I want to truly understand the concept of Laplace transform. The Laplace transform of the cosine is
$$\mathcal L ( \cos (ω_0 t) ) = \frac{s}{s^2 + ω_0^2}$$
but the Fourier transform is two Dirac deltas in two symmetric positions of the $\omega$-axis, meaning it consists of two exponential component of complex frequency. In the $s$-domain, we perceive that cosine is not a pure tone but it's made of lots of tone. We know laplace transform and fourier transform difference is on just real value component added to $w$.
The question is: The Cosine mentioned $x(s) \to s=jw$ must be equal to Fourier$\{x\}$ doesn't ? I want a proof if this is really equal. Also I have seen this question.
It means ($j=\sqrt{-1}$)($s=\sigma+jω$) $$\mathcal B \mathcal L[ cos(ω_0t)] \to \frac{s}{s^2+ω_0^2} (s \to jw) \to \mathcal F[cos(ω_0t)] \to π [ δ (ω − ω_0) + δ (ω + ω_0)]$$
And I expect to $s/s^2+ω_0^2$ becoming equal to $π [ δ (ω − ω_0) + δ (ω + ω_0)]$ but this is not.
This answer assumes the following definitions of the Fourier, Laplace, and two-sided Laplace transforms.
$$\mathcal{F}_t[f(t)](\omega)=\int\limits_{-\infty}^{\infty } f(t)\,e^{-i \omega t}\,dt\tag{1}$$
$$\mathcal{L}_t[f(t)](s)=\int\limits_0^{\infty} f(t)\,e^{-s t}\,dt\tag{2}$$
$$\mathcal{BL}_t[f(t)](s)=\int\limits_{-\infty}^{\infty} f(t)\,e^{-s t}\,\tag{3}dt$$
Note the Fourier transform defined in formula (1) above is equivalent to the bilateral Laplace transform defined in formula (3) above evaluated at $s=i \omega$.
The Fourier and Laplace transforms of $\cos(\omega_0\,t)$ are as follows:
$$\mathcal{F}_t[\cos(\omega_0\,t)](\omega)=\int\limits_{-\infty}^{\infty}\cos (\omega_0\,t)\,e^{-i \omega t}\,dt=\pi\,\delta(\omega+\omega_0)+\pi\,\delta(\omega-\omega_0),\quad\omega\in\mathbb{R}\tag{4}$$
$$\mathcal{L}_t[\cos(\omega_0\,t)](s)=\int\limits_0^{\infty} \cos(\omega_0\,t)\,e^{-s t}\,dt=\frac{s}{s^2+\omega_0^2},\quad\Re(s)>0\tag{5}$$
The Fourier and bilateral Laplace transforms of $\cos(\omega_0\,t)\,\theta(t)$ are
$$\mathcal{F}_t[\cos(\omega_0\,t)\,\theta(t)](\omega)=\int\limits_{-\infty}^{\infty}\cos(\omega_0\,t)\,\theta(t)\,e^{-i \omega t}\,dt$$ $$=\int_0^{\infty}\cos(\omega_0\,t)\,e^{-i \omega t}\,dt=\frac{\pi}{2} \delta(\omega+\omega_0)+\frac{\pi}{2} \delta(\omega-\omega_0)+\frac{i \omega}{\omega_0^2-\omega^2},\quad\omega\in\mathbb{R}\tag{6}$$
$$\mathcal{BL}_t[\cos(\omega_0\,t)\,\theta(t)](s)=\int\limits_{-\infty}^{\infty}\cos(\omega_0\,t)\,\theta(t)\,e^{-s t}\,dt=\int\limits_0^{\infty} \cos(\omega_0\,t)\,e^{-s t}\,dt=\frac{s}{s^2+\omega_0^2},\ \Re(s)>0\tag{7}$$
where
$$\theta(t)=\left\{ \begin{array}{cc} 0 & t<0 \\ 1 & t>0 \\ \end{array}\right.\tag{8}$$
is the Heaviside step function.
