Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 193272 by 073 last updated on 09/Jun/23

Answered by aba last updated on 09/Jun/23

$$\ln \left(\frac{\sqrt{2\pi }}{\mathrm{e}}\right)$$

Answered by aba last updated on 09/Jun/23

$${\int }_0^1\ln \mathrm{\Gamma }(\mathrm{x}+1)\mathrm{dx}={\int }_0^1\mathrm{lnx}\mathrm{\Gamma }\left(\mathrm{x}\right)\mathrm{dx}={\int }_0^1\ln \left(\mathrm{x}\right)+{\int }_0^1\ln \mathrm{\Gamma }\left(\mathrm{x}\right)\mathrm{dx}$$$$=\underset{\mathrm{a}\to 0}{\lim }({\left[\mathrm{xln}\left(\mathrm{x}\right)\right]}_{\mathrm{a}}^1-{\int }_{\mathrm{a}}^1\mathrm{dx})+\frac{1}{2}{\int }_0^{1}2\ln \mathrm{\Gamma }\left(\mathrm{x}\right)\mathrm{dx}$$$$=-1+\frac{1}{2}{\int }_0^{1}2\ln \mathrm{\Gamma }(1-\mathrm{x})\mathrm{dx}$$$$=-1+\frac{1}{2}{\int }_0^1\ln \mathrm{\Gamma }(1-\mathrm{x})+\ln \mathrm{\Gamma }(1-\mathrm{x})\mathrm{dx}$$$$=-1+\frac{1}{2}{\int }_0^1\ln \mathrm{\Gamma }(1-\mathrm{x})+\ln \mathrm{\Gamma }\left(\mathrm{x}\right)\mathrm{dx}$$$$=-1+\frac{1}{2}{\int }_0^1\ln \mathrm{\Gamma }(1-\mathrm{x})\mathrm{\Gamma }\left(\mathrm{x}\right)\mathrm{dx}$$$$=-1+\frac{1}{2}{\int }_0^1\ln \left(\frac{\pi }{\sin (\pi \mathrm{x}}\right)\mathrm{dx}$$$$=-1+\frac{1}{2}{\int }_0^1(\ln \left(\pi \right)-\ln \left(\sin \left(\pi \mathrm{x}\right)\right)\mathrm{dx}$$$$=-1+\frac{1}{2}(\ln \left(\pi \right)-\frac{2}{\pi }{\int }_0^{\frac{\pi }{2}}\ln \left(\sin \left(\mathrm{x}\right)\right)\mathrm{dx})$$$$=-1+\frac{1}{2}(\ln \left(\pi \right)-\frac{2}{\pi }{\int }_0^{\frac{\pi }{2}}\ln \left(2\right)\mathrm{dx})$$$$=-1+\frac{1}{2}(\ln \left(\pi \right)+\frac{2}{\pi }\times \frac{\pi }{2}\ln \left(2\right))$$$$=-1+\frac{1}{2}(\ln \left(\pi \right)+\ln \left(2\right))$$$$=\ln \left(\frac{\sqrt{2\pi }}{\mathrm{e}}\right)$$$$$$$$$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com