Laplace transform: where am I messing up?

1. Apr 16, 2004

faust9

Ok, using the definition of Laplace transforms to find $\L\{f(t)\}$

Given:
$$f(t)=\{^{\sin{t}, 0\le{t}<{\pi}}_{0, t\ge{\pi}}$$

So, this is what I did:

$$\L\{\sin t\}=\int^{\pi}_{0} e^{-st}\sin t dt+\int^{\infty}_{\pi} e^{-st}(0)dt$$

$$=\int^{\pi}_{0} e^{-st}\sin t dt$$

$$=\frac{-e^{-st}\sin{t}}{s}]^{\pi}_{0}+\frac{1}{s}\int^{\pi}_{0} e^{-st}\cos t dt$$

$$=\frac{-e^{-st}\sin{t}}{s}]^{\pi}_{0}+\frac{1}{s}(\frac{-e^{-st}\cos{t}}{s}]^{\pi}_{0}-\frac{1}{s}\int^{\pi}_{0} e^{-st}\sin t dt)$$

$$=\frac{-se^{-s\pi}}{s^2}+\frac{1}{s}(\frac{-1}{s}-\frac{1}{s}\L\{\sin t\})$$

$$\L\{\sin t\}(\frac {s^2+1}{s^2})=\frac{-se^{-s\pi}}{s^2}-\frac{1}{s^2}$$

$$\L\{\sin t\}=\frac {-se^{-s\pi}-1}{s^2+1}$$

Which I know is wrong because the Laplace for sin t should be:

$$\L\{\sin t\}=\frac {1}{s^2+1}$$

I know my limits of integration will affect the problem but the restricted limits should only add a term to the numerator $e^{-s\pi}$ I believe.

Where did I mess up?

Thanks a lot.

2. Apr 16, 2004

arildno

Your mistake is in the 5.line, in the evaluation of cos(t) at the limits indicated.

Last edited: Apr 16, 2004
3. Apr 16, 2004

arildno

Also, the sin(t) evaluation in the 5.line should be zero.

4. Apr 16, 2004

faust9

Got it. Thanks. How silly of me.

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