Question Number 41351 by vajpaithegrate@gmail.com last updated on 06/Aug/18
![∫_0 ^∞ [(5/e^x )]dx=](Q41351.png)
$$\int_{\mathrm{0}} ^{\infty} \left[\frac{\mathrm{5}}{\mathrm{e}^{\mathrm{x}} }\right]\mathrm{dx}= \\ $$
Commented by math khazana by abdo last updated on 06/Aug/18
![let I = ∫_0 ^∞ [(5/e^x )]dx changement (5/e^x ) =t give e^x =(5/t) ⇒ x=ln((5/t)) =ln(5)−ln(t) ⇒dx=−(1/t)dt I =− ∫_5 ^0 [t]((−dt)/t) = −∫_0 ^5 (([t])/t) dt =−Σ_(k=0) ^4 ∫_k ^(k+1) (k/t) dt =−Σ_(k=1) ^4 ∫_k ^(k+1) (k/t)dt =−Σ_(k=1) ^4 k {ln(k+1)−ln(k)} =Σ_(k=1) ^4 k{ln(k)−ln(k+1)} =−ln(2)+2{ln(2)−ln(3)} +3{ln(3)−2ln(2)} +4{2ln(2)−ln(5) =−ln2)+2ln(2)−2ln(3)+3ln(3)−6ln(2) +8ln(2)−4ln(5) I =3ln(2) +3ln(3)−4ln(5) .](Q41362.png)
$${let}\:{I}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\left[\frac{\mathrm{5}}{{e}^{{x}} }\right]{dx}\:{changement}\:\:\frac{\mathrm{5}}{{e}^{{x}} }\:={t}\:{give} \\ $$$${e}^{{x}} =\frac{\mathrm{5}}{{t}}\:\Rightarrow\:{x}={ln}\left(\frac{\mathrm{5}}{{t}}\right)\:={ln}\left(\mathrm{5}\right)−{ln}\left({t}\right)\:\Rightarrow{dx}=−\frac{\mathrm{1}}{{t}}{dt} \\ $$$${I}\:=−\:\int_{\mathrm{5}} ^{\mathrm{0}} \:\:\left[{t}\right]\frac{−{dt}}{{t}}\:=\:−\int_{\mathrm{0}} ^{\mathrm{5}} \:\:\frac{\left[{t}\right]}{{t}}\:{dt} \\ $$$$=−\sum_{{k}=\mathrm{0}} ^{\mathrm{4}} \:\:\int_{{k}} ^{{k}+\mathrm{1}} \:\:\frac{{k}}{{t}}\:{dt}\:=−\sum_{{k}=\mathrm{1}} ^{\mathrm{4}} \:\int_{{k}} ^{{k}+\mathrm{1}} \:\frac{{k}}{{t}}{dt} \\ $$$$=−\sum_{{k}=\mathrm{1}} ^{\mathrm{4}} {k}\:\left\{{ln}\left({k}+\mathrm{1}\right)−{ln}\left({k}\right)\right\} \\ $$$$=\sum_{{k}=\mathrm{1}} ^{\mathrm{4}} {k}\left\{{ln}\left({k}\right)−{ln}\left({k}+\mathrm{1}\right)\right\} \\ $$$$=−{ln}\left(\mathrm{2}\right)+\mathrm{2}\left\{{ln}\left(\mathrm{2}\right)−{ln}\left(\mathrm{3}\right)\right\}\:+\mathrm{3}\left\{{ln}\left(\mathrm{3}\right)−\mathrm{2}{ln}\left(\mathrm{2}\right)\right\} \\ $$$$+\mathrm{4}\left\{\mathrm{2}{ln}\left(\mathrm{2}\right)−{ln}\left(\mathrm{5}\right)\right. \\ $$$$\left.=−{ln}\mathrm{2}\right)+\mathrm{2}{ln}\left(\mathrm{2}\right)−\mathrm{2}{ln}\left(\mathrm{3}\right)+\mathrm{3}{ln}\left(\mathrm{3}\right)−\mathrm{6}{ln}\left(\mathrm{2}\right) \\ $$$$+\mathrm{8}{ln}\left(\mathrm{2}\right)−\mathrm{4}{ln}\left(\mathrm{5}\right) \\ $$$${I}\:=\mathrm{3}{ln}\left(\mathrm{2}\right)\:+\mathrm{3}{ln}\left(\mathrm{3}\right)−\mathrm{4}{ln}\left(\mathrm{5}\right)\:. \\ $$$$ \\ $$
