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 144787 by mnjuly1970 last updated on 29/Jun/21

$$$$$$\text{You can't use 'macro parameter character \#' in math mode}$$$$\mathrm{If}:\mathit{\varphi }\left(n\right):={\int }_0^1\frac{x^{2n}}{1+x^2}\mathrm{d}x$$$$$$$$\mathrm{then}\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}::$$$$$$$$\mathrm{S}:=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n\mathit{\varphi }\left(n\right)}{n}=?$$$$$$$$\mathrm{m}.\mathrm{n}.\mathrm{july}.1970$$

Answered by qaz last updated on 29/Jun/21

$$\mathrm{S}=\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{\mathrm{n}}}{\mathrm{n}}\varphi \left(\mathrm{n}\right)$$$$={\int }_0^1\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{\mathrm{n}}}{\mathrm{n}}\cdot \frac{{\mathrm{x}}^{2\mathrm{n}}}{1+{\mathrm{x}}^2}\mathrm{dx}$$$$={\int }_0^1\frac{\ln (1+{\mathrm{x}}^2)}{1+{\mathrm{x}}^2}\mathrm{dx}$$$$={\int }_0^{\mathrm{\infty }}\frac{\ln (1+{\mathrm{x}}^2)}{1+{\mathrm{x}}^2}\mathrm{dx}-{\int }_1^{\mathrm{\infty }}\frac{\ln (1+{\mathrm{x}}^2)}{1+{\mathrm{x}}^2}\mathrm{dx}$$$$=-2{\int }_0^{\pi /2}\mathrm{lncos}\mathrm{udu}-{\int }_0^1\frac{\ln (1+{\mathrm{x}}^2)-2\mathrm{lnx}}{1+{\mathrm{x}}^2}\mathrm{dx}$$$$=\pi \ln 2-\mathrm{S}-2\mathrm{G}$$$$\Rightarrow \mathrm{S}=\frac{\pi }{2}\ln 2-\mathrm{G}$$

Commented by mnjuly1970 last updated on 29/Jun/21

$$thxmrqaz...$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com