Laplace transform of the cos function

Laplace transform of $ f(t)=\cos t\:h(t) $ is defined by

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

We can notice that $ \cos t=Re(\cos t+i\sin t) =Re(e^{it}) $
Then for real s we have

$$\mathcal{L}(\cos t\:h(t))(s)=Re(\mathcal{L}(e^{it}\:h(t))(s)=Re\frac{1}{s-i}=Re\frac{s+i}{s^2+1}=\frac{s}{s^2+1}$$

So

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

We have also

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

Another method is the following
Function $ f:R\rightarrow R $ , $ f(t)=\cos t\:h(t) $ verifies differential equation

$$x"(t)+x(t)=0\;\:x(0+)=1,\:\:x'(0+)=0$$
$$\mathcal{L}(x(t)(s)=X(s)$$
$$\mathcal{L}(x'(t)(s)=sX(s)-1$$
$$\mathcal{L}(x"\;(t)(s)=s^2X(s)-s$$

We have

$$s^2X(s)-s+X(s)=0$$

so

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

As $ \mathcal{L}(t f(t))(s)=-\mathcal{L}'( f(t))(s) $ we have also

$$\mathcal{L}(t\cos t\:h(t))(s)=-\frac{d}{ds}\frac{s}{s^2+1}=\frac{s^2-1}{(s^2+1)^2}$$

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

$$x"(t)+ \omega^2 x(t)=0\;\:x(0+)=1,\:\:x'(0+)=0$$
$$\mathcal{L}(x(t)(s)=X(s)$$
$$\mathcal{L}(x'(t)(s)=sX(s)-1$$
$$\mathcal{L}(x"\;(t)(s)=s^2X(s)-s$$

We have

$$s^2X(s)-s+ \omega^2 X(s)=0$$

so

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

and also

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