In E. M. Stein's Complex Analysis, Ch 4, pp. 116, proof of the Fourier inversion formula, there is the following step of derivation:
$$\int_0^\infty{\hat f(\xi)e^{2{\pi}ix\xi}d\xi}$$
$$=\int_0^\infty\int_{-\infty}^\infty f(u-ib)e^{-2\pi i(u-x-ib)\xi}dud\xi$$
$$=\int_{-\infty}^\infty f(u-ib)\int_0^\infty e^{-2\pi i(u-x-ib)\xi}d\xi du$$
where $\hat f(\xi)$ is the Fourier transform of $f(x)$. $f(z)$ is assumed to be holomorphic in a strip region $|\text {Im}(z)|<a$ containing the real axis, and $0<b<a$.
My question may be trivial, but I do not know how to justify that the two integration can be exchanged in order. Is there any theorem that provides sufficient conditions for this kind of double integrals to be exchangeable? Thanks.