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Prove: Let *F*(t) be a continuous, piecewise regular function of exponential order. Furthermore,
let *F *'(t) be sectionally continuous. Then if L[*F* (t)] = f(s), the Laplace transform of *F *'(t) is
given by the formula

L[*F *'(t)] = s*f* (s) - *F*(0)

Proof. We will first prove this theorem for the case when *F *'(t) is continuous (as opposed to
sectionally continuous).

Case 1. *F *'(t) is continuous.

We shall now integrate the rightmost member by parts using the formula

where u = e^{-st}, dv = *F *' (t) dt, v = *F*(t), du = -se^{-st} dt

Now because F(t) is of exponential order

for s > α_{0}, the abscissa of convergence of *F*(h). Thus 2) becomes

Case 2. *F *'(t) contains one discontinuity. Assume that *F *'(t) has a discontinuity at t = t_{0}. See
Fig. 1. Then

Integrating the two integrals by parts gives

We now note that in the limit the sum of the two integrals is

The first evaluated portion gives

and the second gives simply

because, since F(t) is of exponential order, the
contribution from the upper limit is zero. Now F(t)
was assumed to be continuous and hence at t = t_{0} its
right- and left-hand limits must be equal. Thus the
terms

cancel, giving as the final result

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