Question and Answers Forum

All Questions      Topic List

Differentiation Questions

Previous in All Question      Next in All Question      

Previous in Differentiation      Next in Differentiation      

Question Number 172359 by mnjuly1970 last updated on 25/Jun/22

Answered by Mathspace last updated on 26/Jun/22

$$J={\int }_0^1{x}^{-\frac{1}{2}}{(1-x)}^{-\frac{1}{2}}{ln}^{2}xdx$$$$letf\left(a\right)={\int }_0^1{x}^{a-\frac{1}{2}}{(1-x)}^{-\frac{1}{2}}dx$$$$wehave{f}^{{}^{″}}\left(a\right)={\int }_0^1{x}^{a-\frac{1}{2}}{(1-x)}^{-\frac{1}{2}}{ln}^{2}xdx$$$$\Rightarrow f^{{}^{″}}\left(0\right)={\int }_0^1{x}^{-\frac{1}{2}}{(1-x)}^{-\frac{1}{2}}{ln}^{2}xdx$$$$f\left(a\right)={\int }_0^1{x}^{a+\frac{1}{2}-1}{(1-x)}^{\frac{1}{2}-1}dx$$$$=B(a+\frac{1}{2},\frac{1}{2})=\frac{\mathrm{\Gamma }(a+\frac{1}{2})\mathrm{\Gamma }\left(\frac{1}{2}\right)}{\mathrm{\Gamma }(a+1)}$$$$=\sqrt{\pi }\times \frac{\mathrm{\Gamma }(a+\frac{1}{2})}{\mathrm{\Gamma }(a+1)}$$$$\Rightarrow f^{{}^{\prime }}\left(a\right)=\sqrt{\pi }\times \frac{{\mathrm{\Gamma }}^{{}^{\prime }}(a+\frac{1}{2})\mathrm{\Gamma }(a+1)-\mathrm{\Gamma }(a+\frac{1}{2}){\mathrm{\Gamma }}^{{}^{\prime }}(a+1)}{{\mathrm{\Gamma }}^2(a+1)}$$$$we/have\psi \left(x\right)=\frac{{\mathrm{\Gamma }}^{{}^{\prime }}\left(x\right)}{\mathrm{\Gamma }\left(x\right)}\Rightarrow$$$$f^{{}^{\prime }}\left(a\right)=\sqrt{\pi }\times \frac{\psi (a+\frac{1}{2})\mathrm{\Gamma }(a+\frac{1}{2})\mathrm{\Gamma }(a+1)-\mathrm{\Gamma }(a+\frac{1}{2})\psi (a+1)\mathrm{\Gamma }(a+1)}{{\mathrm{\Gamma }}^2(a+1)}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com