Can the coefficients for the Fourier transform of the Chebyshev polynomials be further simplified?

Question

In an article by Fokas and Smitheman, it is shown, among others things, how the finite Fourier transform of the Chebyshev polynomials can be computed. They show that, when $T_{m}(x)$ is the $m$'th Chebyshev polynomial of the first order and $$\hat{T}_{m} (\lambda) = \int_{-1}^{1} e^{-i \lambda x} T_{m} (x) dx, \quad \lambda \in \mathbb{C}, \quad m=0,1,2,\dots $$ then $$\hat{T}_{m} (\lambda) = \sum_{n=1}^{m+1} \alpha_{n}^{m} \Bigg{[} \frac{e^{i \lambda}}{(i \lambda)^{n}} + (-1)^{n+m} \frac{e^{i \lambda}}{(i \lambda)^{n}} \Bigg{]} $$ with $\alpha_{1}^{m} = (-1)^{m} $, $\alpha_{2}^{m} = (-1)^{m+1}m^{2}$ and $$ a_{n}^{m} = (-1)^{m+n-1}2^{n-2}m \sum_{k=1}^{m-n+2} \binom{n+k-3}{k-1} \prod_{j=k}^{n+k-3} (m-j) \quad \text{for } n=3, 4, \dots, m+1. $$

Question: I wonder whether the expression for $\alpha_{n}^{m}$ -- when $3 \leq n \leq m+1$ -- can be either simplified further, or be cast into sequences of numbers that are already named or familiar?

Answer 1

Let $m\ge 0$ and $n=3,\cdots,m+1$. Then we have: \begin{eqnarray} &&\sum\limits_{k=1}^{m-n+2} \binom{n+k-3}{k-1} \prod\limits_{j=k}^{n+k-3}(m-j) =\\ &&(n-2)! \sum\limits_{k=1}^{m-n+2} \binom{n+k-3}{k-1} \binom{m-k}{n-2} = \\ && (n-2)! \sum\limits_{k=1}^{m-n+2} (-1)^{k-1} \binom{-n+1}{k-1} \binom{m-k}{n-2} = \\ && (n-2)! \binom{m-1}{n-2}\frac{2^{2-2 n} \Gamma (1-m) \Gamma \left(\frac{1}{2} (3-2 n)\right) }{\sqrt{\pi } \Gamma (-m-n+2)} =\\ && -(n-2)! \frac{2^{3-2 n} \cos (\pi (n-2)) \Gamma \left(\frac{1}{2} (3-2 n)\right) \Gamma (m+n-1)}{\sqrt{\pi } (n-2)! (m-n+1)!}=\\ && -\frac{2^{3-2 n} \cos (\pi (n-2)) \Gamma \left(\frac{1}{2} (3-2 n)\right) (m+n-2)!}{\sqrt{\pi } (m-n+1)!}=\\ &&\frac{2^{3-2 n} (m+n-2)!}{\left(\frac{1}{2}\right)_{n-1} (m-n+1)!} \end{eqnarray} In the second line from the top we wrote the product as a binomial coefficient. In the third line we used the identity $\binom{n}{k} = (-1)^k \binom{k-n-1}{k}$. In the fourth line we did the sum over $k$ owing to the fact that it is a convolution. In the fifth line we removed singularities in the expression by applying the reflection formula for the Gamma function. In the sixth line we simplified and finally in the last line we again applied the reflection formula for the Gamma function and simplified.

In[168]:= m1 = 
 Table[Sum[ 
   Binomial[n + k - 3, k - 1] Product[(m - j), {j, k, n + k - 3}], {k,
     1, m - n + 2}], {m, 0, 10}, {n, 3, m + 1}]
m2 = Table[(n - 2)! Sum[ 
     Binomial[n + k - 3, k - 1] Binomial[m - k, n - 2], {k, 1, 
      m - n + 2}], {m, 0, 10}, {n, 3, m + 1}];
m3 = Table[(n - 2)! Sum[(-1)^(k - 1) Binomial[-n + 1, k - 1] Binomial[
       m - k, n - 2], {k, 1, m - n + 2}], {m, 0, 10}, {n, 3, m + 1}];
m4 = -Table[(n - 2)! (
     2^(3 - 2 n)
       Pi   Gamma[1/2 (3 - 2 n)] Gamma[
       m + n - 1] (Cos[
        Pi (n - 2)]))/((m - n + 1)! (n - 2)! Sqrt[\[Pi]] Pi ), {m, 0, 
     10}, {n, 3, m + 1}];
m5 = -Table[(
    2^(3 - 2 n)
       Gamma[1/2 (3 - 2 n)] (m + n - 2)! (Cos[
       Pi (n - 2)]))/((m - n + 1)! Sqrt[\[Pi]]), {m, 0, 10}, {n, 3, 
     m + 1}];
m6 = Table[(
   2^(3 - 2 n)  (-2 + m + n)!)/((1 + m - n)! Pochhammer[1/2, 
     n - 1]), {m, 0, 10}, {n, 3, m + 1}];
Simplify[{m1 - m2, m1 - m3, m1 - m4, m1 - m5, m1 - m6}]

Out[168]= {{}, {}, {1}, {4, 2}, {10, 12, 6}, {20, 42, 48, 24}, {35, 
  112, 216, 240, 120}, {56, 252, 720, 1320, 1440, 720}, {84, 504, 
  1980, 5280, 9360, 10080, 5040}, {120, 924, 4752, 17160, 43680, 
  75600, 80640, 40320}, {165, 1584, 10296, 48048, 163800, 403200, 
  685440, 725760, 362880}}

Out[174]= {{{}, {}, {0}, {0, 0}, {0, 0, 0}, {0, 0, 0, 0}, {0, 0, 0, 0,
    0}, {0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 
   0, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0}}, {{}, {}, {0}, {0, 0}, {0, 0, 
   0}, {0, 0, 0, 0}, {0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0}, {0, 0, 0, 0,
    0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 
   0}}, {{}, {}, {0}, {0, 0}, {0, 0, 0}, {0, 0, 0, 0}, {0, 0, 0, 0, 
   0}, {0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 
   0, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0}}, {{}, {}, {0}, {0, 0}, {0, 0, 
   0}, {0, 0, 0, 0}, {0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0}, {0, 0, 0, 0,
    0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 
   0}}, {{}, {}, {0}, {0, 0}, {0, 0, 0}, {0, 0, 0, 0}, {0, 0, 0, 0, 
   0}, {0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 
   0, 0}, {0, 0, 0, 0, 0, 0, 0, 0, 0}}}

Answer 2

We also have the finite sum in the form of a couple of 3F1 hypergeometric functions whose inverse Fourier transforms are equal to $T_m$

\begin{equation} \int_{- 1}^1 \frac{ i e^{i y x} }{y} \left( \mathrm{e}^{\textrm{- i} \cdot y} { _3 F_1 } \hspace{-0.17em} \left( [1, m, - m], \left[ \frac{1}{2} \right], \frac{\mathrm{i}}{2y} \right) - \mathrm{e}^{\textrm{i} \cdot y} (-1)^m { _3 F_1 } \textrm{} \hspace{-0.17em} \left( [1, m, - m], \left[ \frac{1}{2} \right], - \frac{\mathrm{i}}{2y} \right) \right) dy = T_m (x) \end{equation}

plug the sum into maple, pass it thru simplify assuming m::integer and you will get the result.