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 208245 by Shrodinger last updated on 08/Jun/24

$$K={\int }_0^{\frac{4}{\pi }}ln\left(cosx\right)dx$$

Answered by mathzup last updated on 09/Jun/24

$$K={\int }_0^{\frac{4}{\pi }}ln\left(\frac{e^{ix}+e^{-ix}}{2}\right)dxlet\lambda =\frac{4}{\pi }$$$$={\int }_0^{\lambda }($$$$ln\left(e^{ix}\right)+ln(1+e^{-2ix})-ln2)dx$$$$={\int }_0^{\lambda }ixdx+{\int }_0^{\lambda }ln(1+e^{-2ix})dx-\lambda ln2$$$$=i\frac{{\lambda }^2}{2}-\lambda ln2+{\int }_0^{\lambda }ln(1+e^{-2ix})dx$$$$(ln{(1+z)}^{{}^{\prime }}=\frac{1}{1+z}=\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{z}^n\Rightarrow$$$$ln(1+z)=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{n+1}{z}^{n+1}+c(c=0)$$$$=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n}{z}^n\Rightarrow$$$${\int }_0^{\lambda }ln(1+e^{-2ix})dx={\int }_0^{\lambda }\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n}{e}^{-2inx}dx$$$$=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n}{\int }_0^{\lambda }{e}^{-2inx}dx$$$$=\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n}{[-\frac{1}{2in}{e}^{-2ix}]}_0^{\lambda }$$$$=\frac{i}{2}\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n^2}(e^{-2i\lambda }-1)$$$$=\frac{i}{2}\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n^2}(e^{-2i\lambda }-1)$$$$=-\frac{i}{2}\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n^2}$$$$+\frac{i}{2}\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n^2}(cos\left(2\lambda \right)-isin\left(2\lambda \right))$$$$orIisrealso$$$$I=-\lambda ln2+\frac{1}{2}\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n^2}sin\left(2\lambda \right)$$$$=-\frac{4}{\pi }ln2+\frac{1}{2}\sum _{n=1}^{\mathrm{\infty }}\frac{{(-1)}^{n-1}}{n^2}sin\left(\frac{8}{\pi }\right)$$$$resttofindthevalueofthisserie$$$$...becontinued...$$

Commented by Shrodinger last updated on 10/Jun/24

$$thankssir$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com