Here $f(x) = \frac{1}{2\sigma} e^{-\frac{|x|}{\sigma}};\quad -\infty < x < \infty,\quad \sigma > 0$.
Now,
$$
\begin{eqnarray}
E(X) &=& \frac{1}{2\sigma} \int_{-\infty}^{\infty} x e^{-\frac{|x|}{\sigma}} dx\\
&=& \frac{1}{2\sigma} \int_{-\infty}^{0} x e^{-\frac{|x|}{\sigma}} dx + \frac{1}{2\sigma} \int_{0}^{\infty} x e^{-\frac{|x|}{\sigma}} dx\\
&=& I_1 + I_2\\
\end{eqnarray}
$$
Note that $$\begin{eqnarray}|x| &=& x; \quad x>0\\
&=& -x; \quad x<0\end{eqnarray}$$
First consider $I_1$:
$$
\begin{eqnarray}
I_1 &=& \frac{1}{2\sigma} \int_{-\infty}^{0} x e^{-\frac{|x|}{\sigma}} dx\\
&=& \frac{1}{2\sigma} \int_{-\infty}^{0} x e^{\frac{x}{\sigma}} dx\\
&=& \frac{1}{2\sigma}\left[\left\{x\int e^{\frac{x}{\sigma}}dx\right\}_{-\infty}^0 - \int_{-\infty}^0 \left\{\frac{d}{dx}x \int e^{\frac{x}{\sigma}}dx\right\}dx\right]; \text{by partial integration}\\
&=&0
\end{eqnarray}
$$
Next consider $I_2$:
$$
\begin{eqnarray}
I_2 &=& \frac{1}{2\sigma} \int_{0}^{\infty} x e^{-\frac{|x|}{\sigma}} dx\\
&=& \frac{1}{2\sigma} \int_{0}^{\infty} x e^{-\frac{x}{\sigma}} dx
\end{eqnarray}
$$
Taking $\frac{x}{\sigma} = u$, this integral becomes simply $\frac{\sigma}{2}$.
Hence the actual expectation is $\frac{\sigma}{2}$.