Commented by vajpaithegrate@gmail.com last updated on 06/Aug/18
$$\mathrm{thank}\:\mathrm{u}\:\mathrm{sir} \\ $$
Commented by math khazana by abdo last updated on 06/Aug/18
$${nevermind}\:{sir} \\ $$
Answered by MJS last updated on 06/Aug/18
![f(x)=[(5/e^x )] ln (5/6)<x≤ln (5/5) =0 ⇒ f(x)=5 0<x≤ln (5/4) ⇒ f(x)=4 ln (5/4)<x≤ln (5/3) ⇒ f(x)=3 ln (5/3)<x≤ln (5/2) ⇒ f(x)=2 ln (5/2)<x≤ln 5 ⇒ f(x)=1 x>ln 5 ⇒ f(x)=0 ⇒ ∫_0 ^∞ [(5/e^x )]dx=4ln (5/4) +3(ln (5/3) −ln (5/4))+2(ln (5/2) −ln (5/3))+ln 5 −ln (5/2)= =ln ((625)/(24)) ≈3.25970](Q41357.png)
$${f}\left({x}\right)=\left[\frac{\mathrm{5}}{\mathrm{e}^{{x}} }\right] \\ $$$$\mathrm{ln}\:\frac{\mathrm{5}}{\mathrm{6}}<{x}\leqslant\mathrm{ln}\:\frac{\mathrm{5}}{\mathrm{5}}\:=\mathrm{0}\:\Rightarrow\:{f}\left({x}\right)=\mathrm{5} \\ $$$$\mathrm{0}<{x}\leqslant\mathrm{ln}\:\frac{\mathrm{5}}{\mathrm{4}}\:\Rightarrow\:{f}\left({x}\right)=\mathrm{4} \\ $$$$\mathrm{ln}\:\frac{\mathrm{5}}{\mathrm{4}}<{x}\leqslant\mathrm{ln}\:\frac{\mathrm{5}}{\mathrm{3}}\:\Rightarrow\:{f}\left({x}\right)=\mathrm{3} \\ $$$$\mathrm{ln}\:\frac{\mathrm{5}}{\mathrm{3}}<{x}\leqslant\mathrm{ln}\:\frac{\mathrm{5}}{\mathrm{2}}\:\Rightarrow\:{f}\left({x}\right)=\mathrm{2} \\ $$$$\mathrm{ln}\:\frac{\mathrm{5}}{\mathrm{2}}<{x}\leqslant\mathrm{ln}\:\mathrm{5}\:\Rightarrow\:{f}\left({x}\right)=\mathrm{1} \\ $$$${x}>\mathrm{ln}\:\mathrm{5}\:\Rightarrow\:{f}\left({x}\right)=\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\underset{\mathrm{0}} {\overset{\infty} {\int}}\left[\frac{\mathrm{5}}{\mathrm{e}^{{x}} }\right]{dx}=\mathrm{4ln}\:\frac{\mathrm{5}}{\mathrm{4}}\:+\mathrm{3}\left(\mathrm{ln}\:\frac{\mathrm{5}}{\mathrm{3}}\:−\mathrm{ln}\:\frac{\mathrm{5}}{\mathrm{4}}\right)+\mathrm{2}\left(\mathrm{ln}\:\frac{\mathrm{5}}{\mathrm{2}}\:−\mathrm{ln}\:\frac{\mathrm{5}}{\mathrm{3}}\right)+\mathrm{ln}\:\mathrm{5}\:−\mathrm{ln}\:\frac{\mathrm{5}}{\mathrm{2}}= \\ $$$$=\mathrm{ln}\:\frac{\mathrm{625}}{\mathrm{24}}\:\approx\mathrm{3}.\mathrm{25970} \\ $$
Commented by vajpaithegrate@gmail.com last updated on 06/Aug/18
$$\mathrm{thank}\:\mathrm{u}\:\mathrm{sir} \\ $$