Laplace transforms of derivatives

where

As shown in the parent entry (http://planetmath.org/LaplaceTransformOfDerivative),the Laplace transform of the first derivative of a Laplace-transformablefunction $f(t)$ is got from the formula

$ℒ\{f^{\prime }(t)\}=sF(s)-\underset{t\to 0+}{lim}f(t).$(1)

The rule can be applied also to the function $f^{\prime }(t)$:

$$ℒ\{f^{\prime \prime }(t)\}=s[sF(s)-\underset{t\to 0+}{lim}f(t)]-\underset{t\to 0+}{lim}{f}^{\prime }(t)=s^{2}F(s)-sf(0^{+})-f^{\prime }(0^{+})$$

Here the short notation $0^{+}$ has been used for the right limits.

Further, one can use the rule to $f^{\prime \prime }(t)$, getting

$$ℒ\{f^{\prime \prime \prime }(t)\}=s[s^{2}F(s)-sf(0^{+})-f^{\prime }(0^{+})]-f^{\prime \prime }(0^{+})=s^{3}F(s)-s^{2}f(0^{+})-s{f}^{\prime }(0^{+})-f^{\prime \prime }(0^{+}).$$

Continuing similarly, one comes to the general formula

$ℒ\{f^{(n)}(t)\}=s^{n}F(s)-s^{n-1}f(0^{+})-s^{n-2}{f}^{\prime }(0^{+})-\mathrm{\dots }-f^{(n-1)}(0^{+}).$(2)

Use of (2) requires that $f(t)$, $f^{\prime }(t)$, $f^{\prime \prime }(t)$, …, $f^{(n)}(t)$ areLaplace-transformable and that $f(t)$, $f^{\prime }(t)$, $f^{\prime \prime }(t)$, …,$f^{(n-1)}(t)$ are continuous when  $t>0$ (not onlypiecewise continuous).

Remark.  Suppose that $f(t)$ and $f^{\prime }(t)$ areLaplace-transformable and that $f(t)$ is continuous for $t>0$  except the point  $t=a$  where the function has afinite jump discontinuity.  Then

$$ℒ\{f^{\prime }(t)\}=sF(s)-f(0^{+})-e^{-as}(\underset{s\to a+}{lim}f(s)-\underset{s\to a-}{lim}f(s)).$$

Application.  Derive the Laplace transform of $\sin at$ using thederivatives of sine (cf. Laplace transform of cosine and sine).

We have

$$f(t):=\sin at,f^{\prime }(t)=a\cos at,f^{\prime \prime }(t)=-a^2\sin at.$$

Using (2) with  $n=2$  we obtain

$$ℒ\{-a^2\sin at\}=s^2ℒ\{\sin at\}-s\sin 0-a\cos 0,$$

i.e.

$$-a^2ℒ\{\sin at\}=s^2ℒ\{\sin at\}-a,$$

which implies

$$ℒ\{\sin at\}=\frac{a}{s^2+a^2}.$$