Question Number 87526 by mathmax by abdo last updated on 04/Apr/20
$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−\left[{nx}\right]} \:{dx} \\ $$
Commented by mathmax by abdo last updated on 05/Apr/20
$${I}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:{e}^{−\left[{nx}\right]} \:{dx}\:\:\:{changement}\:{nx}\:={t}\:{give} \\ $$$${I}_{{n}} =\frac{\mathrm{1}}{{n}}\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−\left[{t}\right]} \:{dt}\:=\frac{\mathrm{1}}{{n}}\sum_{{k}=\mathrm{0}} ^{\infty} \:\int_{{k}} ^{{k}+\mathrm{1}} \:{e}^{−{k}} \:{dt} \\ $$$$=\frac{\mathrm{1}}{{n}}\sum_{{k}=\mathrm{0}} ^{\infty} \:{e}^{−{k}} \:=\frac{\mathrm{1}}{{n}}\sum_{{k}=\mathrm{0}} ^{\infty} \:\left({e}^{−\mathrm{1}} \right)^{{k}} \:=\frac{\mathrm{1}}{{n}}×\frac{\mathrm{1}}{\mathrm{1}−{e}^{−\mathrm{1}} }\:=\frac{\mathrm{1}}{{n}}×\frac{{e}}{{e}−\mathrm{1}} \\ $$$$\Rightarrow\:{I}_{{n}} =\frac{{e}}{{n}\left({e}−\mathrm{1}\right)} \\ $$