∫_0 ^∞ [(5/e^x )]dx=


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Question Number 41351 by vajpaithegrate@gmail.com last updated on 06/Aug/18

$\int_{0}^{\infty} [\frac{5}{e^{x}}] dx =$
Commented by math khazana by abdo last updated on 06/Aug/18
$let I = ∫_0 ^∞ [(5/e^x )]dx changement (5/e^x ) =t give e^x =(5/t) ⇒ x=ln((5/t)) =ln(5)−ln(t) ⇒dx=−(1/t)dt I =− ∫_5 ^0 [t]((−dt)/t) = −∫_0 ^5 (([t])/t) dt =−Σ_(k=0) ^4 ∫_k ^(k+1) (k/t) dt =−Σ_(k=1) ^4 ∫_k ^(k+1) (k/t)dt =−Σ_(k=1) ^4 k {ln(k+1)−ln(k)} =Σ_(k=1) ^4 k{ln(k)−ln(k+1)} =−ln(2)+2{ln(2)−ln(3)} +3{ln(3)−2ln(2)} +4{2ln(2)−ln(5) =−ln2)+2ln(2)−2ln(3)+3ln(3)−6ln(2) +8ln(2)−4ln(5) I =3ln(2) +3ln(3)−4ln(5) .$
$l e t I = \int_{0}^{\infty} [\frac{5}{e^{x}}] d x c h a n g e m e n t \frac{5}{e^{x}} = t g i v e$ $e^{x} = \frac{5}{t} \Rightarrow x = l n (\frac{5}{t}) = l n (5) - l n (t) \Rightarrow d x = - \frac{1}{t} d t$ $I = - \int_{5}^{0} [t] \frac{- d t}{t} = - \int_{0}^{5} \frac{[t]}{t} d t$ $= - \sum_{k = 0}^{4} \int_{k}^{k + 1} \frac{k}{t} d t = - \sum_{k = 1}^{4} \int_{k}^{k + 1} \frac{k}{t} d t$ $= - \sum_{k = 1}^{4} k {l n (k + 1) - l n (k)}$ $= \sum_{k = 1}^{4} k {l n (k) - l n (k + 1)}$ $= - l n (2) + 2 {l n (2) - l n (3)} + 3 {l n (3) - 2 l n (2)}$ $+ 4 {2 l n (2) - l n (5)$ $= - l n 2) + 2 l n (2) - 2 l n (3) + 3 l n (3) - 6 l n (2)$ $+ 8 l n (2) - 4 l n (5)$ $I = 3 l n (2) + 3 l n (3) - 4 l n (5) .$
Commented by vajpaithegrate@gmail.com last updated on 06/Aug/18

$thank u sir$
Commented by math khazana by abdo last updated on 06/Aug/18

$n e v e r m i n d s i r$
	Answered by MJS last updated on 06/Aug/18

	$f (x) = [\frac{5}{e^{x}}]$ $\ln \frac{5}{6} < x ⩽ \ln \frac{5}{5} = 0 \Rightarrow f (x) = 5$ $0 < x ⩽ \ln \frac{5}{4} \Rightarrow f (x) = 4$ $\ln \frac{5}{4} < x ⩽ \ln \frac{5}{3} \Rightarrow f (x) = 3$ $\ln \frac{5}{3} < x ⩽ \ln \frac{5}{2} \Rightarrow f (x) = 2$ $\ln \frac{5}{2} < x ⩽ \ln 5 \Rightarrow f (x) = 1$ $x > \ln 5 \Rightarrow f (x) = 0$ $\Rightarrow$ $\underset{0}{\int^{\infty}} [\frac{5}{e^{x}}] d x = 4 \ln \frac{5}{4} + 3 (\ln \frac{5}{3} - \ln \frac{5}{4}) + 2 (\ln \frac{5}{2} - \ln \frac{5}{3}) + \ln 5 - \ln \frac{5}{2} =$ $= \ln \frac{625}{24} \approx 3.25970$
	Commented by vajpaithegrate@gmail.com last updated on 06/Aug/18

	$thank u sir$
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