Laplace Transform of e^(bx) sin(ax)

Laplace Transform of L {ebx sin 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 sin ax} Solution Putting f (x) = ebx sin ax,s > b in the equation R∞ L {f (x)}=F (s)= 0 e−sx f (x) dx R∞ L {ebx sin ax}=F (s)= 0 e−sx ebx sin ax dx RR L {ebx sin ax} = F (s) = limR−→∞ 0 e−(s−b)x sin 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 sin ax} = − s−b limR−→∞ RR 0 sin ax d(e−(s−b)x ) RR sin ax d(e−(s−b)x ) R R −(s−b)x ax R -(s-b)L {ebx sin ax} = limR−→∞ [ esin d(sin ax) (s−b)x ]0 −limR−→∞ 0 e -(s-b)L {ebx sin ax} = limR−→∞ 0 Now ax R sin aR limR−→∞ [ esin (s−b)x ]0 = limR−→∞ [ e(s−b)R ] = 0 RR -(s-b)L {ebx sin ax} = −a limR−→∞ 0 e−(s−b)x cos ax dx 1 Again putting − s−b d(e−(s−b)x ) = e−(s−b)x dx RR a limR−→∞ 0 cos ax d(e−(s−b)x ) -(s-b)L {ebx sin ax} = s−b 2 2 ax R − (s−b) L {ebx sin ax} = limR−→∞ [ ecos (s−b)x ]0 −limR−→∞ a RR 0 e−(s−b)x d(cos ax) Now again ax R cos aR limR−→∞ [ ecos (s−b)x ]0 = limR−→∞ [ e(s−b)R − 1] = −1 R R −(s−b)x 2 bx − (s−b) L {e sin ax} = −1 − lim d(cos ax) R−→∞ 0 e a R R −(s−b)x 2 bx L {e sin ax} = −1 − (−a) lim − (s−b) sin ax dx R−→∞ 0 e a R R −(s−b)x 2 bx L {e sin ax} = −1 + a lim − (s−b) sin ax dx R−→∞ 0 e a 2 L {ebx sin ax} = −1 + aL {ebx sin ax} − (s−b) a 2 − (s−b) L {ebx sin ax} − aL {ebx sin ax} = −1 a 2 − a]L {ebx sin ax} = −1 [− (s−b) a 2 −[ (s−b) + a]L {ebx sin ax} = −1 a 2 + a]L {ebx sin ax} = 1 [ (s−b) a 2 2 [ (s−b)a +a ]L {ebx sin ax} = 1 L {ebx sin ax} = 1 a (s−b)2 +a2 (s > b) Applications Find area under the curve y = e−7x e4x sin 3x Using the formula 3 Area= a (s−b)2 +a2 (s > b) Here s = 7, b = 4, a = 3 Area= (7−4)32 +32 = 1 6 4