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.

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Laplace transform: where am I messing up?

**Physics Forums | Science Articles, Homework Help, Discussion**