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 :

$\displaystyle F'(s) = \frac{d}{ds} \int_0^\infty f(t) e^{-st} \, dt$

$\displaystyle F'(s) = \int_0^\infty \frac{\partial}{\partial s}  \left[f(t) e^{-st}\right] \, dt$

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

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

$F^{(n+1)}(s) = \dfrac{d}{ds} F^{(n)}(s)$

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

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

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

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

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

$\implies F^{(n+1)}(s) = (-1)^{n+1} F^{(n+1)}(s) \leftrightarrow t^{n+1}f(t)$

• 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

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