Let $S$ and $C$ denote the Fourier sine transform and the Fourier cosine transform, respectively, i.e.,
\begin{align\*}
 S f(k) &= \sqrt{\frac2\pi} \, \int\_0^\infty f(x) \sin(kx)\,dx, \\
 C f(k) &= \sqrt{\frac2\pi} \, \int\_0^\infty f(x) \cos(kx)\,dx.
\end{align\*}
Recall that $S$ and $C$ are unitary selfadjoint operators on $L^2(\mathbb R\_+)$.


Further let $\varphi\_m(x) = e^{-x/2} L\_m(x)$ for $m\in\mathbb N\_0$, where $L\_m$ is the $m$-th Laguerre polynomial. Then $\left\{\varphi\_m\right\}\_{m=0}^\infty$ forms an orthonormal basis of $L^2(\mathbb R\_+)$. I am interested in a reference or a (short) proof of the following fact which is likely to be known:



> 
> $$
>  \left\langle C\varphi\_m,S\varphi\_n\right\rangle = \begin{cases} 
>  \displaystyle
>  \frac2{\left(m+n+1\right)\pi} & \text{if $m\equiv n \pmod{2}$,} \\
>  \displaystyle
>  \frac2{\left(m-n\right)\pi} & \text{if $m\not\equiv n \pmod{2}$.}
>  \end{cases}
> $$
> 
> 
> 


Here $\langle\;,\,\rangle$ is the scalar product on $L^2(\mathbb R\_+)$.



Fourier sine and cosine transforms and Laguerre polynomials
Here is a derivation starting from the Fourier transforms given in [Orthogonal polynomials on the unit circle associated with the Laguerre polynomials.](http://www.ams.org/journals/proc/2001-129-03/S0002-9939-00-05821-4/S0002-9939-00-05821-4.pdf) The Fourier cosine and sine transforms of $\phi\_m$ are equal to the real and imaginary parts of $\sqrt{2/\pi}(1-z)z^m$, with $z=(2k-i)/(2k+i)$ on the unit circle in the complex plane. Notice that $idz/dk=(1-z)^2$ and that $z$ varies over the lower half of the unit circle when $k$ varies over the positive real axis. The desired integral is
$$\langle C\phi\_m,S\phi\_n\rangle=\frac{2}{\pi}\int\_{0}^{\infty}[{\rm Re}\,(1-z)z^m][{\rm Im}\, (1-z)z^{n}]dk$$
$$\qquad = \frac{2i}{\pi}\int\_{|z|=1,{\rm Im}\,z<0}[{\rm Re}\,(1-z)z^m][{\rm Im}\, (1-z)z^{n}]\frac{dz}{(1-z)^2}$$
$$\qquad = -\frac{2}{\pi}\int\_{-\pi}^0[{\rm Re}\,(1-e^{i\phi})e^{im\phi}][{\rm Im}\, (1-e^{i\phi})e^{in\phi}]\frac{e^{i\phi}}{(1-e^{i\phi})^2}d\phi$$
$$\qquad=\frac{(2 m+1) +(-1)^{m+n+1}(2 n+1) }{(m-n) (m+n+1)\pi}$$
$$\qquad = \begin{cases} 
 \displaystyle
 \frac2{\left(m+n+1\right)\pi} & \text{if $m+n$ even,} \\
 \displaystyle
 \frac2{\left(m-n\right)\pi} & \text{if $m+n$ odd.}
 \end{cases}$$


