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Question Number 197365 by mokys last updated on 15/Sep/23

Answered by witcher3 last updated on 15/Sep/23

$$\mathrm{x}=\frac{1-\mathrm{t}}{1+\mathrm{t}}$$$${\int }_0^1\frac{{\left(\frac{1-\mathrm{t}}{1+\mathrm{t}}\right)}^{1-\mathrm{p}}{\left(\frac{2\mathrm{t}}{1+\mathrm{t}}\right)}^{\mathrm{p}}}{{\left(\frac{2}{1+\mathrm{t}}\right)}^3}.\frac{2}{{(1+\mathrm{t})}^2}\mathrm{dt}$$$$={\int }_0^1{(1-\mathrm{t})}^{1-\mathrm{p}}{\mathrm{t}}^{\mathrm{p}}.2^{\mathrm{p}-2}$$$$\text{Missing left or extra right}$$$$=2^{\mathrm{p}-2}\beta (\mathrm{p}+1,2-\mathrm{p})$$$$=2^{\mathrm{p}-2}.\frac{\mathrm{\Gamma }(\mathrm{p}+1)\mathrm{\Gamma }(2-\mathrm{p})}{\mathrm{\Gamma }\left(3\right)}=\frac{2^{\mathrm{p}-2}}{2}.\mathrm{p}\mathrm{\Gamma }\left(\mathrm{p}\right).(1-\mathrm{p})\mathrm{\Gamma }(1-\mathrm{p})$$$$=\frac{2^{\mathrm{p}}\mathrm{p}(1-\mathrm{p}).\mathrm{\Gamma }\left(\mathrm{p}\right)\mathrm{\Gamma }(1-\mathrm{p})}{8}=\frac{2^{\mathrm{p}}\mathrm{p}(1-\mathrm{p})}{8}.\frac{\pi }{\sin \left(\mathrm{p}\pi \right)}$$$$$$

Commented by mokys last updated on 15/Sep/23

$$canyousolvebyresideo$$

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