Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 137439 by mnjuly1970 last updated on 02/Apr/21

$$......nicecalculus.....$$$$provethat::$$$$\mathit{\chi }={\int }_0^1\frac{ln(x+\sqrt{1-x^2})}{x}dx=\frac{{\pi }^2}{16}....$$

Answered by mindispower last updated on 03/Apr/21

$$f\left(t\right)={\int }_0^1\frac{ln(tx+\sqrt{1-x^2})}{x}dx,Wewantf\left(1\right)$$$$f^{\prime }\left(t\right)={\int }_0^1\frac{dx}{tx+\sqrt{1-x^2}}$$$$x=sin\left(r\right)$$$$f^{\prime }\left(t\right)={\int }_0^{\frac{\pi }{2}}\frac{cos\left(r\right)dr}{tsin\left(r\right)+cos\left(r\right)}$$$$cos\left(r\right)=a(tsin\left(r\right)+cos\left(r\right))+b(tcos\left(r\right)-sin\left(r\right))$$$$at-b=0$$$$a+bt=1$$$$b=\frac{t}{t^2+1}$$$$a=\frac{1}{1+t^2}$$$$f^{\prime }\left(t\right)={\int }_0^{\frac{\pi }{2}}\frac{dr}{1+t^2}+\frac{t}{1+t^2}{\int }_0^{\frac{\pi }{2}}\frac{tcos\left(r\right)-sin\left(r\right)}{tsin\left(r\right)+cos\left(r\right)}dr$$$$=\frac{\pi }{2}.\frac{1}{1+t^2}+\frac{t}{1+t^2}ln\left(t\right)$$$$f\left(0\right)=\frac{1}{2}{\int }_0^1\frac{ln(1-t^2)}{t}dt=\frac{1}{2}.-\sum _{n⩾0}{\int }_0^1\frac{t^{2n+1}}{n+1}$$$$=-\frac{1}{4}\sum _{n⩾0}\frac{1}{{(n+1)}^2}=-\frac{{\pi }^2}{24}$$$${\int }_0^1{f}^{\prime }\left(t\right)=f\left(1\right)-f\left(0\right)=\frac{\pi }{2}arctan\left(1\right)+{\int }_0^1\frac{tln\left(t\right)}{1+t^2}dt$$$$=\frac{{\pi }^2}{8}-\sum _{n⩾0}\frac{{(-1)}^n}{{(2n+2)}^2}=\frac{{\pi }^2}{8}-\frac{1}{4}\left(\frac{{\pi }^2}{12}\right)=\frac{5{\pi }^2}{48}$$$$f\left(1\right)=\frac{5{\pi }^2}{48}+f\left(0\right)=\frac{5{\pi }^2}{48}-\frac{{\pi }^2}{24}=\frac{{\pi }^2}{16}$$$$$$$$$$$$$$

