# Thread: Laplace transform

1. ## Laplace transform

What is the laplace transform of $\displaystyle \int_0^t {\frac{\sin (t) \, dt}{t}}$ ?

I got this from an old reviewer and its a multiple choice question. The selections are

a. (arc tan s) / s
b. (arc sin s) / s
c. (arc cos s) / s
d. (arc cot s) / s

sorry for the typo

2. Originally Posted by bibbonacci
What is the laplace transform of $\displaystyle \int_0^\infty {\frac{\sin t \, dt}{t}}$ ?
Is this is typo? $\displaystyle \mathcal{L}\left(\int_0^{\infty}\frac{\sin(t)}{t}\ text{ }dt\right)(s)=\frac{\pi}{2s}$

3. Drexel, thanks for noticing

4. Well, what do you get when you write out the definition of the LT as applied to your function?

5. Originally Posted by bibbonacci
What is the laplace transform of $\displaystyle \int_0^t {\frac{\sin (t) \, dt}{t}}$ ?

I got this from an old reviewer and its a multiple choice question. The selections are

a. (arc tan s) / s
b. (arc sin s) / s
c. (arc cos s) / s
d. (arc cot s) / s

sorry for the typo
Starting from the L-transform of the sine function...

$\displaystyle \mathcal{L} \{\sin t\} = \frac{1}{1+s^{2}}$ (1)

...You first apply the 'division by t' rule...

$\displaystyle \mathcal{L} \{ \frac{f(t)}{t}\} = \int_{s}^{\infty} F(u)\ du \implies \mathcal{L} \{\frac{\sin t}{t} \} = \int_{s}^{\infty} \frac{du}{1+u^{2}} = \frac{\pi}{2} - \tan^{-1} s = \cot^{-1} s$ (2)

... and then the 'transform of integrals' rule...

$\displaystyle \mathcal{L} \{ \int_{0}^{t} f(\tau)\ d\tau\} = \frac{F(s)} {s} \implies \mathcal{L} \{\int_{0}^{t} \frac{\sin \tau}{\tau}\ d\tau \} = \frac{\cot^{-1} s}{s}$ (3)

Merry Christmas from Italy

$\chi$ $\sigma$

6. I was going to go from the definition and use by parts or something like that.