# Taking fourier transform of operators with different exponents

I have a coupled differential equation that I wish to take the Fourier transform (FT) of. However, they consists of different operators which also includes an exponent (more will be shown below). This is taken from my lecture notes where the solution is also given.

So I have an equation defined as

$$\frac{d\hat{a}}{dt}=-\frac{\alpha}{2}\hat{a}-\sqrt{\alpha}\hat{a}_{I}-ig_{1}\hat{b}-ig_{1}\hat{b}^{\dagger}e^{i2\Omega t} \qquad(1)$$

Where $$\hat{a}, \hat{a}_{I},$$ and $$\hat{b}$$ represents three distinct operators, $$i$$ is the complex unit, and $$\alpha, g_{1}, \Omega$$ are constants. After taking the FT, we have that

$$-i\omega \hat{a}[\omega]=-\frac{\alpha}{2}\hat{a}[\omega]-\sqrt{\alpha}\hat{a}_{I}[\omega]-ig_{1}\hat{b}[\omega]-ig_{1}\hat{b}^{\dagger}[\omega+2\Omega] \qquad(2)$$

My question is to understand how to get from equation $$(1)$$ to equation $$(2)$$. The FT that was used seems to be that

$$\hat{X}(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{-i\omega t}\hat{X}[\omega]d\omega \qquad(3)$$

Where $$\hat{X}$$ can be any of the three operators $$\hat{a}, \hat{a}_{I}, \hat{b}$$. I tried applying equation $$(3)$$ onto $$(1)$$ to obtain $$(2)$$ but I don't know how to get the $$-ig_{1}\hat{b}^{\dagger}[\omega+2\Omega]$$ term from the $$-ig_{1}\hat{b}^{\dagger}e^{i2\Omega t}$$ term. Furthermore, how does taking the conjugate of an operator works in the context of FT? In a different manner of speaking, is the conjugate of $$\hat{X}$$ in equation (3) simply

$$\hat{X}^{\dagger} = \frac{1}{2\pi}\int_{-\infty}^{\infty}e^{i\omega t}\hat{X}^{\dagger}[\omega]d\omega$$

? The change of sign in the exponent confuses me as a whole.