Distribution Derivative Question: Find the second derivative of the function $f(x) = e^{-|x|}$ using the sense of Theory of distribution. 
My Attempt: 
$$
\left< f',\phi \right> = -\left< f,\phi' \right> = -\int_{-\infty}^{\infty} f(x) \, \phi'(x) \, dx = 2 \, \phi(0) - \int_{-\infty}^{\infty} e^{-|x|} \, \phi(x) \, dx.
$$
Thus, $f'= 2\delta +h,$ where $h$ is the function $h(x) = e^{-|x|}.$ Now, by definition: $\left< f'', \phi \right> = - \left< f', \phi' \right>,$ but  $\phi' = \psi$ and apply the rule
$$\left< f',\psi \right> = - \left< f,\psi' \right>.$$
This gives:
$$\left< f'',\phi \right>  = \left< f,\phi'' \right>$$
By the preceding computation:
$$- \left< f,\phi' \right> = 2\phi'(0) - \int_{-\infty}^{\infty} e^{-|x|} \, dx$$
After integration by parts:
$$-2\phi'(0) + \int_{-\infty}^{\infty} f(x) \, \phi(x) \, dx.$$
Thus, 
$$\left< f'', \phi \right> =-2\phi'(0)+ \int_{-\infty}^{\infty} f(x) \, \phi(x) \, dx.$$
herefore, in the distribution sense,
$f'' = 2\delta+f,$
where, 
$\left< \delta', \phi \right> = \phi'(0).$
Is my work correct? 
 A: $$\begin{align}
\langle f'', \phi \rangle
&= - \langle f', \phi' \rangle
= \langle f, \phi'' \rangle
= \int_{-\infty}^{\infty} e^{-|x|} \, \phi''(x) \, dx \\
&= \int_{-\infty}^{0} e^{x} \, \phi''(x) \, dx
+ \int_{0}^{\infty} e^{-x} \, \phi''(x) \, dx \\
&= \left( \left[e^{x} \, \phi'(x) \right]_{-\infty}^{0} - \int_{-\infty}^{0} e^{x} \, \phi'(x) \, dx \right) 
+ \left( \left[ e^{-x} \, \phi'(x) \right]_{0}^{\infty} - \int_{0}^{\infty} (-e^{-x}) \, \phi'(x) \, dx \right) \\
&= - \int_{-\infty}^{0} e^{x} \, \phi'(x) \, dx + \int_{0}^{\infty} e^{-x} \, \phi'(x) \, dx \label{middle step}\tag{*} \\
&= - \left( \left[e^{x} \, \phi(x) \right]_{-\infty}^{0} - \int_{-\infty}^{0} e^{x} \, \phi(x) \, dx \right)
+ \left( \left[ e^{-x} \, \phi(x) \right]_{0}^{\infty} - \int_{0}^{\infty} (-e^{-x}) \, \phi(x) \, dx \right) \\
&= -2 \, \phi(0) + \int_{-\infty}^{0} e^{x} \, \phi(x) \, dx + \int_{0}^{\infty} e^{-x} \, \phi(x) \, dx \\
&= \langle -2 \, \delta + f, \phi \rangle
\end{align}$$
A parenthesis: 
Note that \eqref{middle step} can be written using $\operatorname{sign}$ as
$$
\int_{-\infty}^{\infty} \operatorname{sign}(x) e^{-|x|} \, \phi'(x) \, dx
$$
