Is This a Bessel Function? Is the function
$$y(x) = c \int_{-1}^1 \cos(xt)(1-t^2)^{n-\tfrac{1}{2}}dt = c \sum \tfrac{(-1)^m x^{2m}}{(2m)!} \int_{-1}^1 t^{2m}(1-t^2)^{n-\tfrac{1}{2}}dt$$
given here a bessel function? It doesn't look like the one given in 9.1.20 here yet the method of producing it (Laplace transforms, given here) seems rock-solid, what's going on here, why is it different from the standard one?
 A: $\newcommand{\+}{^{\dagger}}
 \newcommand{\angles}[1]{\left\langle #1 \right\rangle}
 \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\down}{\downarrow}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\isdiv}{\,\left.\right\vert\,}
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left( #1 \right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
This is a well known Bessel function identity:
$$
{\rm J}_{\nu}\pars{z}
={\pars{z/2}^{\nu} \over \Gamma\pars{\nu + 1/2}\Gamma\pars{1/2}}
\int_{-1}^{1}\expo{\ic zt}\pars{1 - t^{2}}^{\nu - 1/2}\,\dd t\,,\qquad\Re\pars{\nu} > -\,\half
$$

Then,
  \begin{align}
\color{#00f}{\large{\rm y}\pars{x}} &= c \int_{-1}^{1}\cos\pars{xt}\pars{1 - t^{2}}^{n-1/2}dt
= c\, \Re\int_{-1}^{1}\expo{\ic xt}\pars{1 - t^{2}}^{n-1/2}dt
\\[3mm]&=c\,{\Gamma\pars{n + 1/2}\Gamma\pars{1/2} \over \pars{x/2}^{n}}
\Re\bracks{
{\pars{x/2}^{n} \over \Gamma\pars{n + 1/2}\Gamma\pars{1/2}}\int_{-1}^{1}\expo{\ic xt}\pars{1 - t^{2}}^{n-1/2}dt}
\\[3mm]&=\color{#00f}{\large c\root{\pi}\pars{2 \over x}^{n}\Gamma\pars{n + \half}
\Re\pars{{\rm J}_{n}\pars{x}}\,,\qquad\Re\pars{n} > -\,\half}
\end{align}

A: Use the change of variables $t^2=y$ and the $\beta$ function to evaluate the integral
$$  2\int_{0}^1 t^{2m}(1-t^2)^{n-\tfrac{1}{2}}dt =  
\int _{0}^{1}\!{y}^{m-1/2} \left( 1-y \right) ^{n-1/2}{dy}= {\frac {\Gamma  \left( n+1/2 \right) \Gamma  \left( m+1/2 \right) }{
\Gamma  \left( m+n+1 \right) }}.$$
If you sum the series you will get the result

$$ 2^n \sqrt {\pi }\,c\, \Gamma  \left( n+1/2 \right)\, x^{-n} J_n(x),$$

where $J_n(x)$ is the Bessel function of the first kind.
