Confusion for Part of Bogolyubov Transform Consider the operator (written in terms of a mode expansion) given by
$$\hat P(u) = \int_0^\infty \frac{d\Omega}{\sqrt{2\pi}}\frac{1}{\sqrt{2\Omega}}\left(\hat b_\Omega e^{-i\Omega u} + \hat b_\Omega^\dagger e^{i\Omega u}\right)$$ I would like to compute the Fourier transform
$$\int_{-\infty}^\infty \frac{du}{\sqrt{2\pi}}e^{i\Omega u}\hat P(u) = \frac{1}{\sqrt{2|\Omega|}}\begin{cases}\hat b_\Omega & \Omega > 0\\\hat b_{|\Omega|}^\dagger & \Omega < 0\end{cases}$$
I would like some help getting to that result. I anticipate it is a simple answer, but I do not directly see how to get there. I would greatly appreciate any help for getting to that result. I think it may be as simple as the Fourier Transform of an inverse Fourier Transform, but again, I do not clearly see how to get there.
 A: Let us define $$K(\Omega) \equiv \frac{1}{2\pi} \int\limits_0^\infty \int\limits_{-\infty}^\infty \mathrm{d}\Omega^\prime\, \mathrm{d}u \, \frac{1}{\sqrt{2\Omega^\prime}}\, \left( \hat{b}_{\Omega^\prime} \,e^{-i\Omega^\prime u}+\hat{b}_{\Omega^\prime}^\dagger \,e^{i\Omega^\prime u}\right)\, e^{i\Omega u} \tag{1}$$
and recall that $$ \delta(x-x^\prime) \equiv \frac{1}{2 \pi} \int\limits_{-\infty}^\infty \mathrm{d}k\, e^{ik(x-x^\prime)} \quad . \tag{2}$$
Using equation $(2)$ in $(1)$ and performing the integration with respect to $u$ then yields
$$K(\Omega) = \int\limits_0^\infty \mathrm{d}\Omega^\prime\, \frac{1}{\sqrt{2 \Omega^\prime}}\, \left(\hat{b}_{\Omega^\prime} \, \delta(\Omega - \Omega^\prime)+ \hat{b}_{\Omega^\prime}^\dagger \, \delta(\Omega + \Omega^\prime)\right) \quad .  \tag{3}$$
Now it is easy to see that for $\Omega >0$, only the first term in the integral of equation $(3)$ will contribute. Similarly, for $\Omega <0$ only the second term will contribute.
We eventually find
$$K(\Omega) = \frac{1}{\sqrt{2|\Omega|}}\,\begin{cases}\hat b_\Omega & \Omega > 0\\\hat b_{|\Omega|}^\dagger & \Omega < 0\end{cases} \quad . $$
