Momentum space wave equation of free particle: constant factors I'm trying to solve problem 3.12 in D.J. Griffiths's "Introduction to Quantum Mechanics 3rd ed."; it is as follows:

Find [the momentum space wave equation] $\Phi(p,t)$ for the free particle in terms of $\phi(k)$.

$\phi(k)$ is defined in the free particle's 1D position space wave equation
$$\Psi(x,t) = \frac{1}{\sqrt{2\pi}}\int^{+\infty}_{-\infty}\phi(k)e^{ikx}e^{-i\frac{\hbar k^2}{2m}t}dk$$
as 
$$\phi(k) = \frac{1}{\sqrt{2\pi}}\int^{+\infty}_{-\infty}\Psi(x,0)e^{-ikx}dx$$
That is, if we use the definition of the Fourier transform where $\frac{1}{\sqrt{2\pi}}e^{\mp ikx}$ is used in the integrands for the Fourier and inverse Fourier transforms respectively (I was always taught $e^{\mp 2\pi isx}$, but I'll roll with Griffiths's scaling), then $\phi(k)$ is really just the Fourier transform for the initial state of the wave equation in position space.
Now, the problem I'm running up against is the following: if I use Griffiths's way of converting $\Psi(x,t)$ to $\Phi(p,t)$ (position space to momentum space wave equation), i.e.
$$\Phi(p,t) = \frac{1}{\sqrt{2\pi\hbar}}\int^{+\infty}_{-\infty}\Psi(x,t)e^{-i\frac{p}{\hbar}x}dx$$
I get
$$\Phi(p,t) = \frac{1}{\sqrt{2\pi\hbar}}\int^{+\infty}_{-\infty}\left(\frac{1}{\sqrt{2\pi}}\int^{+\infty}_{-\infty}\phi(k)e^{ikx}e^{-i\frac{\hbar k^2}{2m}t}dk\right)e^{-i\frac{p}{\hbar}x}dx$$
My intuition says that the two exponentials should just cancel out, and so the only way I see of simplifying the expression is if I assume that $p=\hbar k$ (I try to be very careful with this substitution, because it often causes trouble with constant factors). I get:
$$\Phi(p,t) = \frac{1}{\sqrt{\hbar}}\frac{1}{\sqrt{2\pi}}\int^{+\infty}_{-\infty}\left(\frac{1}{\sqrt{2\pi}}\int^{+\infty}_{-\infty}\phi\left(\frac{p}{\hbar}\right)e^{i\frac{p}{\hbar}x}e^{-i\frac{p^2}{2m\hbar}t}d\left(\frac{p}{\hbar}\right)\right)e^{-i\frac{p}{\hbar}x}dx$$
The inner integral performs an inverse Fourier transform, the outer a Fourier transform, so they cancel, to get:
$$\Phi(p,t) = \frac{1}{\sqrt{\hbar}} \phi\left(\frac{p}{\hbar}\right) e^{-i\frac{E}{\hbar}t}$$
This is nice and all, but I've read and been told before that $\phi(k)$ is the time-independent momentum space wave equation similar to $\psi(x)$, not $\frac{1}{\sqrt{\hbar}}\phi(k)$. What should be the scaling factor? I feel like $p=\hbar k$ is either not always applicable, or it may only be done when additional factors are added in front of the Fourier integrals (even if the integration variable is $x$ and the scaling is thus not really due to substitution of $dx$). 
(I've looked here, but it doesn't give me any answers.)
 A: The problem is with your use of mixed variables $k$ and $p$.  First, it is best to think of 
\begin{align}
\langle x\vert p\rangle &=\frac{1}{\sqrt{2\pi\hbar}}e^{-i p x/\hbar}\, ,\qquad 
\langle p\vert x\rangle = \langle x\vert p\rangle^* =\frac{1}{\sqrt{2\pi\hbar}}e^{+i p x/\hbar}
\end{align}
which justifies the symmetric placement of the $\sqrt{2\pi \hbar}$ factor, but  on the other hand 
\begin{align}
\langle x\vert k\rangle &=\frac{1}{\sqrt{2\pi}}e^{-i k x }\, ,\qquad 
\langle k\vert x\rangle = \langle x\vert k\rangle^* =\frac{1}{\sqrt{2\pi}}e^{+i k x}
\end{align}
so that 
\begin{align}
\Psi(p,t)&=\langle p\vert \psi\rangle = \int dx 
\langle p\vert x\rangle \langle x\vert \Psi(t)\rangle = \int dx \frac{1}{\sqrt{2\pi\hbar}}e^{ipx/\hbar}\Psi(x,t)\, ,\\
\Psi(k,t)&=\langle k\vert \psi\rangle = \int dx 
\langle k\vert x\rangle \langle x\vert \Psi(t)\rangle = \int dx \frac{1}{\sqrt{2\pi}}e^{ikx}\Psi(x,t)\, ,\\
&= \sqrt{\hbar} \,\Psi(p,t)\, ,
\end{align}
where the unit operator
\begin{align}
\hat 1=\int dx \vert x\rangle \langle x\vert
\end{align}
has been used.
A: As @Prahar has correctly pointed out, there is a purely mathematical error when equating the integration variable $k$ with the external variable $p$. Using two different symbols (e.g., $k$ and $k'$) would be the correct approach.
Further, the key to the solution is using the Fourier representation of the $\delta$-function (afterr changing the order of integration):
$$\int_{-\infty}^{+\infty}dxe^{i(k-\frac{p}{\hbar})x} = 2\pi\delta(k - \frac{p}{\hbar}).$$
Note
Griffiths scaling in Fourier transform is commonly used in physics, both in space ($k$) and in time ($\omega$) transforms. Also, don't be surprised to see differentials written right after the integration sign, before the integrand - as I have done - although this is more typical of quantum mechanics. 
A: Momentum $p$ and wave number $k$ are in fact related by $p=\hbar k$. Because they just differ by a constant, usually both are considered to describe the momentum of quantum systems (especially if you do your work with $\hbar=1$). 
Note that this follows directly from the de Broglie relation $p=hf=2\pi\hbar/\lambda=\hbar k$, 
