Question Number 156470 by puissant last updated on 11/Oct/21
$${Q}=\int_{\mathrm{0}} ^{\infty} \frac{{x}^{{c}} }{{c}^{{x}} }{dx} \\ $$
Answered by puissant last updated on 11/Oct/21
$${posons}\:{c}^{{x}} ={e}^{{u}} \Rightarrow\:{e}^{{xlnc}} ={e}^{{u}} \Rightarrow\:{x}=\frac{{u}}{{lnc}} \\ $$$$\Rightarrow\:{dx}=\frac{{du}}{{lnc}} \\ $$$$\Rightarrow\:{Q}=\int_{\mathrm{0}} ^{\infty} \frac{\left(\frac{{u}}{{lnc}}\right)^{{c}} }{{e}^{{u}} }×\frac{{du}}{{lnc}}\:=\:\frac{\mathrm{1}}{\left({lnc}\right)^{{c}+\mathrm{1}} }\int_{\mathrm{0}} ^{\infty} \frac{{u}^{{c}} }{{e}^{{u}} }{du} \\ $$$$=\frac{\mathrm{1}}{\left({lnc}\right)^{{c}+\mathrm{1}} }\int_{\mathrm{0}} ^{\infty} {u}^{{c}−\mathrm{1}+\mathrm{1}} {e}^{−{u}} \:{du} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\therefore\because\:{Q}\:=\:\int_{\mathrm{0}} ^{\infty} \frac{{x}^{{c}} }{{c}^{{x}} }{dx}=\frac{\Gamma\left({c}+\mathrm{1}\right)}{\left({lnc}\right)^{{c}+\mathrm{1}} }.. \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:……….\mathscr{L}{e}\:{puissant}……… \\ $$