Question Number 37280 by abdo.msup.com last updated on 11/Jun/18
$${calculate}\:\:\int_{\mathrm{0}} ^{\mathrm{6}} \:\:\:\frac{{e}^{{x}−\left[{x}\right]} }{\mathrm{1}+{e}^{{x}} }{dx}\:. \\ $$
Commented by prof Abdo imad last updated on 16/Jun/18
$${I}\:=\:\sum_{{k}\:=\mathrm{0}} ^{\mathrm{5}} \:\int_{{k}} ^{{k}+\mathrm{1}} \:\:\:\frac{{e}^{{x}−{k}} }{\mathrm{1}+{e}^{{x}} }{dx} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{\mathrm{5}} \:{e}^{−{k}} \:\:\int_{{k}} ^{{k}+\mathrm{1}} \:\:\:\frac{{e}^{{x}} }{\mathrm{1}+{e}^{{x}} }{dx} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{\mathrm{5}} \:{e}^{−{k}} \left[{ln}\left(\mathrm{1}+{e}^{{x}} \right)\right]_{{k}} ^{{k}+\mathrm{1}} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{\mathrm{5}} {e}^{−{k}} \:\left\{{ln}\left(\mathrm{1}+{e}^{{k}+\mathrm{1}} \right)\:−{ln}\left(\mathrm{1}\:+{e}^{{k}} \right)\right\} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{\mathrm{5}} \:{e}^{−{k}} \:{ln}\left(\frac{\mathrm{1}+{e}^{{k}+\mathrm{1}} }{\mathrm{1}+{e}^{{k}} }\right) \\ $$$$={ln}\left(\frac{\mathrm{1}+{e}}{\mathrm{2}}\right)\:\:+{e}^{−\mathrm{1}} {ln}\left(\:\frac{\mathrm{1}+{e}^{\mathrm{2}} }{\mathrm{1}+{e}}\right)\:+{e}^{−\mathrm{2}} {ln}\left(\frac{\mathrm{1}+{e}^{\mathrm{3}} }{\mathrm{1}+{e}^{\mathrm{2}} }\right) \\ $$$$+{e}^{−\mathrm{3}} {ln}\left(\frac{\mathrm{1}+{e}^{\mathrm{4}} }{\mathrm{1}+{e}^{\mathrm{3}} }\right)\:+{e}^{−\mathrm{4}} {ln}\left(\:\frac{\mathrm{1}+{e}^{\mathrm{5}} }{\mathrm{1}\:+{e}^{\mathrm{4}} }\right)\:+{e}^{−\mathrm{5}} {ln}\left(\frac{\mathrm{1}+{e}^{\mathrm{6}} }{\mathrm{1}+{e}^{\mathrm{5}} }\right)\:. \\ $$$$ \\ $$