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Question Number 132741 by mnjuly1970 last updated on 16/Feb/21

...nice calculus... ∫_(0^( ) ) ^( 1) (dx/((x−2)(x^2 (1−x)^3 )^(1/5) )) =?

$$\:\:\:\:\:\:\:\:\:\:\:\:\:...{nice}\:\:\:\:{calculus}... \\ $$$$\:\:\int_{\mathrm{0}^{\:\:\:\:} \:\:} ^{\:\mathrm{1}} \frac{{dx}}{\left({x}−\mathrm{2}\right)\left({x}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)^{\mathrm{3}} \right)^{\frac{\mathrm{1}}{\mathrm{5}}} }\:=? \\ $$

Commented by MJS_new last updated on 16/Feb/21

I get −2^(1/10) π(√(1−((√5)/5)))≈−2.50340750 using the substitution t=(((2(1−x))/x))^(1/5)

$$\mathrm{I}\:\mathrm{get}\:−\mathrm{2}^{\mathrm{1}/\mathrm{10}} \pi\sqrt{\mathrm{1}−\frac{\sqrt{\mathrm{5}}}{\mathrm{5}}}\approx−\mathrm{2}.\mathrm{50340750} \\ $$$$\mathrm{using}\:\mathrm{the}\:\mathrm{substitution}\:{t}=\left(\frac{\mathrm{2}\left(\mathrm{1}−{x}\right)}{{x}}\right)^{\mathrm{1}/\mathrm{5}} \\ $$

Commented by MJS_new last updated on 16/Feb/21

yes of course. forgot the “−”

$$\mathrm{yes}\:\mathrm{of}\:\mathrm{course}.\:\mathrm{forgot}\:\mathrm{the}\:``−'' \\ $$

Commented by Dwaipayan Shikari last updated on 16/Feb/21

Yes sir!But i think the result should be −ve

$${Yes}\:{sir}!{But}\:{i}\:{think}\:{the}\:{result}\:{should}\:{be}\:−{ve} \\ $$

Commented by mnjuly1970 last updated on 16/Feb/21

bravo mr Mj thank you so much...

$${bravo}\:\:{mr}\:{Mj} \\ $$$${thank}\:{you}\:{so}\:{much}... \\ $$

Answered by Dwaipayan Shikari last updated on 16/Feb/21

−∫_0 ^1 t^(−(3/5)) (1−t)^(−(2/5)) (1+t)^(−1) dt 1−x=t _2 F_1 (a,b;c;z)=((Γ(c))/(Γ(c−b)Γ(b)))∫_0 ^1 x^(b−1) (1−x)^(c−b−1) (1−zx)^(−a) dx b−1=−(3/5)⇒b=(2/5) c−b−1=((−2)/5)⇒c=1 a=1 z=−1 −Γ((3/5))Γ((2/5)) _2 F_1 (1,(2/5),1,−1)=−∫_0 ^1 t^(−(3/5)) (1−t)^(−(2/5)) (1+t)^(−1) dt =−((2(√2)π)/( (√(5−(√5))))) _2 F_1 (1,(2/5),1;−1)

$$−\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{−\frac{\mathrm{3}}{\mathrm{5}}} \left(\mathrm{1}−{t}\right)^{−\frac{\mathrm{2}}{\mathrm{5}}} \left(\mathrm{1}+{t}\right)^{−\mathrm{1}} {dt}\:\:\:\:\:\:\mathrm{1}−{x}={t} \\ $$$$\:_{\mathrm{2}} {F}_{\mathrm{1}} \left({a},{b};{c};{z}\right)=\frac{\Gamma\left({c}\right)}{\Gamma\left({c}−{b}\right)\Gamma\left({b}\right)}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{b}−\mathrm{1}} \left(\mathrm{1}−{x}\right)^{{c}−{b}−\mathrm{1}} \left(\mathrm{1}−{zx}\right)^{−{a}} {dx} \\ $$$${b}−\mathrm{1}=−\frac{\mathrm{3}}{\mathrm{5}}\Rightarrow{b}=\frac{\mathrm{2}}{\mathrm{5}} \\ $$$${c}−{b}−\mathrm{1}=\frac{−\mathrm{2}}{\mathrm{5}}\Rightarrow{c}=\mathrm{1} \\ $$$${a}=\mathrm{1}\:\:\:{z}=−\mathrm{1} \\ $$$$−\Gamma\left(\frac{\mathrm{3}}{\mathrm{5}}\right)\Gamma\left(\frac{\mathrm{2}}{\mathrm{5}}\right)\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\mathrm{1},\frac{\mathrm{2}}{\mathrm{5}},\mathrm{1},−\mathrm{1}\right)=−\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{−\frac{\mathrm{3}}{\mathrm{5}}} \left(\mathrm{1}−{t}\right)^{−\frac{\mathrm{2}}{\mathrm{5}}} \left(\mathrm{1}+{t}\right)^{−\mathrm{1}} {dt} \\ $$$$=−\frac{\mathrm{2}\sqrt{\mathrm{2}}\pi}{\:\sqrt{\mathrm{5}−\sqrt{\mathrm{5}}}}\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\mathrm{1},\frac{\mathrm{2}}{\mathrm{5}},\mathrm{1};−\mathrm{1}\right) \\ $$

Commented by mnjuly1970 last updated on 16/Feb/21

mercey grateful... hypergeometrey function...

$${mercey}\: \\ $$$${grateful}... \\ $$$$\:{hypergeometrey}\:{function}... \\ $$

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