Let k = 1, for a start. By definition of the Laplace transform,
[tex]\displaystyle F(s) = \int_0^\infty f(t) e^{-st} \, dt[/tex]
Differentiate both sides with respect to s :
[tex]\displaystyle F'(s) = \frac{d}{ds} \int_0^\infty f(t) e^{-st} \, dt[/tex]
[tex]\displaystyle F'(s) = \int_0^\infty \frac{\partial}{\partial s} \left[f(t) e^{-st}\right] \, dt[/tex]
[tex]\displaystyle F'(s) = \int_0^\infty -t f(t) e^{-st} \, dt[/tex]
so that [tex]t f(t) \leftrightarrow (-1)^1 F^{(1)}(s) = -F'(s)[/tex] is indeed true.
Suppose the claim is true for arbitrary integer k = n, which is to say that [tex]t^n f(t) \leftrightarrow (-1)^n F^{(n)}(s)[/tex]. Then if k = n + 1, we have
[tex]F^{(n+1)}(s) = \dfrac{d}{ds} F^{(n)}(s)[/tex]
Consider the two cases:
• If k = n + 1 is even, then n is odd, so
[tex](-1)^n F^{(n)}(s) = -F^{(n)}(s) \leftrightarrow t^n f(t)[/tex]
and it follows that
[tex]F^{(n+1)}(s) = \displaystyle \frac{d}{ds} \left[-\int_0^\infty t^n f(t) e^{-st} \, dt \right][/tex]
[tex]F^{(n+1)}(s) = \displaystyle -\int_0^\infty \frac{\partial}{\partial s}\left[ t^n f(t) e^{-st} \right] \, dt[/tex]
[tex]F^{(n+1)}(s) = \displaystyle \frac{d}{ds} \left[-\int_0^\infty t^n f(t) e^{-st} \, dt \right][/tex]
[tex]F^{(n+1)}(s) = \displaystyle \int_0^\infty t^{n+1} f(t) e^{-st} \, dt[/tex]
[tex]\implies F^{(n+1)}(s) = (-1)^{n+1} F^{(n+1)}(s) \leftrightarrow t^{n+1}f(t)[/tex]
• Otherwise, if k = n + 1 is odd, then n is even, so
[tex](-1)^n F^{(n)}(s) = F^{(n)}(s) \leftrightarrow t^n f(t)[/tex]
The rest of the proof is the same as the previous case.
So we've proved the claim by induction:
• [tex]t f(t) \leftrightarrow -F(s)[/tex], and
• [tex]\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)[/tex]
![Section 8.1 Introduction to the Laplace Transforms Problem 6.Prove that if [tex]f(t)↔ F(s)[/tex]then [tex] {t}^{k} f(t)↔ {( - 1)}^{k} {F}^{k} (s).[/tex]Hint: As class=](https://us-static.z-dn.net/files/d36/d7b3cad26eac16346a983b8b315a9684.jpg)
