Fourier transform between $x$ and $p$ On this page right at the top they mention two sets of fourier transform. First set is connection between $x$ (position) and $k$ (wave vector) space: 
$$
\begin{split}
f(x) &= \frac{1}{\sqrt{2\pi}} \int\limits_{-\infty}^{\infty} A(k) e^{ikx} dk\\
A(k) &= \frac{1}{\sqrt{2\pi}} \int\limits_{-\infty}^{\infty} f(x) e^{ikx} dx
\end{split}
$$
while the second set is connection between $x$ (position) and $p$ (momentum):
$$
\begin{split}
\psi(x) &= \frac{1}{\sqrt{2\pi \hbar}} \int\limits_{-\infty}^{\infty} \phi(p) e^{i\frac{p}{\hbar}x} dp\\
\phi(p) &= \frac{1}{\sqrt{2\pi \hbar}} \int\limits_{-\infty}^{\infty} \psi(x) e^{-i\frac{p}{\hbar}x} dx\\
\end{split}
$$

Q1: How do i derive the second set out of first one?
I know De Broglie relation $p = k \hbar$. Hence from $\exp[\pm ikx]$ in the first set we get $\exp \left[\pm i \frac{p}{\hbar} x\right]$ in the second set of equations. This is clear to me. What i dont know is how do we get from $1/\sqrt{2\pi}$ in the first set to $1/\sqrt{2 \pi \hbar}$ in the second set. Where does a $\hbar$ come from?
 A: As far as where you put things like the $2 \pi$ and the $\hbar$ in the Fourier transform or Inverse Fourier transform, it doesn't really matter. What really matters is that the operations are the inverse of each other. For example:
$\phi(p) = \frac{1}{\sqrt{2 \pi \hbar}} \int^{\infty}_{-\infty} dx \hspace{2mm} \psi(x) e^{-i \frac{p}{\hbar} x}  $
$ =\frac{1}{2 \pi \hbar} \int^{\infty}_{-\infty}dx \int^{\infty}_{-\infty}dq \hspace{2mm}\phi (q) e^{ i \frac{q}{\hbar} x } e^{-i \frac{p}{\hbar} x}$
$=\frac{1}{2 \pi \hbar} \int^{\infty}_{-\infty} dx \int^{\infty}_{-\infty}dq \hspace{2mm}\phi (q) e^{ i \frac{(q-p)}{\hbar} x } $
$ =\frac{1}{2 \pi \hbar} \int^{\infty}_{-\infty} dq\hspace{2mm} \phi (q)  2 \pi \hbar\delta(q - p)$ 
$= \int^{\infty}_{-\infty} dq \hspace{2mm}\phi (q) \delta(q - p) $
$ = \phi(p)$
This is what matters. Now if you really want to `derive' the second set of relations from the first, simply set $p\rightarrow \frac{p}{\hbar}$ and $A(k)\rightarrow A(k) \sqrt{\hbar}$. This gives
$f(x) = \frac{1}{\sqrt{2 \pi}} \int^{\infty}_{-\infty}A(k) \sqrt{\hbar} e^{i\frac{k}{\hbar} x} \frac{dk}{\hbar} =\frac{1}{\sqrt{2 \pi \hbar}} \int^{\infty}_{-\infty}A(k) e^{i\frac{k}{\hbar} x} dk  $ 
and 
$A(k ) \sqrt{\hbar} = \frac{1}{\sqrt{2 \pi}}  \int^{\infty}_{-\infty}f(x) e^{i \frac{k }{\hbar}x} dx \rightarrow 
A(k )  = \frac{1}{\sqrt{2 \pi\hbar}} \int^{\infty}_{-\infty}f(x) e^{i \frac{k }{\hbar}x} dx$
