1. ## Laplace transform

I'm asked to find the Laplace transform of $\displaystyle f(t)=\dfrac{sin(t)}{t}$.

I begin by applying the definition to the function:

$\displaystyle F(s) = \int_0^{\infty}e^{-st} \dfrac{sin(t)}{t}~dt$

I'm getting the feeling that I need to use integration by parts, but I'm having a hard time identifying good choices for $\displaystyle u$ and $\displaystyle dv$

note: I will continue to add more troublesome Laplace transform problems to this thread as I encounter them.

2. ## A hint

Note that
$\displaystyle F'(s)=\frac{d}{ds}\int_0^{\infty}e^{-st}\frac{\sin{t}}{t}~dt$
$\displaystyle =\int_0^{\infty}\frac{d}{ds}\left(e^{-st}\frac{\sin{t}}{t}\right)dt$
$\displaystyle =\int_0^{\infty}-t\cdot{e^{-st}}\frac{sin(t)}{t}dt$
$\displaystyle =-\int_0^{\infty}e^{-st}\sin{t}dt$
$\displaystyle =-\mathcal{L}[\sin{t}](s)$.
Do you know the Laplace transform of the sine function?

--Kevin C.

3. i do indeed

4. actually, are you sure this is correct?

also, it just occurred to me that perhaps i should expand sin(t) via euler's formula and try it that way.

maple is kicking out arctan(1/s)

5. Originally Posted by jmedsy
I'm asked to find the Laplace transform of $\displaystyle f(t)=\dfrac{sin(t)}{t}$.

I begin by applying the definition to the function:

$\displaystyle F(s) = \int_0^{\infty}e^{-st} \dfrac{sin(t)}{t}~dt$

I'm getting the feeling that I need to use integration by parts, but I'm having a hard time identifying good choices for $\displaystyle u$ and $\displaystyle dv$

note: I will continue to add more troublesome Laplace transform problems to this thread as I encounter them.
Theorem (simple to prove): $\displaystyle LT\left[ \frac{f(t)}{t}\right] = \int_s^{+\infty} F(\tau) \, d \tau$ where $\displaystyle F(s) = LT[f(t)]$.

In your question $\displaystyle f(t) = \sin (t)$.

Originally Posted by jmedsy
actually, are you sure this is correct?

also, it just occurred to me that perhaps i should expand sin(t) via euler's formula and try it that way.

maple is kicking out arctan(1/s)
Note that $\displaystyle \frac{\pi}{2} - \tan^{-1} (s) = \tan^{-1} \left( \frac{1}{s} \right)$.