Commented by mnjuly1970 last updated on 03/Apr/21

$$thanksalotmrpower..$$

Commented by mnjuly1970 last updated on 03/Apr/21

Commented by mindispower last updated on 03/Apr/21

$$pleasursir$$

Answered by mathmax by abdo last updated on 03/Apr/21

$$\chi ={\int }_0^1\frac{\log (\mathrm{x}+\sqrt{1-{\mathrm{x}}^2})}{\mathrm{x}}\mathrm{dx}\mathrm{let}\mathrm{f}\left(\mathrm{a}\right)={\int }_0^1\frac{\log (\mathrm{ax}+\sqrt{1-{\mathrm{x}}^2})}{\mathrm{x}}\mathrm{dx}\mathrm{witha}>0$$$${\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)={\int }_0^1\frac{\mathrm{x}}{\mathrm{x}(\mathrm{ax}+\sqrt{1-{\mathrm{x}}^2})}\mathrm{dx}={\int }_0^1\frac{\mathrm{dx}}{\mathrm{ax}+\sqrt{1-{\mathrm{x}}^2}}$$$${=}_{\mathrm{x}=\mathrm{sint}}{\int }_0^{\frac{\pi }{2}}\frac{\mathrm{cost}}{\mathrm{asint}+\mathrm{cost}}\mathrm{dt}={\int }_0^{\frac{\pi }{2}}\frac{\mathrm{dt}}{\mathrm{atant}+1}$$$${=}_{\mathrm{tant}=\mathrm{y}}{\int }_0^{\mathrm{\infty }}\frac{1}{\mathrm{ay}+1}\times \frac{\mathrm{dy}}{(1+{\mathrm{y}}^2)}\mathrm{let}\mathrm{decompose}$$$$\mathrm{F}\left(\mathrm{y}\right)=\frac{1}{(\mathrm{ay}+1)({\mathrm{y}}^2+1)}\Rightarrow \mathrm{F}\left(\mathrm{y}\right)=\frac{\alpha }{\mathrm{ay}+1}+\frac{\beta \mathrm{y}+\delta }{{\mathrm{y}}^2+1}$$$$\alpha =\frac{1}{\frac{1}{{\mathrm{a}}^2}+1}=\frac{{\mathrm{a}}^2}{{\mathrm{a}}^2+1}$$$${\lim }_{\mathrm{y}\to +\mathrm{\infty }}\mathrm{yF}\left(\mathrm{y}\right)=0=\frac{\alpha }{\mathrm{a}}+\beta \Rightarrow \beta =-\frac{\mathrm{a}}{{\mathrm{a}}^2+1}$$$$\mathrm{F}\left(0\right)=1=\alpha +\delta \Rightarrow \delta =1-\frac{{\mathrm{a}}^2}{{\mathrm{a}}^2+1}=\frac{1}{{\mathrm{a}}^2+1}\Rightarrow$$$$\mathrm{F}\left(\mathrm{y}\right)=\frac{{\mathrm{a}}^2}{({\mathrm{a}}^2+1)(\mathrm{ay}+1)}+\frac{-\frac{\mathrm{a}}{1+{\mathrm{a}}^2}\mathrm{y}+\frac{1}{1+{\mathrm{a}}^2}}{{\mathrm{y}}^2+1}\Rightarrow$$$${\int }_0^{\mathrm{\infty }}\mathrm{F}\left(\mathrm{y}\right)\mathrm{dy}=\frac{1}{{\mathrm{a}}^2+1}{\int }_0^{\mathrm{\infty }}(\frac{{\mathrm{a}}^2\mathrm{dy}}{\mathrm{ay}+1}-\frac{\mathrm{ay}-1}{{\mathrm{y}}^2+1})\mathrm{dy}$$$$=\frac{1}{{\mathrm{a}}^2+1}{\int }_0^{\mathrm{\infty }}(\frac{{\mathrm{a}}^2}{\mathrm{ay}+1}-\frac{\mathrm{a}}{2}\frac{2\mathrm{y}}{{\mathrm{y}}^2+1}+\frac{1}{{\mathrm{y}}^2+1})\mathrm{dy}$$$$=\frac{1}{{\mathrm{a}}^2+1}{[\mathrm{aln}(\mathrm{ay}+1)-\mathrm{aln}\sqrt{{\mathrm{y}}^2+1}]}_0^{\mathrm{\infty }}+\frac{\pi }{2(1+{\mathrm{a}}^2)}$$$$=\frac{\mathrm{a}}{{\mathrm{a}}^2+1}{\left[\ln \left(\frac{\mathrm{ay}+1}{\sqrt{{\mathrm{y}}^2+1}}\right)\right]}_0^{\mathrm{\infty }}+\frac{\pi }{2(1+{\mathrm{a}}^2)}$$$$=\frac{\mathrm{aloga}}{{\mathrm{a}}^2+1}+\frac{\pi }{2({\mathrm{a}}^2+1)}={\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)\Rightarrow$$$$\mathrm{f}\left(\mathrm{a}\right)=\int \frac{\mathrm{aloga}}{{\mathrm{a}}^2+1}\mathrm{da}+\frac{\pi }{2}\arctan \left(\mathrm{a}\right)+\mathrm{C}$$$$\mathrm{f}\left(1\right)-\mathrm{f}\left(0\right)={\int }_0^1{\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)\mathrm{da}={\int }_0^1\frac{\mathrm{aloga}}{1+{\mathrm{a}}^2}\mathrm{da}+\frac{\pi }{2}{\int }_0^1\frac{\mathrm{da}}{1+{\mathrm{a}}^2}$$$${\int }_0^1\frac{\mathrm{da}}{1+{\mathrm{a}}^2}={\left[\mathrm{arctana}\right]}_0^1=\frac{\pi }{4}$$$${\int }_0^1\frac{\mathrm{aloga}}{1+{\mathrm{a}}^2}\mathrm{da}={\left[\frac{1}{2}\log (1+{\mathrm{a}}^2)\mathrm{loga}\right]}_0^1-\frac{1}{2}{\int }_0^1\frac{\log (1+{\mathrm{a}}^2)}{\mathrm{a}}\mathrm{da}$$$$=-\frac{1}{2}{\int }_0^1\frac{\log (1+{\mathrm{a}}^2)}{\mathrm{a}}\mathrm{da}$$$$(\log {(1+\mathrm{u})}^{{}^{\prime }}=\frac{1}{1+\mathrm{u}}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\mathrm{u}}^{\mathrm{n}}\Rightarrow \log (1+\mathrm{u})=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}\frac{{\mathrm{u}}^{\mathrm{n}+1}}{\mathrm{n}+1}$$$$=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}-1}\frac{{\mathrm{u}}^{\mathrm{n}}}{\mathrm{n}}\Rightarrow \log (1+{\mathrm{x}}^2)=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}-1}\frac{{\mathrm{x}}^{2\mathrm{n}}}{\mathrm{n}}\Rightarrow$$$${\int }_0^1\frac{\log (1+{\mathrm{x}}^2)}{\mathrm{x}}\mathrm{dx}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}-1}}{\mathrm{n}(2\mathrm{n}+1)}={\mathrm{u}}_{\mathrm{n}}$$$$\frac{{\mathrm{u}}_{\mathrm{n}}}{2}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}-1}}{2\mathrm{n}(2\mathrm{n}+1)}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}-1}(\frac{1}{2\mathrm{n}}-\frac{1}{2\mathrm{n}+1})$$$$=\frac{1}{2}\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}-1}}{\mathrm{n}}-\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{2\mathrm{n}+1}$$$$=\frac{\log 2}{2}-(\frac{\pi }{4}-1)\Rightarrow \mathrm{f}\left(1\right)-\mathrm{f}\left(0\right)=-\frac{\log 2}{4}+\frac{1}{2}(\frac{\pi }{4}-1)$$$$=-\frac{\log 2}{4}+\frac{\pi }{8}-\frac{1}{2}\Rightarrow \mathrm{f}\left(1\right)=\chi =-\frac{\log 2}{4}+\frac{\pi }{8}-\frac{1}{2}+\mathrm{f}\left(0\right)$$$$\mathrm{f}\left(0\right)={\int }_0^1\frac{\log \left(\sqrt{1-{\mathrm{x}}^2}\right)}{\mathrm{x}}\mathrm{dx}\mathrm{rest}\mathrm{to}\mathrm{find}\mathrm{f}\left(0\right)....\mathrm{be}\mathrm{continued}...$$$$$$

Commented by mnjuly1970 last updated on 03/Apr/21

$$thanksalotmrmax....$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com