Laplace Transform of ebx cos ax
shahbaz ahmed
June 2024
Introduction to Laplace Transform
Let f(x) be defined for 0 ≤ x < ∞ and let s denote an arbitrary real
variable. The Laplace transform of f (x), designated by either L or
F (s), is
R∞
L {f (x)}=F (s)= 0 e−sx f (x)dx
for all values of s for which the improper integral converges.
Convergence occur when the limit
RR
limR−→∞ 0 e−sx f (x)dx
exists.
1 Find L {ebx cos ax}
Solution
Putting f (x) = ebx cos ax,s > b in the equation
R∞
L {f (x)}=F (s)= 0 e−sx f (x) dx
R∞
L {ebx cos ax}=F (s)= 0 e−sx ebx cos ax dx
RR
L {ebx cos ax} = F (s) = limR−→∞ 0 e−(s−b)x cos ax dx
Now
d(e−(s−b)x ) = e−(s−b)x d(−(s − b)x)
d(e−(s−b)x ) = −(s − b)e−(s−b)x dx
1
− s−b
d(e−(s−b)x ) = e−(s−b)x dx
1
L {ebx cos ax} = − s−b
limR−→∞
RR
0
cos ax d(e−(s−b)x )
RR
cos ax d(e−(s−b)x )
R R −(s−b)x
ax R
-(s-b)L {ebx cos ax} = limR−→∞ [ ecos
d(cos ax)
(s−b)x ]0 −limR−→∞ 0 e
-(s-b)L {ebx cos ax} = limR−→∞
0
Now
ax R
cos aR
limR−→∞ [ ecos
(s−b)x ]0 = limR−→∞ [ e(s−b)R − 1] = −1
RR
-(s-b)L {ebx cos ax} = −1 − limR−→∞ 0 e−(s−b)x (−a sin ax) dx
RR
-(s-b)L {ebx cos ax} = −1 + a limR−→∞ 0 e−(s−b)x sin ax dx
RR
1
d(e−(s−b)x ]
-(s-b)L {ebx cos ax} = −1 + a limR−→∞ 0 sin ax [− s−b
2 RR
limR−→∞ 0 sin ax d(e−(s−b)x )
h
i
R R −(s−b)x
sin ax R
a
bx
d(sin ax)
-(s-b)L {e cos ax} = −1− s−b limR−→∞ [ e(s−b)x ]0 − limR−→∞ 0 e
-(s-b)L {ebx cos ax} = −1 −
a
s−b
Now
ax R
sin aR
limR−→∞ [ esin
(s−b)x ]0 = limR−→∞ [ e(s−b)R ] = 0
RR
-(s-b)L {ebx cos ax} = −1 +
a2
s−b
-(s-b)L {ebx cos ax} = −1 +
a2
bx
s−b L {e cos ax}
-(s-b)L {ebx cos ax} −
limR−→∞
a2
bx
s−b L {e cos ax}
[−(s − b) −
a2
bx
s−b ]L {e cos ax}
= −1
−[(s − b) +
a2
bx
s−b ]L {e cos ax}
= −1
[(s − b) +
2
a2
bx
s−b ]L {e cos ax}
=1
2
+a
]L {ebx cos ax} = 1
[ (s−b)
s−b
L {ebx cos ax} =
s−b
(s−b)2 +a2
(s > b)
3
0
e−(s−b)x cos ax dx
= −1
Laplace Transform of e^(bx) cos(ax)
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