Let $X$ be a random variable with pdf $$f_x(x)=\frac{1}{2\sigma}e^\dfrac{-|x-\mu|}{\sigma}$$, $-\infty< x<\infty$, $-\infty< \mu<\infty$, and $\sigma>0$. I have to find the mgf of $X$?.
What I got was this. MGF of $X =$ $$E(e^{tx})=\frac{1}{2\sigma}\int_{-\infty}^{\infty}e^{tx}e^\dfrac{-|x-\mu|}{\sigma} dx$$ $$=\frac{1}{2\sigma}\left[\int_{-\infty}^{\mu}e^{tx}e^\dfrac{x-\mu}{\sigma} dx +\int_{u}^{\infty} e^{tx}e^\dfrac{-x+\mu}{\sigma} dx\right] $$
$$=\frac{1}{2\sigma} \left[e^{\dfrac{-\mu}{\sigma}} \int_{-\infty}^{\mu} e^\dfrac{tx\sigma+x}{\sigma} dx+e^\dfrac{\mu}{\sigma} \int_{\mu}^{\infty} e^\dfrac{tx\sigma-x}{\sigma} dx\right] $$
$$=\frac{1}{2\sigma} [e^{\dfrac{-\mu}{\sigma}} \int_{-\infty}^\mu e^{\dfrac{x(t\sigma+1)}{\sigma}}dx +e^{\dfrac{\mu}{\sigma}} \int_\mu^\infty e^\dfrac{-x(1-t\sigma)}{\sigma} dx$$
$$=\dfrac{1}{2\sigma}[e^{\dfrac{-\mu}{\sigma}} (\dfrac{\sigma}{t\sigma+1})e^{\dfrac{\mu t\sigma+\mu}{\sigma}}+ \dfrac{\sigma}{1-\sigma t}e^\dfrac{\mu}{\sigma}e^{\dfrac{-\mu(1-\sigma t)}{\sigma}}] $$ $$=\dfrac{1}{2(t\sigma+1)}e^{\mu t} +\dfrac{1}{2(1-\sigma t)} e^{\mu t}$$ $$=e^{\mu t}\dfrac{1-\sigma t+1+\sigma t}{2(t\sigma+1)(1-\sigma t)}$$ $$=e^{\mu t}\dfrac{1}{1-\sigma^2 t^2}$$ (two's cancel out). Would this be right? I think im sure I made a mistake somewhere but not sure.
\begin{align} \mathbb E(e^{tX}) & = \int_{-\infty}^\infty e^{tx} \frac 1 \sigma e^{-\left|(x-\mu)/\sigma\right|} \, dx = \int_{-\infty}^\infty e^{t(\mu+\sigma w)} e^{-|w|}\, dw \\[10pt] & = e^{t\mu} \int_{-\infty}^\infty e^{(t\sigma)w} e^{-|w|} \, dw \\[10pt] & = e^{t\mu} \left( \int_{-\infty}^0 e^{(t\sigma+1)w} \,dw + \int_0^\infty e^{(t\sigma-1)w}\,dw \right) \\[10pt] & = e^{t\mu} \left( \frac1{1+t\sigma} + \frac1{1-t\sigma} \right) \\[10pt] & = e^{t\mu}\frac2{1-t^2\sigma^2}. \end{align}