Integration Questions

Question Number 152142 by Ar Brandon last updated on 26/Aug/21

$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}}{\left(\mathrm{1}−{x}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\mathrm{ln}\left(\mathrm{ln}\frac{\mathrm{1}}{{x}}\right){dx}=−\frac{\gamma}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln}\frac{\mathrm{6}\sqrt{\mathrm{3}}}{\pi}+\frac{\pi\sqrt{\mathrm{3}}}{\mathrm{27}}\left(\mathrm{5ln2}\pi−\mathrm{6ln}\Gamma\left(\frac{\mathrm{1}}{\mathrm{6}}\right)\right) \\$$

Answered by mindispower last updated on 28/Aug/21

$$\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−\mathrm{2}{t}} {ln}\left({t}\right){dt}}{\left(\mathrm{1}−{e}^{−{t}} +{e}^{−\mathrm{2}{t}} \right)^{\mathrm{2}} } \\$$$$\int_{\mathrm{0}} ^{\infty} \frac{{t}^{{s}} {e}^{−\mathrm{2}{t}} }{\left(\mathrm{1}−{e}^{−{t}} +{e}^{−\mathrm{2}{t}} \right)^{\mathrm{2}} }{dt} \\$$$${f}\left({s}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{t}^{{s}} {e}^{−\mathrm{2}{t}} \left(\mathrm{1}+\mathrm{2}{e}^{−{t}} +{e}^{−\mathrm{2}{t}} \right)}{\left(\mathrm{1}+{e}^{−\mathrm{3}{t}} \right)^{\mathrm{2}} }{dt} \\$$$$=\underset{{m}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{{m}} .{m}\int_{\mathrm{0}} ^{\infty} {t}^{{s}} \left({e}^{−{t}\left(\mathrm{2}+\mathrm{3}{m}\right)} +{e}^{−{t}\left(\mathrm{3}+\mathrm{3}{m}\right)} +{e}^{−{t}\left(\mathrm{4}+\mathrm{3}{m}\right)} \right){dt} \\$$$$=\underset{{m}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\overset{{m}} {\right)}.{m}\Gamma\left(\mathrm{1}+{s}\right)\left(\frac{\mathrm{1}}{\left(\mathrm{2}+\mathrm{3}{m}\right)^{{s}+\mathrm{1}} }+\frac{\mathrm{1}}{\left(\mathrm{3}+\mathrm{3}{m}\right)^{{s}+\mathrm{1}} }+\frac{\mathrm{1}}{\left(\mathrm{4}+\mathrm{3}{m}\right)^{{s}+\mathrm{1}} }\right) \\$$$$=\frac{\Gamma\left(\mathrm{1}+{s}\right)}{\mathrm{3}^{{s}+\mathrm{1}} }\underset{{m}\geqslant\mathrm{1}} {\sum}\left(−\mathrm{1}\right)^{{m}} .{m}\left(\frac{\mathrm{1}}{\left({m}+\frac{\mathrm{2}}{\mathrm{3}}\right)^{{s}+\mathrm{1}} }+\frac{\mathrm{1}}{\left(\mathrm{1}+{m}\right)^{{s}+\mathrm{1}} }+\frac{\mathrm{1}}{\left({m}+\frac{\mathrm{4}}{\mathrm{3}}\right)^{{s}+\mathrm{1}} }\right) \\$$$$=\frac{\Gamma\left(\mathrm{1}+{s}\right)}{\mathrm{3}^{{s}+\mathrm{1}} }\underset{{m}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{{m}} \left(\frac{\mathrm{1}}{\left({m}+\frac{\mathrm{2}}{\mathrm{3}}\right)^{{s}} }+\frac{\mathrm{1}}{\left(\mathrm{1}+{m}\right)^{{s}} }+\frac{\mathrm{1}}{\left({m}+\frac{\mathrm{4}}{\mathrm{3}}\right)^{{s}} }\right. \\$$$$\left.−\frac{\mathrm{2}}{\mathrm{3}\left({m}+\frac{\mathrm{2}}{\mathrm{3}}\right)^{{s}+\mathrm{1}} }−\frac{\mathrm{1}}{\left(\mathrm{1}+{m}\right)^{{s}+\mathrm{1}} }−\frac{\mathrm{4}}{\mathrm{3}\left({m}+\frac{\mathrm{4}}{\mathrm{3}}\right)^{{s}+\mathrm{1}} }\right) \\$$$$=\frac{\Gamma\left(\mathrm{1}+{s}\right)}{\mathrm{3}^{{s}+\mathrm{1}} }\underset{{m}\geqslant\mathrm{0}} {\sum}\left(\zeta\left(\frac{\mathrm{1}}{\mathrm{3}},{s}\right)−\zeta\left(\frac{\mathrm{5}}{\mathrm{6}},{s}\right)+\eta\left({s}\right)−\eta\left(\mathrm{1}+{s}\right)\right. \\$$$$+\zeta\left(\frac{\mathrm{2}}{\mathrm{3}},{s}\right)−\zeta\left(\frac{\mathrm{7}}{\mathrm{6}},{s}\right)−\frac{\mathrm{2}}{\mathrm{3}}\left(\zeta\left(\frac{\mathrm{1}}{\mathrm{3}},{s}+\mathrm{1}\right)−\zeta\left(\frac{\mathrm{5}}{\mathrm{6}},{s}+\mathrm{1}\right)\right) \\$$$$−\frac{\mathrm{4}}{\mathrm{3}}\left(\zeta\left(\frac{\mathrm{2}}{\mathrm{3}},\mathrm{1}+{s}\right)−\zeta\left(\frac{\mathrm{7}}{\mathrm{6}},\mathrm{1}+{s}\right)\right) \\$$$${we}\:{want}\:,{f}'\left(\mathrm{0}\right) \\$$$${too}\:{bee}\:{continuede}\:{too}\:{many}\:{calculation} \\$$$$\:{just}\:{using}\:{steiljes}\:{constante} \\$$$${of}\:{laurent}\:{expesion}\:{zeta}\:{hurwitz}\:{fomction} \\$$

Commented by Ar Brandon last updated on 28/Aug/21

$$\mathrm{wow}\:\mathrm{thanks}\:\mathrm{sir} \\$$