find-the-laplace-transform-of-f-t-ln-t-

Question Number 177493 by ali009 last updated on 06/Oct/22

f i n d t h e l a p l a c e t r a n s f o r m o f

f (t) = l n (t)

Commented by JDamian last updated on 06/Oct/22

Is it suitable for Laplace Transform?

Commented by ali009 last updated on 06/Oct/22

y e s i w i l l p u t m y s o l u t i o n l a t e r

Answered by ali009 last updated on 06/Oct/22

i n g e n e r a l L [t^{v} l n (t)] = \int_{0}^{\infty} t^{v} l n (t) e^{- s t} d t

l e t x = s t d x = s d t

L [t^{v} l n (t)] = \frac{1}{s} \int_{0}^{\infty} e^{- x} {(\frac{x}{s})}^{v} l n (\frac{x}{s}) d x

= \frac{1}{s^{v + 1}} \int_{0}^{\infty} e^{- x} x^{v} (l n (x) - l n (s)) d x

= \frac{1}{s^{v + 1}} (\int_{0}^{\infty} e^{- x} l n (x) x^{v} d x - l n (s) \int_{0}^{\infty} e^{- x} x^{v})

= \frac{1}{s^{v + 1}} (\frac{d}{d v} \int e^{- x} x^{v} d x - l n (s) Γ (v + 1))

= \frac{1}{s^{v + 1}} (\frac{d}{d v} Γ (v + 1) - l n (s) Γ (v + 1))

= \frac{1}{s^{v + 1}} Γ (v + 1) (\frac{Γ^{'} (v + 1)}{Γ (v + 1)} - l n (s))

= \frac{1}{s^{v + 1}} Γ (v + 1) (ψ (v + 1) - l n (s))

w h e n t = 0

L [l n (t)] = \frac{1}{s^{v + 1}} Γ (1) (ψ (1) - l n (s))

L [l n (t)] = \frac{- γ - l n (s)}{s}

γ i s {E u l e r}^{'} s c o n s t a n t

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