How Do I Prove the Laplace Transform Formula for Functions with Finite Jumps?

[cabellos]

how do i go about solving the laplace transform of sint * cost ?

i know the answer becomes 1/(s^2 + 4) but what is the method?

thanks.

 

[murshid_islam]

do you know that
\mathcal{L}\{f(t)*g(t)\} = \mathcal{L}\{f(t)\} \cdot \mathcal{L}\{g(t)\} = F(s) \cdot G(s)

and i think the answer comes out to be \frac{s}{{\left(s^2+1\right)}^2}

 

[cabellos]

yes i did know that but i didnt think it was as simple as that. From reading the certain rules we can apply to problems e.g the shift rule to solve L(sin2t * e^3t) do we HAVE to use the shift rule or can we separate each part and solve then multiply them together?

 

[murshid_islam]

i am sorry. do you mean convolution by the asterisk? or do you mean multiplication? what i said in my last post is only correct if the asterisk means convolution.

 

[cabellos]

i meant multiplication

 

[HallsofIvy]

Then use the definition
\mathcal{L}(sin(x)cos(x)}= \int_0^{\infnty}e^{-st}sin(t)cos(t)dt
which can be done by integration by parts.

 

[d_leet]

Mybe it would help to notice that sin(t)cos(t) is equal to \frac{1}{2}sin(2t)

 

[I_power2]

I want to ask about a proof of this transform: L{(f(t)}= sF(s) - f(0) - f(a^+) - f(a^-) - exp^(-as) inwhich f(t) is continuous except for an ordinary discontinuity (finite jump) at t=a, a>0