Question Number 87526 by mathmax by abdo last updated on 04/Apr/20
![calculate ∫_0 ^∞ e^(−[nx]) dx](https://www.tinkutara.com/question/Q87526.png)
$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−\left[{nx}\right]} \:{dx} \\ $$
Commented by mathmax by abdo last updated on 05/Apr/20
![I_n =∫_0 ^∞ e^(−[nx]) dx changement nx =t give I_n =(1/n) ∫_0 ^∞ e^(−[t]) dt =(1/n)Σ_(k=0) ^∞ ∫_k ^(k+1) e^(−k) dt =(1/n)Σ_(k=0) ^∞ e^(−k) =(1/n)Σ_(k=0) ^∞ (e^(−1) )^k =(1/n)×(1/(1−e^(−1) )) =(1/n)×(e/(e−1)) ⇒ I_n =(e/(n(e−1)))](https://www.tinkutara.com/question/Q87620.png)
$${I}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:{e}^{−\left[{nx}\right]} \:{dx}\:\:\:{changement}\:{nx}\:={t}\:{give} \\ $$$${I}_{{n}} =\frac{\mathrm{1}}{{n}}\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−\left[{t}\right]} \:{dt}\:=\frac{\mathrm{1}}{{n}}\sum_{{k}=\mathrm{0}} ^{\infty} \:\int_{{k}} ^{{k}+\mathrm{1}} \:{e}^{−{k}} \:{dt} \\ $$$$=\frac{\mathrm{1}}{{n}}\sum_{{k}=\mathrm{0}} ^{\infty} \:{e}^{−{k}} \:=\frac{\mathrm{1}}{{n}}\sum_{{k}=\mathrm{0}} ^{\infty} \:\left({e}^{−\mathrm{1}} \right)^{{k}} \:=\frac{\mathrm{1}}{{n}}×\frac{\mathrm{1}}{\mathrm{1}−{e}^{−\mathrm{1}} }\:=\frac{\mathrm{1}}{{n}}×\frac{{e}}{{e}−\mathrm{1}} \\ $$$$\Rightarrow\:{I}_{{n}} =\frac{{e}}{{n}\left({e}−\mathrm{1}\right)} \\ $$