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Here is a derivation starting from the Fourier transforms given in Orthogonal polynomials on the unit circle associated with the Laguerre polynomials. 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 $dz/dk=-i(1-z)^2$ and that $z$ varies over the lower half of the unit circle when $k$ varies over the positive real axis. The desired integral is $$\langle C\phi_m,S\phi_n\rangle=\frac{2}{\pi}\int_{0}^{\infty}[{\rm Re}\,(1-z)z^m][{\rm Im}\, (1-z)z^{n}]dk$$ $$\qquad = \frac{2i}{\pi}\int_{|z|=1,{\rm Im}\,z<0}[{\rm Re}\,(1-z)z^m][{\rm Im}\, (1-z)z^{n}]\frac{dz}{(1-z)^2}$$ $$\qquad = -\frac{2}{\pi}\int_{-\pi}^0[{\rm Re}\,(1-e^{i\phi})e^{im\phi}][{\rm Im}\, (1-e^{i\phi})e^{in\phi}]\frac{e^{i\phi}}{(1-e^{i\phi})^2}d\phi$$ $$\qquad=\frac{(2 m+1) +(-1)^{m+n+1}(2 n+1) }{(m-n) (m+n+1)\pi}$$ $$\qquad = \begin{cases} \displaystyle \frac2{\left(m+n+1\right)\pi} & \text{if $m+n$ even,} \\ \displaystyle \frac2{\left(m-n\right)\pi} & \text{if $m+n$ odd.} \end{cases}$$

Here is a derivation starting from the Fourier transforms given in Orthogonal polynomials on the unit circle associated with the Laguerre polynomials. 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 $$\langle C\phi_m,S\phi_n\rangle=\frac{2}{\pi}\int_{0}^{\infty}[{\rm Re}\,(1-z)z^m][{\rm Im}\, (1-z)z^{n}]dk$$ $$\qquad = \frac{2i}{\pi}\int_{|z|=1,{\rm Im}\,z<0}[{\rm Re}\,(1-z)z^m][{\rm Im}\, (1-z)z^{n}]\frac{dz}{(1-z)^2}$$ $$\qquad = -\frac{2}{\pi}\int_{-\pi}^0[{\rm Re}\,(1-e^{i\phi})e^{im\phi}][{\rm Im}\, (1-e^{i\phi})e^{in\phi}]\frac{e^{i\phi}}{(1-e^{i\phi})^2}d\phi$$ $$\qquad=\frac{(2 m+1) +(-1)^{m+n+1}(2 n+1) }{(m-n) (m+n+1)\pi}$$ $$\qquad = \begin{cases} \displaystyle \frac2{\left(m+n+1\right)\pi} & \text{if $m+n$ even,} \\ \displaystyle \frac2{\left(m-n\right)\pi} & \text{if $m+n$ odd.} \end{cases}$$

The identity follows directly from the Fourier transforms given in Orthogonal polynomials on the unit circle associated with the Laguerre polynomials. The Fourier cosine and sine transforms of $\phi_m$ are equal to the real and imaginary parts of $(1-z)z^m$, with $z=(2k-i)/(2k+i)$. The desired integral is $$\langle C\phi_m,S\phi_n\rangle=\frac{2}{\pi}\int_{0}^{\infty}[{\rm Re}\,(1-z)z^m][{\rm Im}\, (1-z)z^{n}]dk$$ $$\qquad = \frac{2i}{\pi}\int_{|z|=1,{\rm Im}\,z<0}[{\rm Re}\,(1-z)z^m][{\rm Im}\, (1-z)z^{n}]\frac{dz}{(1-z)^2}$$ $$\qquad = -\frac{2}{\pi}\int_{-\pi}^0[{\rm Re}\,(1-e^{i\phi})e^{im\phi}][{\rm Im}\, (1-e^{i\phi})e^{in\phi}]\frac{e^{i\phi}}{(1-e^{i\phi})^2}d\phi$$ $$\qquad=\frac{(2 m+1) +(-1)^{m+n+1}(2 n+1) }{(m-n) (m+n+1)\pi}$$ $$\qquad = \begin{cases} \displaystyle \frac2{\left(m+n+1\right)\pi} & \text{if $m+n$ even,} \\ \displaystyle \frac2{\left(m-n\right)\pi} & \text{if $m+n$ odd.} \end{cases}$$

