Question Number 48715 by Abdo msup. last updated on 27/Nov/18
$$\left.\mathrm{1}\right)\:{find}\:\:{f}\left(\lambda\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dx}}{\mathrm{2}+{e}^{−\lambda{x}} }\:\:{with}\:\lambda>\mathrm{0} \\ $$$$\left.\mathrm{2}\right){calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{x}}{\left(\mathrm{2}+{e}^{−\lambda{x}} \right)^{\mathrm{2}} }{dx} \\ $$$$\left.\mathrm{3}\right)\:{find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{dx}}{\mathrm{2}\:+{e}^{−{x}\sqrt{\mathrm{3}}} }{dx}\:{and}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{x}}{\left(\mathrm{2}+{e}^{−{x}\sqrt{\mathrm{3}}} \right)^{\mathrm{2}} }{dx} \\ $$
Commented by maxmathsup by imad last updated on 01/Dec/18
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{f}\left(\lambda\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dx}}{\mathrm{2}+{e}^{−\lambda{x}} }\:\:{changement}\:\:{e}^{\lambda{x}\:} ={t}\:{give}\:\lambda{x}={ln}\left({t}\right)\:\Rightarrow \\ $$$${x}=\frac{\mathrm{1}}{\lambda}{ln}\left({t}\right)\:\Rightarrow{f}\left(\lambda\right)=\int_{\mathrm{1}} ^{{e}^{\lambda} } \:\:\:\:\:\frac{\mathrm{1}}{\lambda\left(\mathrm{2}+{t}^{−\mathrm{1}} \right)}\:\frac{{dt}}{{t}}\:=\frac{\mathrm{1}}{\lambda}\:\int_{\mathrm{1}} ^{{e}^{\lambda} } \:\:\:\frac{{dt}}{\mathrm{2}{t}\:+\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\lambda}\left[{ln}\left(\mathrm{2}{t}+\mathrm{1}\right)\right]_{\mathrm{1}} ^{{e}^{\lambda} } \:\:=\:\frac{\mathrm{1}}{\mathrm{2}\lambda}\left\{{ln}\left(\mathrm{2}{e}^{\lambda} \:+\mathrm{1}\right)−{ln}\left(\mathrm{3}\right)\right\}\:\:\:\Rightarrow \\ $$$${f}\left(\lambda\right)=\frac{\mathrm{1}}{\mathrm{2}\lambda}\left\{{ln}\left(\mathrm{2}\:{e}^{\lambda} \:+\mathrm{1}\right)−{ln}\left(\mathrm{3}\right)\right\}\:. \\ $$
Commented by maxmathsup by imad last updated on 01/Dec/18
$$\left.\mathrm{2}\right)\:{we}\:{have}\:{f}^{'} \left(\lambda\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \:−\frac{−{x}\:{e}^{−\lambda{x}} }{\left(\mathrm{2}+{e}^{−\lambda{x}} \right)^{\mathrm{2}} }\:{dx}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{x}\:{e}^{−\lambda{x}} }{\left(\mathrm{2}+{e}^{−\lambda{x}} \right)^{\mathrm{2}} }{dx}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{x}\:{e}^{−\lambda{x}} }{\left(\mathrm{2}+{e}^{−\lambda{x}} \right)^{\mathrm{2}} }{dx}\:={f}^{'} \left(\lambda\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\left\{−\frac{\mathrm{1}}{\lambda^{\mathrm{2}} }\left({ln}\left(\mathrm{2}{e}^{\lambda} \:+\mathrm{1}\right)−{ln}\left(\mathrm{3}\right)\:+\frac{\mathrm{1}}{\lambda}\:\frac{\mathrm{2}\:{e}^{\lambda} }{\mathrm{2}{e}^{\lambda\:} \:+\mathrm{1}}\right\}\right. \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}\lambda^{\mathrm{2}} }\left\{\left(\:{ln}\left(\mathrm{2}\:{e}^{\lambda} \:+\mathrm{1}\right)−{ln}\left(\mathrm{3}\right)\right)\:−\frac{\mathrm{2}\lambda\:{e}^{\lambda} }{\mathrm{2}{e}^{\lambda} \:+\mathrm{1}}\right\} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{x}\:{e}^{−{x}\sqrt{\mathrm{3}}} }{\left(\mathrm{2}+{e}^{−{x}\sqrt{\mathrm{3}}} \right)^{\mathrm{2}} }{dx}\:=\:{f}^{'} \left(\sqrt{\mathrm{3}}\right)\:=−\frac{\mathrm{1}}{\mathrm{6}}\left\{\:\left({ln}\left(\mathrm{2}\:{e}^{\sqrt{\mathrm{3}}} +\mathrm{1}\right)−{ln}\left(\mathrm{3}\right)\right)\:−\frac{\mathrm{2}\sqrt{\mathrm{3}}{e}^{\sqrt{\mathrm{3}}} }{\mathrm{2}\:{e}^{\sqrt{\mathrm{3}}} +\mathrm{1}}\right\}\:. \\ $$
Commented by maxmathsup by imad last updated on 01/Dec/18
$$\left.\mathrm{2}\right)\:{the}\:{Q}\:.{is}\:{calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{x}\:{e}^{−\lambda{x}} }{\left(\mathrm{2}+{e}^{−\lambda{x}} \right)^{\mathrm{2}} }{dx} \\ $$
Commented by maxmathsup by imad last updated on 01/Dec/18
$$\left.\right)\:{we}\:{have}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dx}}{\mathrm{2}+{e}^{−{x}\sqrt{\mathrm{3}}} }\:={f}\left(\sqrt{\mathrm{3}}\right)\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\left\{{ln}\left(\mathrm{2}\:{e}^{\sqrt{\mathrm{3}}} +\mathrm{1}\right)−{ln}\left(\mathrm{3}\right)\right\}\: \\ $$