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Question Number 87526 by mathmax by abdo last updated on 04/Apr/20

calculate ∫_0 ^∞  e^(−[nx])  dx

$${calculate}\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−\left[{nx}\right]} \:{dx} \\ $$

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)))

$${I}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:{e}^{−\left[{nx}\right]} \:{dx}\:\:\:{changement}\:{nx}\:={t}\:{give} \\ $$$${I}_{{n}} =\frac{\mathrm{1}}{{n}}\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−\left[{t}\right]} \:{dt}\:=\frac{\mathrm{1}}{{n}}\sum_{{k}=\mathrm{0}} ^{\infty} \:\int_{{k}} ^{{k}+\mathrm{1}} \:{e}^{−{k}} \:{dt} \\ $$$$=\frac{\mathrm{1}}{{n}}\sum_{{k}=\mathrm{0}} ^{\infty} \:{e}^{−{k}} \:=\frac{\mathrm{1}}{{n}}\sum_{{k}=\mathrm{0}} ^{\infty} \:\left({e}^{−\mathrm{1}} \right)^{{k}} \:=\frac{\mathrm{1}}{{n}}×\frac{\mathrm{1}}{\mathrm{1}−{e}^{−\mathrm{1}} }\:=\frac{\mathrm{1}}{{n}}×\frac{{e}}{{e}−\mathrm{1}} \\ $$$$\Rightarrow\:{I}_{{n}} =\frac{{e}}{{n}\left({e}−\mathrm{1}\right)} \\ $$

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