1-calculate-0-e-at-dt-with-a-gt-0-2-by-using-fubinni-theorem-find-the-value-of-0-e-t-e-xt-t-dt-with-x-gt-0-

Question Number 53270 by Abdo msup. last updated on 19/Jan/19

1) c a l c u l a t e \int_{0}^{\infty} e^{- a t} d t w i t h a > 0

2) b y u s i n g f u b i n n i t h e o r e m f i n d t h e v a l u e o f

\int_{0}^{\infty} \frac{e^{- t} - e^{- x t}}{t} d t w i t h x > 0 .

Commented by maxmathsup by imad last updated on 20/Jan/19

1) \int_{0}^{\infty} e^{- a t} d t = {[- \frac{1}{a} e^{- a t}]}_{0}^{+ \infty} = \frac{1}{a}

2) \Rightarrow \int_{1}^{x} \frac{d a}{a} = l n (x) w e t a k e x > 0 b u t

\int_{1}^{x} \frac{d a}{a} = \int_{1}^{x} (\int_{0}^{\infty} e^{- a t} d t) d a = \int_{0}^{\infty} (\int_{1}^{x} e^{- a t} d a) d t (f u b i n i t h e o r e m)

= \int_{0}^{\infty} ({[- \frac{1}{t} e^{- a t}]}_{a = 1}^{a = x}) d t = \int_{0}^{\infty} \frac{e^{- t} - e^{- x t}}{t} d t \Rightarrow

\int_{0}^{\infty} \frac{e^{- t} - e^{- x t}}{t} d t = l n (x) w i t h x > 0

Answered by kaivan.ahmadi last updated on 19/Jan/19

1) u = - at \Rightarrow du = - adt

\frac{- 1}{a} \int e^{u} du = \frac{- 1}{a} e^{u} = \frac{- 1}{a} e^{- at} ∣_{0}^{\infty} = \frac{- 1}{a} (e^{- \infty} - e^{0}) =

\frac{- 1}{a} (0 - 1) = \frac{1}{a}