The Laplace transform transforms a function from the time domain to the frequency domain. But I am having trouble understanding what this really means. I assume this picture is correct
Then, a frequency domain - representation of the function $\sin(t)$, should be a line of height = amplitude of $\sin(x) = 1$, and it should have no width. On the $x-$axis, the line should be located at $x=\frac{1}{2\pi}$, since that is the frequency of $\sin(x)$.
But the Laplace transform of $\sin(t)$ is $\frac{1}{s^2+1}$. This is not a straight line. Why not? I dont understand how $\sin(t)$ can be equal to $\frac{1}{s^2+1}$ in the frequency domain?
Any help would be much apreciated!!
I believe your confusing the unilateral Laplace transform
$$\mathcal{L}_t[f(t)](s)=\int\limits_0^{\infty} f(t)\, e^{-s t}\,dt\tag{1}$$
with the Fourier transform
$$F(\omega)=\mathcal{F}_t[f(t)](\omega)=\int\limits_{-\infty}^{\infty} f(t)\, e^{-i \omega t}\,dt\tag{2}$$
which is equivalent to the bilateral Laplace transform
$$\mathcal{BL}_t[f(t)](s)=\int\limits_{-\infty}^{\infty} f(t)\, e^{-s t}\,dt\tag{3}$$
evaluated at $s=i \omega$.
The unilateral Laplace transform of $\sin(t)$ is $$\mathcal{L}_t[\sin(t)](s)=\int\limits_0^{\infty} \sin(t)\, e^{-s t}\,dt=\frac{1}{s^2+1}\,,\quad\Re(s)>0\tag{4}$$
whereas the Fourier transform of $\sin(t)$ is
$$\mathcal{F}_t[\sin(t)](\omega)=\int\limits_{-\infty}^{\infty} \sin(t)\, e^{-i \omega t}\,dt=i \pi\, \delta(\omega+1)-i \pi\, \delta(\omega-1)\tag{5}$$
which evaluates to $0$ for $\omega\in\mathbb{R}\land\omega\ne\pm1$ and is undefined at $\omega=\pm1$.
The inverse Fourier transform associated with formula (2) above is
$$f(t)=\mathcal{F}_{\omega}^{-1}[F(\omega)](t)=\frac{1}{2 \pi} \int\limits_{-\infty}^{\infty} F(\omega)\, e^{i t \omega}\,d\omega\tag{6}.$$
Formula (5) above can be validated via the inverse Fourier transform
$$\mathcal{F}_{\omega}^{-1}[i \pi\, \delta(\omega+1)-i \pi\, \delta(\omega-1)](t)=$$ $$\frac{1}{2 \pi} \int\limits_{-\infty}^{\infty} (i \pi\, \delta(\omega+1)-i \pi\, \delta(\omega-1))\, e^{i t \omega}\,d\omega=\frac{1}{2} i\, e^{-i t}-\frac{1}{2} i\, e^{i t}=\sin (t)\tag{7}$$
which can be derived from the following property of the Dirac delta function $\delta(\omega)$:
$$\int\limits_{-\infty}^{\infty} \delta(\omega-\alpha)\, F(\omega)\,d\omega=F(\alpha)\,,\quad\alpha\in\mathbb{R}\tag{8}.$$
It is not true that $\delta(0)=1$ (actually $\delta(0)$ is undefined), rather it is true that
$$\int\limits_{-\beta}^{\beta} \delta(\omega)\,d\omega=1\tag{9}$$
for all $\beta>0$.
The Fourier and inverse Fourier transforms can also be defined as
$$F(\omega)=\mathcal{F}_t[f(t)](\omega)=\int\limits_{-\infty}^{\infty} f(t)\, e^{-i 2 \pi \omega t}\,dt\tag{10}$$
and
$$f(t)=\mathcal{F}_{\omega}^{-1}[F(\omega)](t)=\int\limits_{-\infty}^{\infty} F(\omega)\, e^{i 2 \pi t \omega}\ d\omega\tag{11}$$
which for $f(t)=\sin(t)$ leads to
$$\mathcal{F}_t[\sin(t)](\omega)=\int\limits_{-\infty}^{\infty} \sin(t)\, e^{-i 2 \pi \omega t}\,dt=\frac{1}{2} i\, \delta\left(\omega+\frac{1}{2 \pi}\right)-\frac{1}{2} i\, \delta\left(\omega-\frac{1}{2 \pi}\right)\tag{12}$$
and
$$\mathcal{F}_{\omega}^{-1}\left[\frac{1}{2} i\, \delta\left(\omega+\frac{1}{2 \pi}\right)-\frac{1}{2} i\, \delta\left(\omega-\frac{1}{2 \pi}\right)\right](t)=$$ $$\int\limits_{-\infty}^{\infty} \left(\frac{1}{2} i\, \delta\left(\omega+\frac{1}{2 \pi}\right)-\frac{1}{2} i\, \delta\left(\omega-\frac{1}{2 \pi}\right)\right)\, e^{i 2 \pi t \omega}\,d\omega=\frac{1}{2} i\, e^{-i t}-\frac{1}{2} i\, e^{i t}=\sin(t)\tag{13}$$
which seem more consistent with some of the expected results outlined in the question, and for $f(t)=\cos(t)$ leads to
$$\mathcal{F}_t[\cos(t)](\omega)=\int_{-\infty}^{\infty} \cos(t)\, e^{-i 2 \pi \omega t}\,dt=\frac{1}{2}\, \delta\left(\omega+\frac{1}{2 \pi}\right)+\frac{1}{2}\, \delta\left(\omega-\frac{1}{2 \pi}\right)\tag{14}$$
and
$$\mathcal{F}_{\omega}^{-1}\left[\frac{1}{2}\, \delta\left(\omega+\frac{1}{2 \pi}\right)+\frac{1}{2}\, \delta\left(\omega-\frac{1}{2 \pi}\right)\right](t)=$$ $$\int\limits_{-\infty}^{\infty} \left(\frac{1}{2}\, \delta\left(\omega+\frac{1}{2 \pi}\right)+\frac{1}{2}\, \delta\left(\omega-\frac{1}{2 \pi}\right)\right)\, e^{i 2 \pi t \omega } \, d\omega=\frac{1}{2}\, e^{-i t}+\frac{1}{2}\, e^{i t}=\cos(t)\tag{15}.$$