Laplace transform of the sin function

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

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

so

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

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

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

so

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