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Question-100450




Question Number 100450 by 175 last updated on 26/Jun/20
Answered by maths mind last updated on 26/Jun/20
∫_0 ^1 (((1−x)^a )/( (√x).(√(1−x)).(√(1+x))))dx=f(a)  =∫_0 ^1 x^(−(1/2)) (1−x)^(a−(1/2)) (1+x)^(−(1/2)) dx  ∫_0 ^1 x^((1/2)−1) (1−x)^(a+1−(1/2)−1) .(1−(−1)x)^(−(1/2)) dx=f(a)  f(a)=β((1/2),a+(1/2))._2 F_1 ((1/2),(1/2);a+1;−1)  we can see f′(a)=∫_0 ^1 ((ln(1−x)(1−x)^a )/( (√(x.))(√(1−x^2 ))))dx  we want f′(0)  ∂_a β((1/2),a+(1/2))=β((1/2),a+(1/2))(Ψ((1/2))−Ψ(a+1))  too bee continued
$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(\mathrm{1}−{x}\right)^{{a}} }{\:\sqrt{{x}}.\sqrt{\mathrm{1}−{x}}.\sqrt{\mathrm{1}+{x}}}{dx}={f}\left({a}\right) \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{−\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}−{x}\right)^{{a}−\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}+{x}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} {dx} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} \left(\mathrm{1}−{x}\right)^{{a}+\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} .\left(\mathrm{1}−\left(−\mathrm{1}\right){x}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} {dx}={f}\left({a}\right) \\ $$$${f}\left({a}\right)=\beta\left(\frac{\mathrm{1}}{\mathrm{2}},{a}+\frac{\mathrm{1}}{\mathrm{2}}\right)._{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{2}};{a}+\mathrm{1};−\mathrm{1}\right) \\ $$$${we}\:{can}\:{see}\:{f}'\left({a}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{x}\right)\left(\mathrm{1}−{x}\right)^{{a}} }{\:\sqrt{{x}.}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{dx} \\ $$$${we}\:{want}\:{f}'\left(\mathrm{0}\right) \\ $$$$\partial_{{a}} \beta\left(\frac{\mathrm{1}}{\mathrm{2}},{a}+\frac{\mathrm{1}}{\mathrm{2}}\right)=\beta\left(\frac{\mathrm{1}}{\mathrm{2}},{a}+\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\Psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)−\Psi\left({a}+\mathrm{1}\right)\right) \\ $$$${too}\:{bee}\:{continued} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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