I am stuck on this statistics problem:
If the pdf of a measurement error $ X $ is $f(x)=0.5e^{-|x|}$, $ -\infty<x<\infty $, show that $ M_X(t)=\frac{1}{1-t^2} $ for $ |t|<1. $
I've split this into 2 parts of a piece wise. I get for $ x\geq0 $, $ M_{X}(t)=-\frac{1}{2(t-1)} $ and for $ x<0 $, $ M_{X}(t)=\frac{1}{2(t+1)} $ by integrating. I don't know how to relate this to the $ \frac{1}{1-t^2} $.
\begin{align} M_X(t) &= E(\exp(Xt)) \\ &= \int_{-\infty}^\infty \exp(xt) f(x) \, dx \\ &= 0.5 \int_{-\infty}^\infty \exp(xt-|x|) \, dx \\ &= 0.5 \left(\int_{-\infty}^0 \exp(xt+x) \, dx + \int_0^{\infty} \exp(xt-x) \, dx\right)\\ &= 0.5 \left( \frac{1}{t+1} - \frac{1}{t-1} \right)\\ &= 0.5\left( \frac{t-1-(t+1)}{(t+1)(t-1)}\right) \\ &= 0.5\left(\frac{-2}{t^2-1} \right)\\ &= \frac{1}{1-t^2} \end{align}