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Question Number 44318 by abdo.msup.com last updated on 26/Sep/18

l e t f (x) = \int_{x}^{+ \infty} \frac{e^{- t}}{t} d t

1) c a l c u l a t e f^{^{'}} (x)

2) f i n d a e q u i v a l e n t o f f (x) w h e n

x \to + \infty .

Commented by tanmay.chaudhury50@gmail.com last updated on 27/Sep/18

Commented by maxmathsup by imad last updated on 27/Sep/18

w e h a v e f (x) = \int_{x}^{1} \frac{e^{- t}}{t} d t + \int_{1}^{+ \infty} \frac{e^{- t}}{t} d t = \int_{1}^{+ \infty} \frac{e^{- t}}{t} d t - \int_{1}^{x} \frac{e^{- t}}{t} d t \Rightarrow

f^{^{'}} (x) = - \frac{e^{- x}}{x} (x > 0)

2) l e t i n t e g r a t e b y p a r t s

f (x) = {[- \frac{e^{- t}}{t}]}_{x}^{+ \infty} - \int_{x}^{+ \infty} \frac{e^{- t}}{t^{2}} d t = \frac{e^{- x}}{x} - \int_{x}^{+ \infty} \frac{e^{- t}}{t^{2}} d t \Rightarrow

∣ f (x) - \frac{e^{- x}}{x} ∣ =∣ \int_{x}^{+ \infty} \frac{e^{- t}}{t^{2}} d t ∣ ⩽ \int_{x}^{+ \infty} \frac{d t}{t^{2}} = {[- \frac{1}{t}]}_{x}^{+ \infty} = \frac{1}{x} \to 0 (x \to + \infty) \Rightarrow

f (x) \sim \frac{e^{- x}}{x} (x \to + \infty) .

Answered by tanmay.chaudhury50@gmail.com last updated on 27/Sep/18

\frac{d f (x)}{d x} = - \frac{e^{- x}}{x} \times \frac{d x}{d x}

f^{^{'}} (x) = - \frac{e^{- x}}{x}