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Question Number 129988 by mnjuly1970 last updated on 21/Jan/21

... nice calculus ... evaluate: φ = ∫_0 ^( 1) (dx/((x−x^3 )^(1/2) )) =?

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...\:{nice}\:\:{calculus}\:... \\ $$$$\:\:{evaluate}:\: \\ $$$$\:\:\:\phi\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{dx}}{\left({x}−{x}^{\mathrm{3}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} }\:=? \\ $$$$ \\ $$

Answered by Dwaipayan Shikari last updated on 21/Jan/21

∫_0 ^1 (1/( (√x).(√(1+x^2 )))) x^2 =u⇒1=(1/(2x)).(du/dx) =(1/2)∫_0 ^1 u^(−(3/4)) (1+u)^(−(1/2)) (1−u)^0 du _2 F_1 (a,b;c;−1)=((Γ(c))/(Γ(c−b)Γ(b)))∫_0 ^1 t^(b−1) (1−t)^(c−b−1) (1+t)^(−a) dt c−b=1 (Here) a=(1/2) b=(1/4) c=(5/4) So the integral is 4 _2 F_1 ((1/2),(1/4);(5/4),−1)

$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt{{x}}.\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:\:\:\:\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} ={u}\Rightarrow\mathrm{1}=\frac{\mathrm{1}}{\mathrm{2}{x}}.\frac{{du}}{{dx}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} {u}^{−\frac{\mathrm{3}}{\mathrm{4}}} \left(\mathrm{1}+{u}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}−{u}\right)^{\mathrm{0}} {du} \\ $$$$\:\:_{\mathrm{2}} {F}_{\mathrm{1}} \left({a},{b};{c};−\mathrm{1}\right)=\frac{\Gamma\left({c}\right)}{\Gamma\left({c}−{b}\right)\Gamma\left({b}\right)}\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{{b}−\mathrm{1}} \left(\mathrm{1}−{t}\right)^{{c}−{b}−\mathrm{1}} \left(\mathrm{1}+{t}\right)^{−{a}} {dt} \\ $$$${c}−{b}=\mathrm{1}\:\:\:\:\left({Here}\right) \\ $$$${a}=\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:{b}=\frac{\mathrm{1}}{\mathrm{4}}\:\:\:\:\:\:{c}=\frac{\mathrm{5}}{\mathrm{4}} \\ $$$${So}\:\:{the}\:{integral}\:{is}\: \\ $$$$\mathrm{4}\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{4}};\frac{\mathrm{5}}{\mathrm{4}},−\mathrm{1}\right) \\ $$

Answered by Dwaipayan Shikari last updated on 21/Jan/21

∫_0 ^1 (1/((x−x^3 )^(1/2) ))dx=(1/2)∫_0 ^1 u^(−(3/4)) (1−u)^(−(1/2)) du=((Γ((1/4))Γ((1/2)))/(2Γ((3/4)))) =((√(π^3 /2))/(Γ^2 ((3/4))))

$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\left({x}−{x}^{\mathrm{3}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} }{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} {u}^{−\frac{\mathrm{3}}{\mathrm{4}}} \left(\mathrm{1}−{u}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} {du}=\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)} \\ $$$$=\frac{\sqrt{\frac{\pi^{\mathrm{3}} }{\mathrm{2}}}}{\Gamma^{\mathrm{2}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)} \\ $$

Commented by mnjuly1970 last updated on 22/Jan/21

thanks alot...

$${thanks}\:{alot}... \\ $$

Answered by mathmax by abdo last updated on 21/Jan/21

Φ=∫_0 ^1 (dx/((x−x^3 )^(1/2) )) ⇒Φ =∫_0 ^1 (dx/( (√x)(1−x^2 )^(1/2) ))dx =_((√x)=t) ∫_0 ^1 ((2tdt)/(t(1−t^4 )^(1/2) )) =2∫_0 ^1 (dt/((1−t^4 )^(1/2) )) =_(t=u^(1/4) ) 2×(1/4)∫_0 ^1 (1−u)^(−(1/2)) u^((1/4)−1) du =(1/2)∫_0 ^1 u^((1/4)−1) (1−u)^((1/2)−1) du=(1/2)B((1/4),(1/2)) =(1/2)((Γ((1/4)).Γ((1/2)))/(Γ((1/4)+(1/2))))=((√π)/2)((Γ((1/4)))/(Γ((3/4))))

$$\Phi=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{dx}}{\left(\mathrm{x}−\mathrm{x}^{\mathrm{3}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} }\:\Rightarrow\Phi\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{dx}}{\:\sqrt{\mathrm{x}}\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} }\mathrm{dx}\:=_{\sqrt{\mathrm{x}}=\mathrm{t}} \:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{2tdt}}{\mathrm{t}\left(\mathrm{1}−\mathrm{t}^{\mathrm{4}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} } \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{dt}}{\left(\mathrm{1}−\mathrm{t}^{\mathrm{4}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} }\:=_{\mathrm{t}=\mathrm{u}^{\frac{\mathrm{1}}{\mathrm{4}}} } \:\:\:\mathrm{2}×\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\left(\mathrm{1}−\mathrm{u}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \:\mathrm{u}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} \:\mathrm{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{u}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} \left(\mathrm{1}−\mathrm{u}\right)^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} \:\mathrm{du}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{B}\left(\frac{\mathrm{1}}{\mathrm{4}},\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right).\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\right)}=\frac{\sqrt{\pi}}{\mathrm{2}}\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)}{\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)} \\ $$$$ \\ $$

Commented by mnjuly1970 last updated on 22/Jan/21

grateful sir max ...

$${grateful}\:{sir}\:{max}\:... \\ $$

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