Laplace Transform xsin(ax)

Laplace Transform x 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. Find L {x sin ax} Solution 1 Since R∞ L {f (x)}=F (s)= 0 e−sx f (x)dx Putting f (x) = x sin ax L {x sin ax}=F (s)=lim R −→ ∞ RR e−sx x sin ax dx RRR∞ L {x sin ax}=F (s)= lim R −→ ∞ 0 0 x sin ax (e−sx dx) 0 d(e−sx ) = e−sx d(−sx) = −se−sx dx d(e−sx ) = −se−sx dx − 1s d(e−sx ) = e−sx dx R∞ L {x sin ax}= 0 x sin ax [− 1s d(e−sx )] RR L {x sin ax}=− 1s lim R −→ ∞ 0 x sin ax d(e−sx ) RR −sL {x sin ax}=lim R −→ ∞ 0 x sin ax d(e−sx ) R R −sx ax −sL {x sin ax} = x sin d(x sin ax) − lim R −→ ∞ sx 0 e e R R −sx ax −sL {x sin ax} = x sin (sin ax+ax cos ax)dx −lim R −→ ∞ sx 0 e e R R −sx ax −sL {x sin ax} = x sin sin ax dx − lim R −→ ∞ sx 0 e e RR − a lim R −→ ∞ 0 e−sx x cos ax dx R R −sx ax a −sL {x sin ax} = x sin − − a lim R −→ ∞ x cos ax dx sx 2 2 0 e e s +a RR ax a a −sx − + lim R −→ ∞ ) −sL {x sin ax} = x sin 0 x cos ax d(e esx s2 +a2 s R R −sx a a x cos ax ax −sL {x sin ax} = x sin − + [ −lim R −→ ∞ d(x cos ax)] sx 2 2 sx 0 e e s +a s e R R −sx a a x cos ax a ax −sL {x sin ax} = x sin − + − lim R −→ ∞ (cos ax− 0 e esx s2 +a2 s esx s 2 ax sin ax)]dx ax a a x cos ax a −sL {x sin ax} = x sin esx − s2 +a2 + s esx − s lim R −→ ∞ R R −sx a2 lim R −→ ∞ x sin ax dx 0 e s a x sin ax esx − s2 +a2 −sL {x sin ax} = 0 e−sx cos ax dx+ 2 ax a s a + as x cos esx − s × s2 +a2 + s L {x sin ax} 2 −sL {x sin ax} − as L {x sin ax} = −sL {x sin ax} − RR a2 s L {x sin ax} = x sin ax a esx − s2 +a2 x sin ax esx − ax a s + as x cos esx − s × s2 +a2 a s2 +a2 + a x cos ax s esx − 2 a s2 +a2 x sin ax a x cos ax R −sL {x sin ax}− as L {x sin ax}+ s22a +a2 = limR−→∞ [ esx + s esx ]0 [−s − 2 [− s a2 s ]L {x sin ax} +a2 s ]L {x sin ax} + + 2a s2 +a2 2a s2 +a2 aR = limR−→∞ [ R sin + esR = limR−→∞ eRsR [sin aR + as cos aR] Using L’Hospital’s Rule limR−→∞ eRsR = limR−→∞ Re1sR = 0 2 +a2 s ]L {x sin ax} + 2 +a2 s ]L {x sin ax} = − s22a +a2 [− s [− s 2 [s +a2 s ]L {x sin ax} L {x sin ax} = = 2a s2 +a2 2a s2 +a2 =0 2a s2 +a2 × s s2 +a2 a R cos aR s esR ] = 2as (s2 +a2 )2 3