Find density given characteristic function

Question

Given,

Characteristic Function $\displaystyle G(k) = \frac{\sin (ka)}{ka}$

Find the corresponding density, I know the answer is

$\displaystyle p(x) = \frac{1}{2a} \ \ \ \ \ \ \ \ \ x \in (-a,a) $

I tried writing $\displaystyle \frac{\sin (ka)}{ka}$ as $\displaystyle\frac{e^{ika} - e^{-ika}}{2iak}$

And then using the inverse function but i am unable to integrate

I know,

$\displaystyle p(x) = \frac{1}{2\pi}\int_{-\infty}^{\infty} G(k) e^{-ikx} dk$

Can anyone pls help

Answer 1

You're right that that last integral is quite hard to do. To show that $p$ is the corresponding density, you just need to calculate $\int_\mathbb{R} p(x) e^{it x} \,dx$ and show that it equals $\frac{\sin(ta)}{ta}$.