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Question Number 38726 by maxmathsup by imad last updated on 28/Jun/18
$${let}\:{f}\left({x}\right)=\frac{{x}+\mathrm{1}}{\mathrm{2}\:+{e}^{−\mathrm{2}{x}} }\:\:\:{developp}\:{f}\:{at}\:{integr}\:{serie}. \\ $$
Commented by math khazana by abdo last updated on 01/Jul/18
$${f}\left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{2}}\left({x}+\mathrm{1}\right)\:\frac{\mathrm{1}}{\mathrm{1}+\left(\frac{{e}^{−\mathrm{2}{x}} }{\mathrm{2}}\right)}\:=\frac{\mathrm{1}}{\mathrm{2}}\left({x}+\mathrm{1}\right)\sum_{{n}=\mathrm{0}} ^{\infty} \frac{{e}^{−\mathrm{2}{nx}} }{\mathrm{2}^{{n}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({x}+\mathrm{1}\right)\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\:\left(\sum_{{p}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{2}{nx}\right)^{{p}} }{{p}!}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({x}+\mathrm{1}\right)\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\left(\sum_{{p}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{2}{n}\right)^{{p}} }{{p}!}\:{x}^{{p}} \right)... \\ $$