Find the Laplace transform of f(t)=te^at.
Answer: F(s)=1/(s-a)^2, s>a
Find the Laplace transform of f(t)=te^at.
Answer: F(s)=1/(s-a)^2, s>a
Lf(t) = ∫_{0∞} t e^{t}e^{-st}dt = ∫te ^{ -(s-a)t} dt (from 0 to ∞).
Divide and multiply t and dt in the integral by (s-a), introduce new variable z = (s-a)t to come up with the following integral:
Lf(t) = (1/(s-a)^{2}) ∫(0,∞) ze^{-z} dz.
You can make sure that the integral (if integrated in parts) equals 1 (-e^{-z} with upper limit z=∞ minus -^{e-z} with z=0). Thus, your answwer is
Lf(t) = 1/(s-a)^{2}