quantum mechanics - Infinite square well in momentum space

As we know the eigenfunctions for a particle of mass $m$ in an infinite square well defined by $V(x) = 0$ if $0 \leq x \leq a$ and $V(x) = \infty$ otherwise are:

$$\psi_n (x) = \sqrt{\frac{2}{a}} \sin \left(\frac{n\pi x}{a} \right).$$

How does the ground state wave function look like in momentum space? As far as I recall I have to integrate $\psi_n(x)$ over the whole of space with the extra factor $\frac{e^{(-i p x / \hbar)}} {\sqrt {2 \pi \hbar}}$ (everything for $n = 1$).

In the solutions to this problem they integrated over $-a \leq x \leq a$ while I would've integrated from $0$ to $a$. Am I somehow missing something or is this solution just plain wrong?

A further question: How would I check whether or not my resulting $\psi(p)$ is an eigenstate of the momentum operator? Just slap the momentum operator in front of my function and see if I get something of the form $c \psi(p)$, where $c$ is some constant? Or how does this work?

Answer

Seems good to me. You are right integrating only from 0 to $a$ because $\psi$ is zero in the region of infinite potential. The solution would be $\psi(p)=\frac{1}{\sqrt{a\pi h}}\int_o^ae^{-ipx/\hbar}\sin(\pi x/a) dx=\frac{\sqrt{a\pi \hbar^3 }}{\pi^2 \hbar^2 - p^2 a^2 } (e^{-ipa/ \hbar }+1)$

As for the other question, that's what one typically does