Rewriting a double integral with complex exponential function

Question

Why can we write $$ \begin{align} I_T &= \int_\mathbb{R}\int_{-T}^{T}\frac{e^{-ita}-e^{-itb}}{it}e^{itx}dtdF(x)\\ &= \int_\mathbb{R}\left[\int_{-T}^{T}\frac{\sin(t(x-a))}{t}dt - \int_{-T}^{T}\frac{\sin(t(x-b))}{t}dt\right]dF(x) \end{align} $$ if we consider the addition theorems and symmetry properties of the circular functions?

Answer 1

We have

$$\begin{align} \frac{e^{-ita}-e^{-itb}}{it}e^{itx} &= \frac{e^{it(x-a)}-e^{it(x-b)}}{it}\\ &= \frac{\cos\left( t(x-a)\right) - \cos \left(t(x-b)\right)}{it} + \frac{\sin \left(t(x-a)\right) - \sin \left(t(x-b)\right)}{t} \end{align}$$

by the addition theorem for the exponential function and Euler's formula $e^{iz} = \cos z + i\sin z$.

The first term, $\frac{\cos\left( t(x-a)\right) - \cos \left(t(x-b)\right)}{it}$ is an odd function, so its integral over the symmetric interval $[-T,T]$ is zero, and we are left with the integral involving only the sine terms.