm{1}}}\right) \\ $$$$....{be}\:{continued}... \\ $$$$ \\ $$
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 ))))}](Q67550.png)
$${case}\:\mathrm{2}\:\:{if}\:{a}>\mathrm{0}\:\:\Rightarrow\Delta^{'} >\mathrm{0}\:\:\Rightarrow{t}_{\mathrm{1}} =\frac{\sqrt{{a}^{\mathrm{2}} +{a}}}{{a}}\:\:\:{and}\:{t}_{\mathrm{2}} =−\frac{\sqrt{{a}^{\mathrm{2}} +{a}}}{{a}} \\ $$$${g}\left({a}\right)\:=\:\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dt}}{\sqrt{−{a}\left({t}−{t}_{\mathrm{1}} \right)\left({t}−{t}_{\mathrm{2}} \right)}}\:=\frac{\mathrm{2}}{\sqrt{{a}}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dt}}{\sqrt{\left({t}_{\mathrm{1}} −{t}\right)\left({t}−{t}_{\mathrm{2}} \right)}} \\ $$$${we}\:{use}\:{the}\:{changement}\:\:\sqrt{{t}_{\mathrm{1}} −{t}}={u}\:\Rightarrow{t}_{\mathrm{1}} −{t}\:={u}^{\mathrm{2}} \:\Rightarrow{t}\:={t}_{\mathrm{1}} −{u}^{\mathrm{2}} \\ $$$${g}\left({a}\right)\:=\frac{\mathrm{2}}{\sqrt{{a}}}\:\int_{\sqrt{{t}_{\mathrm{1}} }} ^{\sqrt{{t}_{\mathrm{1}} −\mathrm{1}}} \:\:\:\:\:\:\:\:\frac{\left(−\mathrm{2}{u}\right){du}}{{u}\sqrt{{t}_{\mathrm{1}} −{u}^{\mathrm{2}} −{t}_{\mathrm{2}} }}\:=\frac{−\mathrm{4}}{\sqrt{{a}}}\:\int_{\sqrt{{t}_{\mathrm{1}} }} ^{\sqrt{{t}_{\mathrm{1}} −\mathrm{1}}} \:\:\frac{{du}}{\sqrt{{t}_{\mathrm{1}} −{t}_{\mathrm{2}} −{u}^{\mathrm{2}} }} \\ $$$$=\frac{−\mathrm{4}}{\sqrt{{a}}}\:\int_{\sqrt{{t}_{\mathrm{1}} }} ^{\sqrt{{t}_{\mathrm{1}} −\mathrm{1}}} \:\:\:\:\:\:\frac{{du}}{\sqrt{\left(\sqrt{{t}_{\mathrm{1}} −{t}_{\mathrm{2}} }\right)^{\mathrm{2}} −{u}^{\mathrm{2}} }} \\ $$$$=_{{u}=\sqrt{{t}_{\mathrm{1}} −{t}_{\mathrm{2}} }{z}} \:\:\:\:\:\:\frac{−\mathrm{4}}{\sqrt{{a}}}\:\:\int_{\frac{\sqrt{{t}_{\mathrm{1}} }}{\sqrt{{t}_{\mathrm{1}} −{t}_{\mathrm{2}} }}} ^{\frac{\sqrt{{t}_{\mathrm{1}} −\mathrm{1}}}{\sqrt{{t}_{\mathrm{1}} −{t}_{\mathrm{2}} }}} \:\:\:\:\:\:\:\frac{\sqrt{{t}_{\mathrm{1}} −{t}_{\mathrm{2}} }\:{dz}}{\sqrt{{t}_{\mathrm{1}} −{t}_{\mathrm{2}} }\left(\sqrt{\mathrm{1}−{z}^{\mathrm{2}} }\right)} \\ $$$$=\frac{−\mathrm{4}}{\sqrt{{a}}}\left[{arcsin}\left({z}\right)\right]_{\sqrt{\frac{{t}_{\mathrm{1}} }{{t}_{\mathrm{1}} −{t}_{\mathrm{2}} }}} ^{\sqrt{\frac{{t}_{\mathrm{1}} −\mathrm{1}}{{t}_{\mathrm{1}} −{t}_{\mathrm{2}} }}} \:\:\:\:=\frac{−\mathrm{4}}{\sqrt{{a}}}\left\{\:{arcsin}\left(\sqrt{\frac{{t}_{\mathrm{1}} −\mathrm{1}}{{t}_{\mathrm{1}} −{t}_{\mathrm{2}} }}\right)−{arcsin}\left(\sqrt{\frac{{t}_{\mathrm{1}} }{{t}_{\mathrm{1}} −{t}_{\mathrm{2}} }}\right)\right\} \\ $$