Laplace transform of the sin function

Posted December 8th, 2007 by Structure

Laplace transform of sinus function is defined by

$\mathcal{L}(\sin t \:h(t))(s)=\int_{0}^{+\infty}\sin t\: e^{-st}dt$

Integral is convergent for

$Re\:(s)>0$

As we have

$\sin t=Im\:(\cos t+i\sin t)=Im\: e^{it}$

we can use it to get the sinus function transform.
We have

$\mathcal{L}(\cos t+i\sin t)\: h(t)(s)=\int_{0}^{+\infty}e^{it}e^{-st}dt=\int_{0}^{+\infty}e^{-(s-i)t}dt=\frac{1}{s-i}=\frac{s+i}{s^2+1}$

Taking real and imaginary parts we have

$\mathcal{L}(\sin t\: h(t))(s)=\int_{0}^{+\infty}\sin t e^{-st}dt=\frac{1}{s^2+1}$

$\mathcal{L}(\cos t\: h(t))(s)=\int_{0}^{+\infty}\cos e^{-st}dt=\frac{s}{s^2+1}$

Now we give another method to have the same result using properties of Laplace transform.

Function $f:R\rightarrow R$ , $f(t)=\sin t\:h(t)$ verifies differential equation

$x"(t)+x(t)=0\;\:x(0+)=0,\:\:x'(0+)=1$

$\mathcal{L}(x(t)(s)=X(s)$

$\mathcal{L}(x'(t)(s)=sX(s)$

$\mathcal{L}(x"\;(t)(s)=s^2X(s)-1$

We have

$\mathcal{L}(x"(t)+x(t))(s)=s^2X(s)-1+X(s)=0$

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

Function $f:R\rightarrow R$ , $f(t)=\sin \omega t\:h(t)$ verifies differential equation

$x"(t)+\omega^2x(t)=0\;\:x(0+)=0,\:\:x'(0+)=\omega$

$\mathcal{L}(x(t)(s)=X(s)$

$\mathcal{L}(x'(t)(s)=sX(s)$

$\mathcal{L}(x"\;(t)(s)=s^2X(s)-\omega$

We have

$\mathcal{L}(x"(t)+\omega^2x(t))(s)=s^2X(s)-\omega+\omega^2X(s)=0$

$X(s)=\mathcal{L}(\sin \omega t\:h(t))(s)=\frac{\omega}{s^2+\omega^2}$

Instant start