Proof of Inversion formula for characteristic function

Question

I have a question about the proof to the inversion formula for characteristic function. The Theorem is stated as following:

$\lim_{T\rightarrow\infty}\frac{1}{2\pi}\int_{-T}^T \frac{e^{-ita} - e^{-itb}}{it}\phi(t)dt = \mathbb{P}(a,b) + \frac{1}{2}\mathbb{P}(\{a,b\})$,

where $\phi_{X}(t)$ is the characteristic function of a random variable. In the proof of Chung in his book "A course in probability theory" on page 162 there is the following identity:

\begin{align} \int_{-T}^{T}\frac{e^{-ita} - e^{-itb}}{2\pi it}e^{itx}dt &= \int_{-T}^{T}\frac{e^{it(x - a)} - e^{it(x-b)}}{2 \pi it}dt \\ &= \frac{1}{\pi}\int_{0}^{T}\frac{\sin(t(x-a))}{t}dt - \frac{1}{\pi}\int_{0}^{T}\frac{\sin(t(x-b))}{t}dt \end{align}

I don't know how to show this. My attempt is the following: \begin{align} \frac{e^{-ita} - e^{-itb}}{it}e^{itx} &= \frac{e^{it(x - a)} - e^{it(x-b)}}{it}\\&=\frac{-i\left(e^{it(x - a)} - e^{it(x-b)}\right)}{t}\\ &=\frac{\sin(t(x-a)) - \sin(t(x-b)) + i\left(\cos(t(x-b)) - \cos(t(x-a)) \right)}{t}. \end{align}

Has anyone an idea?

Answer 1

Hint: Use that

$$t \mapsto \frac{\cos(t \alpha)-\cos(t \beta)}{t}$$

is an uneven function for any constants $\alpha,\beta \in \mathbb{R}$.

Answer 2

Thanks for the hint. Then I have the following:

\begin{align} \int_{-T}^{T}\frac{\cos(t\alpha) - \cos(t\beta)}{t}dt &= \int_{0}^{T}\frac{\cos(t\alpha) - \cos(t\beta)}{t}dt + \int_{-T}^{0}\frac{\cos(t\alpha) - \cos(t\beta)}{t}dt\\ &= \int_{0}^{T}\frac{\cos(t\alpha) - \cos(t\beta)}{t} + \int_{0}^{T}\frac{\cos(t\beta) - \cos(t\alpha)}{t}\\ &= 0 \end{align}

Correct?