Laplace transform

**bibbonacci** · December 22nd 2010, 08:15 PM

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

**Drexel28** · December 22nd 2010, 08:23 PM

Originally Posted by bibbonacci

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

Is this is typo? $\displaystyle \mathcal{L}\left(\int_0^{\infty}\frac{\sin(t)}{t}\ text{ }dt\right)(s)=\frac{\pi}{2s}$

**bibbonacci** · December 22nd 2010, 08:32 PM

Drexel, thanks for noticing

**Ackbeet** · December 23rd 2010, 02:52 AM

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

**chisigma** · December 23rd 2010, 11:23 AM

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

$\chi$ $\sigma$

**Ackbeet** · December 23rd 2010, 11:35 AM

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

Thread: Laplace transform

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