Question

Laplace of integral 0 to t (1-e^-2t) /t dt

Answer 1

Answer

Step-by-step explanation:

L(1- e^-2) = 1/s - 1/s+2

Now :

L(1- e^-2) /t = integration from s to infinity (1/s - 1/s+2) ds

= [log s - log s+2] limit s to infinity

=[log (s/s+2)] limit s to infinity

=[log 1/(1+2/s)] limit s to infinity (dividing numerator and denominator by s)

above expression can be written as

= [log1 - log (1+2/s)] limit s to infinity

= -[log (1+2/s)] limit s to infinity ( as log 1= 0)

on putting limits we get

log[(s+2)/s]

Now:

L (integral 0 to t (1-e^-2t) /t dt) = 1/s × log[(s+2)/s] (ANS)