Relation between Laplace and Fourier transforms I know that  $$X_L(s=j\omega)=X_F(\omega)$$ if $x(t)$ is one sided and absolutely integrable and hence the imaginary axis of the Laplace transform is the Fourier transform. 

But Fourier transform also has imaginary and real parts. So how could this be right?
The Laplace transform evaluated at $s=j\omega$ is equal to the Fourier transform if its region of convergence (ROC) contains the imaginary axis. This is also true for the bilateral (two-sided) Laplace transform, so the function need not be one-sided.

As for real and imaginary parts, since $s$ is a complex variable, both the Laplace and the Fourier transform generally have real and imaginary parts. Take as a simple example the function $x(t)=e^{-at}u(t)$, with $a&gt;0$, where $u(t)$ is the unit step function. The Laplace transform is

$$X_L(s)=\frac{1}{s+a}\tag{1}$$

Since $a&gt;0$, the ROC of $X_L(s)$ contains the imaginary axis, and the Fourier transform of $x(t)$ is simply obtained by evaluating $X_L(s)$ on the imaginary axis $s=j\omega$:

$$X_F(\omega)=X_L(j\omega)=\frac{1}{j\omega+a}\tag{2}$$

Since $s=\sigma+j\omega$ is generally complex, not only the Fourier transform but also the Laplace transform $(1)$ has a real and an imaginary part:

$$X_L(\sigma+j\omega)=\frac{1}{\sigma+j\omega+a}=\frac{\sigma+a}{(\sigma+a)^2+\omega^2}-j\frac{\omega}{(\sigma+a)^2+\omega^2}\tag{3}$$

Only when evaluated on the real axis $s=\sigma$ ($\omega=0$) does the imaginary part vanish.