# Finding the characteristic function of $Y \sim U(-1,1)$

I know that $\phi_Y(t) = E(e^{itY})=E(\cos(tY))+iE(\sin(tY))$

After integration I have found that $E(\cos(tY))= \frac{\sin(t)}{t}$ and $E(\sin(tY))=0$. So is the characteristic function just $\frac{\sin(t)}{t}$?

• This calculation is illustrated for the $U(0,1)$ distribution near the beginning of my post at stats.stackexchange.com/a/43075. Comparing your approach to that one might increase your confidence in your result.
– whuber
Oct 20, 2015 at 19:33

Here is another way to approach this. The characteristic function might be viewed as the moment generating function evaluated at $it$, in case the latter exists of course. But it does for this distribution (a bounded distribution you see), so let's find it. We have
$$\psi (t) = \int_{-1}^1 \frac{e^{tx}}{2}dx =\frac{1}{2t} \left[e^{t}- e^{-t} \right]$$
And evaluating at $it$ we get,
\begin{align} \psi (it)=\frac{1}{2it} \left[ e^{it}-e^{-it} \right] &= \frac{1}{2it}\left[ \cos t + i \sin t -\left( \cos t- i\sin t \right) \right] \\ &= \frac{1}{2it} \left[ 2i \sin t\right] \\ &= \frac{\sin t}{t} \end{align}