I want to solve the integral:
\begin{align} F_{\alpha}(\omega) = \frac{1}{\sqrt{2 \pi}} \cdot \sqrt{1 - j \text{cot}(\alpha)} \cdot e^{j \frac{1}{2} \text{cot}(\alpha) \omega^{2} } \int_{-\infty}^{\infty} e^{-j \text{csc}(\alpha) \omega t + j \frac{1}{2} \text{cot}(\alpha)t^{2} } e^{ j \omega_{0} t} dt \end{align}
where $j$ is the imaginary number. The problem is, the integral I get is not the answer I am expecting.
Worked Solution:
Combining the exponents gives us: \begin{align} F_{\alpha}(\omega) = \frac{1}{\sqrt{2 \pi}} \cdot \sqrt{1 - j \text{cot}(\alpha)} \cdot e^{j \frac{1}{2} \text{cot}(\alpha) \omega^{2} } \int_{-\infty}^{\infty} e^{- \frac{1}{2} \left[ -j \text{cot}(\alpha)t^{2} + 2j( \omega\text{csc}(\alpha) - \omega_{0} ) t \right]} dt \end{align} Completing the square gives us: \begin{align} F_{\alpha}(\omega) &= \frac{1}{\sqrt{2 \pi}} \cdot \sqrt{1 - j \text{cot}(\alpha)} \cdot e^{j \frac{1}{2} \text{cot}(\alpha) \omega^{2} } \int_{-\infty}^{\infty} e^{-\frac{1}{2} \left[ -j\text{cot}(\alpha) \left( t + \frac{\omega_{0} - \omega \text{csc}(\alpha)}{\text{cot}(\alpha)} \right)^{2} + j \frac{\left( \omega \text{csc}(\alpha) - \omega_{0} \right)^{2}}{\text{cot}(\alpha)} \right] } dt \\ F_{\alpha}(\omega) &= \frac{1}{\sqrt{2 \pi}} \cdot \sqrt{1 - j \text{cot}(\alpha)} \cdot e^{j \frac{1}{2} \text{cot}(\alpha) \omega^{2} } e^{-j \frac{1}{2} \frac{\left( \omega \text{csc}(\alpha) - \omega_{0} \right)^{2}}{\text{cot}(\alpha)}} \int_{-\infty}^{\infty} e^{-\frac{1}{2} \left[ -j\text{cot}(\alpha) \left( t + \frac{\omega_{0} - \omega \text{csc}(\alpha)}{\text{cot}(\alpha)} \right)^{2} \right] } dt \end{align} Now we let $z = t + \frac{\omega_{0} - \omega \text{csc}(\alpha)}{\text{cot}(\alpha)} $, which means $dz = dt$, and the bounds don't change. This means we have: \begin{align} F_{\alpha}(\omega) &= \frac{1}{\sqrt{2 \pi}} \cdot \sqrt{1 - j \text{cot}(\alpha)} \cdot e^{j \frac{1}{2} \text{cot}(\alpha) \omega^{2} } e^{-j \frac{1}{2} \frac{\left( \omega \text{csc}(\alpha) - \omega_{0} \right)^{2}}{\text{cot}(\alpha)}} \int_{-\infty}^{\infty} e^{j \frac{1}{2} \text{cot}(\alpha) z^{2} } dz \end{align} Well this is just a Gaussian integral, so we have: \begin{align} F_{\alpha}(\omega) &= \frac{1}{\sqrt{2 \pi}} \cdot \sqrt{1 - j \text{cot}(\alpha)} \cdot e^{j \frac{1}{2} \text{cot}(\alpha) \omega^{2} } e^{-j \frac{1}{2} \frac{\left( \omega \text{csc}(\alpha) - \omega_{0} \right)^{2}}{\text{cot}(\alpha)}} \sqrt{\frac{2 \pi}{-j \text{cot}(\alpha)}} \\ F_{\alpha}(\omega) &= \sqrt{ \frac{ 1 - j \text{cot}(\alpha)}{-j \text{cot}(\alpha)} } \cdot e^{j \frac{1}{2} \text{cot}(\alpha) \omega^{2} } e^{-j \frac{1}{2} \frac{\left( \omega \text{csc}(\alpha) - \omega_{0} \right)^{2}}{\text{cot}(\alpha)}} \end{align} Now we will work on simplifiying this expression. We start with the radical: \begin{align} F_{\alpha}(\omega) &= \sqrt{ 1 + j \text{tan}(\alpha) } \cdot e^{j \frac{1}{2} \text{cot}(\alpha) \omega^{2} } e^{-j \frac{1}{2} \frac{\left( \omega \text{csc}(\alpha) - \omega_{0} \right)^{2}}{\text{cot}(\alpha)}} \end{align} Now we combine the exponentials: \begin{align} F_{\alpha}(\omega) &= \sqrt{ 1 + j \text{tan}(\alpha) } \cdot e^{j \frac{1}{2} \left[ \text{cot}(\alpha) \omega^{2} - \frac{\left( \omega \text{csc}(\alpha) - \omega_{0} \right)^{2}}{\text{cot}(\alpha)} \right]} \end{align} Now lets expand the exponent: \begin{align} \text{cot}(\alpha) \omega^{2} - \frac{\left( \omega \text{csc}(\alpha) - \omega_{0} \right)^{2}}{\text{cot}(\alpha)} & = \text{cot}(\alpha) \omega^{2} - \frac{1}{\text{cot}(\alpha)} \cdot \left( \omega^{2} \text{csc}^{2}(\alpha) - 2 \omega_{0} \omega \text{csc}(\alpha) + \omega_{0}^{2} \right)\\ &= \frac{\text{cot}^{2}(\alpha) - \text{csc}^{2}(\alpha)}{\text{cot}(\alpha)} \omega^{2} + 2 \frac{\text{csc}(\alpha)}{\text{cot}(\alpha)} \omega_{0} \omega - \frac{1}{\text{cot}(\alpha)} \omega_{0}^{2} \end{align} Since $\text{cot}(\alpha) = \frac{1}{\text{tan}(\alpha)} = \frac{\text{cos}(\alpha)}{\text{sin}(\alpha)}$, and $\text{csc}(\alpha) = \frac{1}{\text{sin}(\alpha)}$, we have: \begin{align} &= \left( \text{cot}(\alpha) - \text{tan}(\alpha) \right) \omega^{2} + 2 \text{sec}(\alpha) \omega_{0} \omega - \text{tan}(\alpha) \omega_{0}^{2} \\ &= -(\omega^{2} + \omega_{0}^{2}) \text{tan}(\alpha) + 2 \text{sec}(\alpha) \omega_{0} \omega + \text{cot}(\alpha) \omega^{2} \\ \end{align} So we have: \begin{align} \boxed{ F_{\alpha}(\omega) = \sqrt{ 1 + j \text{tan}(\alpha) } \cdot e^{j\frac{1}{2} \text{cot}(\alpha) \omega^{2}} e^{-j \frac{1}{2} (\omega^{2} + \omega_{0}^{2}) \text{tan}(\alpha) + j \text{sec}(\alpha) \omega_{0} \omega} } \end{align}
The Problem:
According to the paper "The Fractional Fourier Transform and Time-Frequency Representations" by Luis B Almeida, the answer should be:
\begin{align} \boxed{ F_{\alpha}(\omega) = \sqrt{ 1 + j \text{tan}(\alpha) } e^{-j \frac{1}{2} (\omega^{2} + \omega_{0}^{2}) \text{tan}(\alpha) + j \text{sec}(\alpha) \omega_{0} \omega} } \end{align}
In other words, my answer has an extra $e^{-j\frac{1}{2} \text{cot}(\alpha) \omega^{2}}$ term.
Question:
What have I done wrong? Am I correct, or is the paper incorrect?
Edit 1: I have tried plugging it into Mathematica, but it is of no help, even after trying all the simplify commands.
The Mathematica Answer:
(0.707107 E^((0. + 0.5 I) w^2 Cot[a] - (0. + 0.5 I) (1. v - 1. wCsc[a])^2 Tan[a]) Sqrt[1 - I Cot[a]])/Sqrt[(0. - 0.5 I) Cot[a]]
which is different from everyone's....
Edit 2: Here is another twist in the saga.
In the paper, it says that Almeida's result holds if $\alpha - \frac{pi}{2} \neq n \cdot pi$ where $n \in \mathbb{Z}$
However, when $\alpha = n \cdot \pi + \frac{pi}{2} = \frac{2n+1}{2} \cdot \pi$, then $\text{cot}(\alpha) = \text{cot}\left( \frac{2n+1}{2} \cdot \pi \right) = 0$, so that extra complex exponential term is 1, which makes the my results agree with Almeida's.
What in the world??