The Laplace transform of the integral of f(t) is given by the following formula L{∫0t f(u) du} = F(s)/s. Here we prove this Laplace formula of integrals.
Table of Contents
Statement
The Laplace transform of an integral is given by
$L\Big\{ \int_0^t f(u) du \Big\} = \dfrac{F(s)}{s}$
where L{f(t)} = F(s) denotes the Laplace transform of f(t).
Proof
Let us put
g(t) = $\int_0^t f(u) du$.
Note that g(0)=0. By the fundamental theorem of Calculus, we have that
g$’$(t)=f(t).
Now, using the Laplace Transform of derivatives, we have that
L{g$’$(t)} = -g(0) + s L{g(t)}
⇒ L{f(t)} = -0 + s L{g(t)} as we have g(0)=0 and g$’$(t)=f(t).
⇒ F(s) = s L{g(t)}
⇒ L{g(t)} = $\dfrac{F(s)}{s}$
This proves that L{∫0t f(u) du} = F(s)/s and this is the Laplace transform of integrals.
Have You Read These?
Laplace Transform: Definition, Table, Formulas, Properties
Solved problems of Laplace transforms
Laplace Transform of derivatives
FAQs
Answer: If L{f(t)} = F(s), then the Laplace of the integral of f(t) is given by L{∫0t f(u) du} = F(s)/s.
