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Is $e^{-\frac{|x|}{a}}$ an eigenfunction of momentum?
If we apply the momentum operator $\hat{P}=-i\hbar\frac{\partial }{\partial x}$ we get:
$$ -i\hbar\frac{\partial }{\partial x}e^{\frac{|x|}{a}}=\cases{i\hbar e^{-x/a} \ \ \ \ (x>0)\\-i\hbar e^{x/a}\ \ \ (x<0)} $$
Which is a constant times the function, however the constant depends on $x$ and so I would say it is not an eigenfunction.
It is not an eigenstate of momentum in infinite 1D space [$x\in(-\infty;\infty)$]. The reason is that the absolute value is not analytic in that space. You can think of it in general terms: Let $\psi(x)=A\mathrm{e}^{if(x)}$, with dimensionless $A$ and $f(x)$. Then $$-i\hbar\partial_x\psi(x)=\hbar A\frac{\partial f(x)}{\partial x}\mathrm{e}^{if(x)}=C\psi(x)\Leftrightarrow C=\hbar \frac{\partial f(x)}{\partial x}.$$
If $f$ is not differentiable everywhere inside the domain, then $\psi$ isn't an eigenfunction. You could redefine your domain to be only half space [$(-\infty,0]$ or $[0,\infty)$], but then what you have is simply $\mathrm{e}^{\pm ikx}$.
It is an eigenfunction of $p~=~-i\hbar\partial_x$ except at $x~=~0$. The derivative of the function $f(x)~=~e^{|x|/a}$ jumps discontinuously at $x~=~0$. We may approximate this with $f(x)~=~e^{\sqrt{\epsilon~+~x^2 }/a}$, for $\epsilon$ small. The derivative with respect to $x$ is $$ \frac{d}{dx}~=~\frac{x}{\sqrt{\epsilon~+~x^2}}e^{\sqrt{\epsilon~+~x^2 }/a}. $$ In the limit $\epsilon~\rightarrow~0$ this is $\frac{x}{|x|}e^{\sqrt{\epsilon~+~x^2 }/a}$ or $sgn(x)e^{\sqrt{\epsilon~+~x^2 }/a}$.
