Question

Section 8.1 Introduction to the Laplace Transforms

Problem 6.

Prove that if

`f(t)↔ F(s)`
then

`{t}^(k) f(t)↔ {( - 1)}^(k) {F}^(k) (s).`
Hint: Assume that it's permissable to the differentiate the integral

`F(s)={∫}^( \infty ) _(0) {e}^( - st) f(t)dt`
with respect to s under the integral sign.

​

Answer 1

Let k = 1, for a start. By definition of the Laplace transform,

`\displaystyle F(s) = \int_0^\infty f(t) e^(-st) \, dt`

Differentiate both sides with respect to s :

`\displaystyle F'(s) = (d)/(ds) \int_0^\infty f(t) e^(-st) \, dt`

`\displaystyle F'(s) = \int_0^\infty (\partial)/(\partial s)  \left[f(t) e^(-st)\right] \, dt`

`\displaystyle F'(s) = \int_0^\infty -t f(t) e^(-st) \, dt`

so that
`t f(t) \leftrightarrow (-1)^1 F^((1))(s) = -F'(s)` is indeed true.

Suppose the claim is true for arbitrary integer k = n, which is to say that
`t^n f(t) \leftrightarrow (-1)^n F^((n))(s)`. Then if k = n + 1, we have

`F^((n+1))(s) = (d)/(ds) F^((n))(s)`

Consider the two cases:

• If k = n + 1 is even, then n is odd, so

`(-1)^n F^((n))(s) = -F^((n))(s) \leftrightarrow t^n f(t)`

and it follows that

`F^((n+1))(s) = \displaystyle (d)/(ds) \left[-\int_0^\infty t^n f(t) e^(-st) \, dt \right]`

`F^((n+1))(s) = \displaystyle -\int_0^\infty (\partial)/(\partial s)\left[ t^n f(t) e^(-st) \right] \, dt`

`F^((n+1))(s) = \displaystyle (d)/(ds) \left[-\int_0^\infty t^n f(t) e^(-st) \, dt \right]`

`F^((n+1))(s) = \displaystyle \int_0^\infty t^(n+1) f(t) e^(-st) \, dt`

`\implies F^((n+1))(s) = (-1)^(n+1) F^((n+1))(s) \leftrightarrow t^(n+1)f(t)`

• Otherwise, if k = n + 1 is odd, then n is even, so

`(-1)^n F^((n))(s) = F^((n))(s) \leftrightarrow t^n f(t)`

The rest of the proof is the same as the previous case.

So we've proved the claim by induction:

•
`t f(t) \leftrightarrow -F(s)`, and

•
`\bigg(t^n f(t) \leftrightarrow (-1)^n F^((n))(s)\bigg) \implies \bigg(t^(n+1) f(t) \leftrightarrow (-1)^(n+1) F^((n+1))(s)\bigg)`