I will illustrate this with a simple example:
Consider the exponential decay function $$f(t)=\begin{cases} 0 & \ t\lt 0 , \\ A e^{-\lambda t} & \ t\ge 0 \end{cases}$$ Where $\lambda \gt 0$ and $A$ is a constant.
Using the formula for the Fourier transform of $f(t)$: $$\mathcal{F}(\omega)=\frac{1}{\sqrt{2\pi}}\int_{t=-\infty}^{\infty}f(t)e^{-i\omega t}\mathrm{d}t= \frac{1}{\sqrt{2\pi}}\int_{t=-\infty}^{0}0e^{-i\omega t}\mathrm{d}t+\frac{A}{\sqrt{2\pi}}\int_{t=0}^{\infty}\color{#F80}{e^{-\left(\lambda+i\omega \right)t}\mathrm{d}t}$$ $$=0 + \frac{A}{\sqrt{2\pi}}\left[\frac{e^{-\left(\lambda+i\omega \right)t}}{\lambda + i \omega}\right]_{t=0}^{\infty}=\color{#180}{\frac{A}{\sqrt{2\pi}(\lambda+i\omega )}}$$
My question is how was this integration carried out? Since the $\color{#F80}{\mathrm{integrand}}$ is a function of 2 variables ($\omega$ and $t$).
Thank you kindly.
You integrate with respect to $t$ so $\omega$ is considered as a constant here.