Consider the following limit representation of the bilateral Laplace transform of $\cos(\omega_0\,t)$.
$$\mathcal{BL}_t[\cos(\omega_0\,t)](s)=\underset{T\to\infty}{\text{lim}}\left(\int\limits_{-T}^{T}\cos(\omega_0\,t)\,e^{-s t}\,dt\right)=\underset{T\to\infty}{\text{lim}}\left(\frac{2 (s \sinh(s T) \cos(T \omega_0)+\omega_0\cosh (s T) \sin (T \omega_0))}{s^2+\omega_0^2}\right)\tag{9}$$
Substituting $T=2 \pi f$ and $s=i \omega$ in formula (9) above and accounting for the resulting equivalence to formula (4) above leads to the limit representation
$$\delta(\omega+\omega_0)+\delta(\omega-\omega_0)=\underset{f\to\infty}{\text{lim}}\left(\frac{2}{\pi} \frac{\omega \cos(2 \pi f \omega_0) \sin (2 \pi f \omega )-\omega_0 \sin(2 \pi f \omega_0) \cos(2 \pi f \omega )}{\omega ^2-\omega_0^2}\right),\quad\omega\in\mathbb{R}\tag{10}$$
which is equivalent to the limit representation
$$\delta(\omega+\omega_0)+\delta(\omega-\omega_0)=\underset{f\to\infty}{\text{lim}}\left(2 f\,\text{sinc}(2 \pi f (\omega+\omega_0))+2 f\,\text{sinc}(2 \pi f (\omega -\omega_0))\right),\quad\omega\in\mathbb{R}.\tag{11}$$
The relevant question seems to be whether $\mathcal{F}_t[\cos(\omega_0\,t)\,\theta(t)](\omega)$ can be approximated by $\underset{\epsilon\to 0^+}{\text{lim}}\left(\mathcal{BL}_t[\cos(\omega_0\,t)\,\theta(t)](\epsilon+i \omega)\right)=\underset{\epsilon\to 0^+}{\text{lim}}\left(\mathcal{L}_t[\cos(\omega_0\,t)](\epsilon+i \omega)\right)$ for $\omega\in\mathbb{R}$ which, based on formulas (5) to (7) above, is equivalent to the following.
$$\frac{\pi}{2} \delta(\omega+\omega_0)+\frac{\pi}{2} \delta(\omega-\omega_0)+\frac{i \omega}{\omega_0^2-\omega^2}\approx\underset{\epsilon\to 0^+}{\text{lim}}\left(\frac{\epsilon+i \omega}{(\epsilon+i \omega)^2+\omega_0^2}\right),\quad\omega\in\mathbb{R}\tag{12}$$
Figure (1) below illustrates the imaginary part of the right side of formula (12) in orange overlaid on the imaginary part of the left-side of formula (12) in blue where both sides are evaluated at $\omega_0=1$ and the right side is evaluated at $\epsilon=10^{-6}$. Note for $\epsilon=0$ the right-side of formula (12) above becomes equivalent to the last term on the left-side of formula (12) above (which is the imaginary part of the left-side of formula (12) above), so clearly $\mathcal{F}_t[\cos(\omega_0\,t)\,\theta(t)](\omega)\ne\mathcal{L}_t[\cos(\omega_0\,t)](i \omega)$.
Figure (1): Illustration of imaginary parts of right and left sides of formula (12) (orange overlaid on blue) evaluated at $\omega_0=1$ and $\epsilon=10^{-6}$
Figures (2) below illustrates the real part of the right-side of formula (12) above evaluated at $\omega_0=1$ and $\epsilon=10^{-6}$. Note the evaluation seems to suggest the real part of the right-side of formula (12) above may perhaps approximate a limit representation for the real-part of the left side of formula (12) above (i.e. the two Dirac delta terms).
Figure (2): Illustration of real part of right-side of formula (12) evaluated at $\omega_0=1$ and $\epsilon=10^{-6}$
The analysis and results illustrated above seem to suggest the left-side of formula (12) above can be approximated by the limit representation on the right-side of formula (12) above, but the following integral which assumes $\omega_0=1$ provides further evidence.
$$\int\limits_{-2}^2 \left(\frac{\pi}{2} \delta(\omega+1)+\frac{\pi}{2} \delta(\omega-1)+\frac{i \omega}{1-\omega^2}\right)\,d\omega =\pi\approx \underset{\epsilon\to 0^+}{\text{lim}}\left(\frac{i}{2} (\log(\epsilon (\epsilon-4 i)-3)-\log(\epsilon (\epsilon+4 i)-3))\right)=\underset{\epsilon\to 0^+}{\text{lim}}\left(\int\limits_{-2}^2 \frac{\epsilon+i \omega}{(\epsilon+i \omega)^2+1}\,d\omega\right)\tag{13}$$
Note for the integral of the imaginary part of the left-side of formula (12) above the Cauchy principal value $\mathcal{P}\int\limits_{-2}^2 \frac{i \omega}{1-\omega^2}\,d\omega=0$ is consistent with the fact that $\frac{i \omega}{1-\omega^2}$ is an odd function of $\omega$ as illustrated in Figure (1) above.
For $\epsilon=10^{-6}$ the right side of formula (13) above evaluates to $3.14159\,+0.i$ which is very close to the value of $\pi$.