I'm struggling with a discrepancy that Wolfram and Maple are giving me. To make everything comprehensible, I describe in the following what I did step by step.
When I let Wolfram calculate the Fourier Transform of the function:
$$H(z)=\sqrt{\frac{2}{\pi z}}e^{i \left(z-\frac{\pi}{4}\right)}$$
by using H[z_]:=Sqrt[2/(Pi*z)]*E^(I(z-Pi/4)); FourierTransform[H[z],z,w], I obtain:
$$
\frac{(1+i) e^{-\frac{1}{4} (i \pi )} (\text{sgn}(w+1)+1)}{\sqrt{2 \pi } \sqrt{\left| w+1\right| }}
$$
But when I use Matlab with the following code:
syms x v A w;
% We make the following substitutions:
% A=sqrt(2/pi) and
f = A/sqrt(x) * exp(i*(x-pi/4));
f_FT = fourier(f)
then Matlab yields f_FT=(2^(1/2)*(-1i)^(1/2)*A*pi^(1/2)*(sign(w-1)+1)*(1/2-1i/2))/abs(w-1)^(1/2) which after resubstitution of my variable $A$ leads to:
$$
\frac{2\sqrt{-i}\cdot(\text{sgn}(w+1)+1)\cdot\left(\frac{1}{2}-\frac{i}{2}\right)}{\sqrt{\left| w+1\right| }}
$$
Plotting both results using
w = 0:1:1000;
H1 = ((1+i)*exp(-i*pi/4).*(sign(w+1)+1))./(sqrt(2*pi).*sqrt(abs(w+1)));
H2 = (2*sqrt(-i).*(sign(w-1)+1)).*(1/2-i/2)./(sqrt(abs(w-1)));
figure; hold on;
plot(w, abs(H1), 'LineWidth', 1.5);
plot(w, abs(H2), 'LineWidth', 1.5);
xlim([0 1000])
ylim([0 3])
legend('Wolfram', 'Matlab'); grid on;
lead to the following chart:
Why we have such a kind of discrepancy?
Remark (inspired by Steven Clark):
Dividing the result that I obtained from Matlab by $\sqrt{2\pi}$ and when I accordingly define H2 = (2*sqrt(-i).*(sign(w-1)+1)).*(1/2-i/2)./(sqrt(2*pi).*sqrt(abs(w-1))), then both charts are graphically identical:
Best Answer
Another answer has been posted which provides some valuable theoretical insight, but this answer more specifically addresses the actual question.
The relationship $\sqrt{\frac{2}{\pi\ z}}=\sqrt{\frac{2}{\pi}}\frac{1}{\sqrt{z}}$ is only valid for $\Re(z)>0$. The correct relationship is $\sqrt{\frac{2}{\pi\ z}}=\sqrt{\frac{2}{\pi}}\sqrt{\frac{1}{z}}$ which for $z<0$ is equivalent to $\sqrt{\frac{2}{\pi\ z}}=-\sqrt{\frac{2}{\pi}}\frac{1}{\sqrt{z}}$.
Therefore there appears to be a problem in the assumptions for the Matlab evaluation of the Fourier transform of $H(z)=\sqrt{\frac{2}{\pi z}}e^{i \left(z-\frac{\pi}{4}\right)}$.
Using the same false assumption Mathematica evaluates the following statement as True where $H2(\omega)$ is the result of the Matlab Fourier transform evaluation as defined in the question above. The selection of Fourier parameters is described further below.
$\text{Assuming}\left[\omega \in \mathbb{R},\text{FullSimplify}\left[\sqrt{\frac{2}{\pi }} \text{FourierTransform}\left[\frac{e^{i \left(z-\frac{\pi }{4}\right)}}{\sqrt{z}},z,\omega ,\text{FourierParameters}\to \{1,-1\}\right]=\text{H2}(\omega )\right]\right]$
Mathematica defines the Fourier transform as illustrated in formula (1) below (see Details and Options at Wolfram FourierTransform) where the default Fourier parameters $\{a,b\}=\{0,1\}$ result in Formula (2) below and the Fourier parameters $\{a,b\}=\{1,-1\}$ result in Formula (3) below.
$$\mathcal{F}_x[f(x)](\omega)=\sqrt{\frac{| b| }{(2 \pi )^{1-a}}} \int\limits_{-\infty }^{\infty } f(x) e^{i b x \omega } \, dx\tag{1}$$
$$\mathcal{F}_x[f(x)](\omega )=\sqrt{\frac{1}{2 \pi }} \int\limits_{-\infty }^{\infty } f(x) e^{i x \omega } \, dx\ ,\quad\{a,b\}=\{0,1\}\tag{2}$$
$$\mathcal{F}_x[f(x)](\omega )=\int\limits_{-\infty }^{\infty } f(x) e^{-i x \omega } \, dx\ ,\quad\{a,b\}=\{1,-1\}\tag{3}$$
Matlab defines the Fourier transform as follows (see Matlab Fourier Transform) where the $fourier$ function uses the parameters $\{c,s\}=\{1,-1\}$.
$$\mathcal{F}_x[f(x)](\omega)=c \int\limits_{-\infty}^{\infty} f(x) e^{i s x \omega } \, dx\tag{4}$$
Therefore using the Fourier parameters $\{a,b\}=\{1,-1\}$ in Mathematica is equivalent to using the Fourier parameters $\{c,s\}=\{1,-1\}$ in Matlab.