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 130167 by Lordose last updated on 23/Jan/21

$${\int }_0^{\mathrm{\infty }}\frac{\ln \left(\mathrm{u}\right){\mathrm{e}}^{-\mathrm{u}}}{{(1+{\mathrm{e}}^{-\mathrm{u}})}^2}\mathrm{du}$$$$$$$$$$

Answered by mindispower last updated on 26/Jan/21

$${\int }_0^{\mathrm{\infty }}\frac{x^s}{(1+e^{-x})}{e}^{-x}dx=f\left(s\right)$$$$\frac{1}{1+x}=\mathrm{\Sigma }{(-1)}^k{x}^k$$$$\Rightarrow \frac{x}{{(1+x)}^2}=\sum _{k⩾1}k{(-1)}^{k+1}{x}^k$$$${\int }_0^{\mathrm{\infty }}{x}^s\sum _{k⩾1}k{(-1)}^{k+1}{e}^{-kx}dx$$$$kx=t$$$${\int }_0^{\mathrm{\infty }}\sum _{k⩾1}\frac{t^s{(-1)}^{k+1}{e}^{-x}dx}{k^s}$$$$=\sum _{k⩾1}\frac{{(-1)}^{k+1}}{k^s}{\int }_0^{\mathrm{\infty }}{t}^s{e}^{-x}dx$$$$=\sum _{k⩾1}\frac{{(-1)}^{k+1}}{k^s}\mathrm{\Gamma }(s+1)=\mathrm{\Gamma }(s+1)(1-\frac{1}{2^{s-1}})\zeta \left(s\right)$$$$f\left(s\right)=\mathrm{\Gamma }(s+1)(1-\frac{1}{2^{s-1}})\zeta \left(s\right)$$$$f^{\prime }\left(s\right)=(\mathrm{\Psi }(s+1)(1-\frac{1}{2^{s-1}})+\mathrm{\Gamma }(s+1)\left(ln\left(2\right)2^{1-s}\right))\zeta \left(s\right)$$$$+{\zeta }^{\prime }\left(s\right)\left(\mathrm{\Gamma }(s+1)(1-\frac{1}{2^{s-1}})\right)$$$$\mathrm{\Gamma }\left(1\right)=1,\mathrm{\Psi }\left(1\right)=-\gamma$$$$\zeta \left(0\right)=-\frac{1}{2},{\zeta }^{\prime }\left(0\right)=-\frac{ln\left(2\pi \right)}{2}$$$$weget$$$$(-\gamma (-1)+2ln\left(2\right)).-\frac{1}{2}+-\frac{ln\left(2\pi \right)}{2}(-2)$$$$=\frac{\gamma }{2}+ln\left(2\pi \right)-ln\left(2\right)=\frac{\gamma }{2}+ln\left(\frac{2\pi }{2}\right)=\frac{\gamma }{2}+ln\left(\pi \right)$$$$$$$$$$$$$$$$$$$$$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com