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Question Number 100450 by 175 last updated on 26/Jun/20

Answered by maths mind last updated on 26/Jun/20

$${\int }_0^1\frac{{(1-x)}^a}{\sqrt{x}.\sqrt{1-x}.\sqrt{1+x}}dx=f\left(a\right)$$$$={\int }_0^1{x}^{-\frac{1}{2}}{(1-x)}^{a-\frac{1}{2}}{(1+x)}^{-\frac{1}{2}}dx$$$${\int }_0^1{x}^{\frac{1}{2}-1}{(1-x)}^{a+1-\frac{1}{2}-1}.{(1-(-1)x)}^{-\frac{1}{2}}dx=f\left(a\right)$$$$f\left(a\right)=\beta (\frac{1}{2},a+\frac{1}{2}){.}_2{F}_1(\frac{1}{2},\frac{1}{2};a+1;-1)$$$$wecansee{f}^{\prime }\left(a\right)={\int }_0^1\frac{ln(1-x){(1-x)}^a}{\sqrt{x.}\sqrt{1-x^2}}dx$$$$wewant{f}^{\prime }\left(0\right)$$$${\partial }_a\beta (\frac{1}{2},a+\frac{1}{2})=\beta (\frac{1}{2},a+\frac{1}{2})(\mathrm{\Psi }\left(\frac{1}{2}\right)-\mathrm{\Psi }(a+1))$$$$toobeecontinued$$$$$$$$$$$$$$$$$$$$$$

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