Question Number 160349 by mr W last updated on 28/Nov/21
$${find}\:\int\frac{{dx}}{{x}+{e}^{{x}} }=? \\ $$
Answered by mathmax by abdo last updated on 28/Nov/21
$$\int\:\:\frac{\mathrm{dx}}{\mathrm{x}+\mathrm{e}^{\mathrm{x}} }=\int\:\:\frac{\mathrm{e}^{−\mathrm{x}} }{\mathrm{1}+\mathrm{xe}^{−\mathrm{x}} }\mathrm{dx}\:=\int\:\mathrm{e}^{−\mathrm{x}} \sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{x}^{\mathrm{n}} \:\mathrm{e}^{−\mathrm{nx}} \mathrm{dx}\:\:\mathrm{if}\:\mid\mathrm{x}\mid<\mathrm{1} \\ $$$$=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{\mathrm{n}} \:\int\:\mathrm{x}^{\mathrm{n}} \:\mathrm{e}^{−\left(\mathrm{n}+\mathrm{1}\right)\mathrm{x}} \mathrm{dx} \\ $$$$=_{\left(\mathrm{n}+\mathrm{1}\right)\mathrm{x}=\mathrm{t}} \sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{\mathrm{n}} \:\int\:\frac{\mathrm{t}^{\mathrm{n}} }{\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{n}} }\mathrm{e}^{−\mathrm{t}} \:\frac{\mathrm{dt}}{\left(\mathrm{n}+\mathrm{1}\right)} \\ $$$$=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty\:} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{n}+\mathrm{1}} }\:\int\:\mathrm{t}^{\mathrm{n}} \:\mathrm{e}^{−\mathrm{t}} \:\mathrm{dt}\:\:=\Gamma_{\mathrm{inc}} \left(\mathrm{n}+\mathrm{1}\right).\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{n}+\mathrm{1}} } \\ $$
Commented by mr W last updated on 29/Nov/21
$${thanks}\:{sir}! \\ $$$$\Gamma_{{inc}} \left({n}+\mathrm{1}\right)\:{seems}\:{to}\:{be}\:{dependent}\:{from} \\ $$$${n}.\:{but}\:{how}\:{can}\:{you}\:{take}\:{it}\:{out}\:{of}\:{the} \\ $$$${summation}\:{about}\:{n}\:{and}\:{treat}\:{it}\:{as} \\ $$$${independent}\:{from}\:{n}? \\ $$