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?
Best Answer
Hint: Use that
$$t \mapsto \frac{\cos(t \alpha)-\cos(t \beta)}{t}$$
is an uneven function for any constants $\alpha,\beta \in \mathbb{R}$.