Ok, using the definition of Laplace transforms to find [itex]\L\{f(t)\}[/itex](adsbygoogle = window.adsbygoogle || []).push({});

Given:

[tex]f(t)=\{^{\sin{t}, 0\le{t}<{\pi}}_{0, t\ge{\pi}}[/tex]

So, this is what I did:

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

[tex]=\int^{\pi}_{0} e^{-st}\sin t dt[/tex]

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

[tex]=\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)[/tex]

[tex]=\frac{-se^{-s\pi}}{s^2}+\frac{1}{s}(\frac{-1}{s}-\frac{1}{s}\L\{\sin t\})[/tex]

[tex]\L\{\sin t\}(\frac {s^2+1}{s^2})=\frac{-se^{-s\pi}}{s^2}-\frac{1}{s^2}[/tex]

[tex]\L\{\sin t\}=\frac {-se^{-s\pi}-1}{s^2+1}[/tex]

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

[tex]\L\{\sin t\}=\frac {1}{s^2+1}[/tex]

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

Where did I mess up?

Thanks a lot.

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# Laplace transform: where am I messing up?

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