Table of Common Laplace Transforms

L {c} = cs, c a constant L {tn} = n!

sn+1 , n ≥ 1

L{t−1/2

}=

√πs

L{t1/2

}=

√π

2s3/2

L {sin kt} = ks2+k2

L {cos kt} = ss2+k2

L{sin2 kt

}= 2k2

s(s2+4k2)L {cos2 kt} = s2+2k2

s(s2+4k2)

L {eat} = 1s−a

L {sinh kt} = ks2−k2

L {cosh kt} = ss2−k2

L{sinh2 kt

}= 2k2

s(s2−4k2)

L{cosh2 kt

}= s2−2k2

s(s2−4k2)L {tneat} = n!

(s−a)n+1 , n a positive integer

L {eat sin kt} = k(s−a)2+k2

L {eat cos kt} = s−a(s−a)2+k2

L {eat sinh kt} = k(s−a)2−k2

L {eat cosh kt} = s−a(s−a)2−k2

L {t sin kt} = 2ks(s2+k2)2

L {t cos kt} = s2−k2

(s2+k2)2

L {sin kt+ kt cos kt} = 2ks2

(s2+k2)2L {sin kt− kt cos kt} = 2k3

(s2+k2)2

L {t sinh kt} = 2ks(s2−k2)2

L {t cosh kt} = s2+k2

(s2−k2)2

L{

eat−ebt

a−b

}= 1

(s−a)(s−b)L

{aeat−bebt

a−b

}= s

(s−a)(s−b)

L {1− cos kt} = k2

s(s2+k2)L {kt− sin kt} = k3

s2(s2+k2)

L{

a sin bt−b sin atab(a2−b2)

}= 1

(s2+a2)(s2+b2)L

{cos bt−cos at

a2−b2

}= s

(s2+a2)(s2+b2)

L {sin kt sinh kt} = 2k2ss4+4k4

L {sin kt cosh kt} = k(s2+2k2)s4+4k4

L {cos kt sinh kt} = k(s2−2k2)s4+4k4

L {cos kt cosh kt} = s3

s4+4k4

L{

ebt−eat

t

}= ln s−a

s−bL

{2(1−cos kt)

t

}= ln s2+k2

s2

L{

2(1−cosh kt)t

}= ln s2−k2

s2L

{sin at

t

}= arctan

(as

)

L {eatf(t)} = F (s− a) L {U (t− a)} = e−as

s

L {f(t− a)U (t− a)} = e−asF (s) L {g(t)U (t− a)} = e−asL {g(t+ a)}

L{∫ t

0f(τ)g(t− τ) dτ

}= F (s)G(s) L

{f (n)(t)

}= snF (s)−sn−1f(0)−sn−2f ′(0)−· · ·−f (n−1)(0)

Department of Mathematics, Sinclair Community College, Dayton, OH

Derivatives of Transforms

If F (s) = L {f(t)} and n = 1, 2, 3, . . ., then L {tnf(t)} = (−1)ndn

dsnF (s).

General Formulas for the Unit Step Function

f(t) =

{g(t), 0 ≤ t < ah(t), t ≥ a

= g(t)− g(t)U (t− a) + h(t)U (t− a)

f(t) =

⎧⎨⎩

g(t), 0 ≤ t < aj(t), a ≤ t < bh(t), t ≥ b

= g(t) [1− U (t− a)] + j(t) [U (t− a)− U (t− b)] + h(t)U (t− b)

Transform of an Integral

When g(t) = 1 and L {g(t)} = G(s) = 1s, then L

{∫ t

0

f(τ) dτ

}=

F (s)

s.

Inverse Laplace Transform Formulas for Rational FunctionsWhen taking the inverse Laplace transform of a rational function, we can use the values where

the function is undefined to calculate the transform. If the function F (s) is undefined at b, then thepiece of the inverse Laplace transform at b is

lims→b

[(s− b)F (s)est

]If the value b has multiplicity of m, then the limit becomes

1

(m− 1)!lims→b

[dm−1

dsm−1

[(s− b)mF (s)est

]]

We do this for every value (including imaginary) where F (s) is undefined. Then we add up thesepieces to obtain the final inverse Laplace transform of F (s).

Example: Find L −1

{1

(s− 3)2(s+ 4)

}.

Here, F (s) = 1(s−3)2(s+4)

, which is undefined at 3 (with multiplicity 2) and at -4. So, for 3,

1

1!lims→3

[d

ds

[(s− 3)2F (s)est

]]= lim

s→3

[test(s+ 4)− est

(s+ 4)2

]=

7te3t − e3t

49

For -4,

lims→−4

[(s+ 4)F (s)est

]= lim

s→−4

[est

(s− 3)2

]=

e−4t

49

Thus,

L −1

{1

(s− 3)2(s+ 4)

}=

7te3t − e3t

49+

e−4t

49=

1

49

(7te3t − e3t + e−4t

)

Department of Mathematics, Sinclair Community College, Dayton, OH