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Laplace Transform Basics
.

by Euler’s formula. Performing a double integration by parts on both these

integrals gives

∫ e xt
st
e dt = ((x cos yt + y sin yt) + i(x sin yt − y cos yt)) .
2
x + y 2
Now the right-hand side of (1.3) can be expressed as

xt
e st e (x+iy)t e (cos yt + i sin yt)(x − iy)
= = =
2
s x + iy x + y 2
e xt
= ((x cos yt + y sin yt) + i(x sin yt − y cos yt)) ,
2
x + y 2

which equals the left-hand side, and (1.3) follows.
Furthermore, we obtain the result of (1.2) for s complex if we take Re(s) =
x > 0, since then

lim |e −sτ | = lim e −xτ = 0,
τ→∞ τ→∞

. . . . killing off the limit term in (1.2).
.

at
Example 1.2, f(t) = e . Evaluate Lf(t) for s > a.
. . . . .
.
Solution.

∞ ∞
∫ ∫ ∞
1 1
F(s) = e −st at e −(s−a)t dt = − e −(s−a)t = .
e dt =
s − a s − a
0
0 0
. . . .
.
n
Example 1.3, For a positive integer n, f(t) = t . Evaluate Lf(t).
. . . . .
.
Solution.

∞ ∞
∫ ∫ ( ) ′
−1
n
n
t dt =
F(s) = L[t ] = e −st n e −st t dt =
s
0 0
∞ ∞
( ) ( ) ∫ ∫
−1 ∞ −n n n
n
= e −st t − e −st n−1 dt = e −st n−1 dt = L[t n−1 ].
t
t
s 0 s s s
0 0
n
n
So we’ve found that L[t ] = L[t n−1 ]. We can now use this formula over
. . . . s
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