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If it does, $$\int_{-\infty}^{\infty} \cos(bt)\,u(t)e^{-j\omega t} dt = \int_{0}^{\infty} \cos(bt)\,e^{-j\omega t} dt = \int_{0}^{\infty} \frac{e^{jbt} + e^{-jbt}}{2}\,e^{-j\omega t} dt$$
Then how do we carry on from this point?
$$ \frac{1}{2} \int_{0}^{\infty} \left[e^{j(b-\omega )t} + e^{-j(b+\omega )t}\right]dt$$
Thank you.
The integral doesn't converge in the conventional sense, so you can't solve it with standard methods. Assuming that you know (or can look up) the Fourier transform of the unit step function $u(t)$, it is straightforward to compute the Fourier transform of $\cos(\omega_0t)\,u(t)$ using the modulation property:
$$\mathcal{F}\big\{u(t)\big\}=U(\omega)=\pi\delta(\omega)+\frac{1}{j\omega}\tag{1}$$
$$\begin{align}\mathcal{F}\big\{\cos(\omega_0t)\,u(t)\big\}&=\frac12\big[U(\omega-\omega_0)+U(\omega+\omega_0)\big]\\&=\frac{\pi}{2}\big[\delta(\omega-\omega_0)+\delta(\omega+\omega_0)\big]+\frac{1}{j}\frac{\omega}{\omega^2-\omega_0^2}\tag{2}\end{align}$$
