# The Laplace transform of the product of two functions is the product of the Laplace transforms of each given function. True of False? Explain why.

The Laplace transform of the product of two functions is the product of the Laplace transforms of each given function. True of False? Explain why.
You can still ask an expert for help

## Want to know more about Laplace transform?

• Live experts 24/7
• Questions are typically answered in as fast as 30 minutes
• Personalized clear answers

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

unett

Answer: False. The Laplace transform of the product of two functions is not the product of the Laplace transforms of the each given function.
Explanation: Laplace transform of a function f(t) is given by:
$F\left(s\right)={\int }_{0}^{\mathrm{\infty }}{e}^{-st}f\left(t\right)dt$
Then, for any two functions f(t) and g(t) the Laplace transform of the product of two functions is:
$L\left\{f\left(t\right)\cdot g\left(t\right)\right\}$
Integral form is: ${\int }_{0}^{\mathrm{\infty }}{e}^{-st}\left[f\left(t\right)\cdot g\left(t\right)\right]dt$
Which is not equal to ${\int }_{0}^{\mathrm{\infty }}{e}^{-st}f\left(t\right)dt\cdot {\int }_{0}^{\mathrm{\infty }}{e}^{-st}g\left(t\right)dt$
Therefore, the given statement if false.
But, it is equal to the product of the convolution of the Laplace transforms of the each function.
If F(s) and G(s) are the Laplace transforms of the functions f(t) and g(t), then,
Laplace transform of $f\left(t\right)\cdot g\left(t\right)$ is:
$L\left\{f\left(t\right)\cdot g\left(t\right)\right\}=F\left(s\right)\star G\left(s\right)$
Where, indicates convolution of the two functions, given by
$f\star g\left(t\right)={\int }_{u=0}^{t}f\left(u\right)g\left(t-u\right)du$
Hence, the given statement is false.