Chapter 6 Laplace Transform 365
Then by definition
L(
f′(t))
=
∫∞
0e−stf′(t)dt.Integrating by parts, we get
L(
f′(t))
=e−stf(t)∣∣∞
0 −
∫∞
0(−s)e−stf(t)dt.Iff(t)is of exponential order,e−stf(t)must tend to 0 asttends to infinity (for
large enoughs), so the foregoing equation becomes
L(
f′(t))
=−f( 0 )+s∫∞
0e−stf(t)dt=−f( 0 )+sL(
f(t))
.
(Iff(t)has a jump att=0,f( 0 )is to be interpreted asf( 0 +).)
Similarly, iffandf′are continuous,f′′is sectionally continuous; and if all
three functions are exponential order, then
L(
f′′(t))
=−f( 0 )+sL(
f′(t))
=−f( 0 )−sf( 0 )+s^2 L(
f(t))
.
An easy generalization extends this formula to thenth derivative,
L[
f(n)(t)]
=−f(n−^1 )( 0 )−sf(n−^2 )( 0 )−···−sn−^1 f( 0 )+snL(
f(t))
, (4)
on the assumption thatfand its firstn−1 derivatives are continuous,f(n)is
sectionally continuous, and all are of exponential order.
We may apply Eq. (4) to the functionf(t)=tk,kbeing a nonnegative inte-
ger. Here we have
f( 0 )=f′( 0 )=···=f(k−^1 )( 0 )= 0 , f(k)( 0 )=k!, f(k+^1 )(t)= 0.Thus, Eq. (4) withn=k+1yields
0 =−k!+sk+^1 L(
tk)
,
or
L(
tk)
= k!
sk+^1.
A different application of the derivative rule is used to transform integrals. If
f(t)is sectionally continuous, then
∫t
0 f(t′)dt′is a continuous function, equal
to zero att=0, and has derivativef(t).Hence
L(
f(t))
=sL[∫t0f(t′)dt′