Here is a derivation starting from the Fourier transforms given in Orthogonal polynomials on the unit circle associated with the Laguerre polynomials. 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 $dz/dk=-i(1-z)^2$ and that $z$ varies over the lower half of the unit circle when $k$ varies over the positive real axis. The desired integral is $$\langle C\phi_m,S\phi_n\rangle=\frac{2}{\pi}\int_{0}^{\infty}[{\rm Re}\,(1-z)z^m][{\rm Im}\, (1-z)z^{n}]dk$$ $$\qquad = \frac{2i}{\pi}\int_{|z|=1,{\rm Im}\,z<0}[{\rm Re}\,(1-z)z^m][{\rm Im}\, (1-z)z^{n}]\frac{dz}{(1-z)^2}$$ $$\qquad = -\frac{2}{\pi}\int_{-\pi}^0[{\rm Re}\,(1-e^{i\phi})e^{im\phi}][{\rm Im}\, (1-e^{i\phi})e^{in\phi}]\frac{e^{i\phi}}{(1-e^{i\phi})^2}d\phi$$ $$\qquad=\frac{(2 m+1) +(-1)^{m+n+1}(2 n+1) }{(m-n) (m+n+1)\pi}$$ $$\qquad = \begin{cases} \displaystyle \frac2{\left(m+n+1\right)\pi} & \text{if $m+n$ even,} \\ \displaystyle \frac2{\left(m-n\right)\pi} & \text{if $m+n$ odd.} \end{cases}$$

The identity follows directly from the Fourier transforms given in Orthogonal polynomials on the unit circle associated with the Laguerre polynomials. The Fourier cosine and sine transforms of $\phi_m$ are equal to the real and imaginary parts of $(1-z)z^m$, with $z=(2k-i)/(2k+i)$. The desired integral is $$\langle C\phi_m,S\phi_n\rangle=\frac{2}{\pi}\int_{0}^{\infty}[{\rm Re}\,(1-z)z^m][{\rm Im}\, (1-z)z^{n}]dk$$ $$\qquad = \frac{2i}{\pi}\int_{|z|=1,{\rm Im}\,z<0}[{\rm Re}\,(1-z)z^m][{\rm Im}\, (1-z)z^{n}]\frac{dz}{(1-z)^2}$$ $$\qquad = -\frac{2}{\pi}\int_{-\pi}^0[{\rm Re}\,(1-e^{i\phi})e^{im\phi}][{\rm Im}\, (1-e^{i\phi})e^{in\phi}]\frac{e^{i\phi}}{(1-e^{i\phi})^2}d\phi$$ $$\qquad=\frac{2}{\pi}\frac{(2 m+1) +(-1)^{m+n+1}(2 n+1) }{2 (m-n) (m+n+1)}$$

which is a different way to write the answer in the OP.

The identity follows directly from the Fourier transforms given in Orthogonal polynomials on the unit circle associated with the Laguerre polynomials. The Fourier cosine and sine transforms of $\phi_m$ are equal to the real and imaginary parts of $(1-z)z^m$, with $z=(2k-i)/(2k+i)$. The desired integral is $$\langle C\phi_m,S\phi_n\rangle=\frac{2}{\pi}\int_{0}^{\infty}[{\rm Re}\,(1-z)z^m][{\rm Im}\, (1-z)z^{n}]dk$$ $$\qquad = \frac{2i}{\pi}\int_{|z|=1,{\rm Im}\,z<0}[{\rm Re}\,(1-z)z^m][{\rm Im}\, (1-z)z^{n}]\frac{dz}{(1-z)^2}$$ $$\qquad = -\frac{2}{\pi}\int_{-\pi}^0[{\rm Re}\,(1-e^{i\phi})e^{im\phi}][{\rm Im}\, (1-e^{i\phi})e^{in\phi}]\frac{e^{i\phi}}{(1-e^{i\phi})^2}d\phi$$ $$\qquad=\frac{(2 m+1) +(-1)^{m+n+1}(2 n+1) }{(m-n) (m+n+1)\pi}$$ $$\qquad = \begin{cases} \displaystyle \frac2{\left(m+n+1\right)\pi} & \text{if $m+n$ even,} \\ \displaystyle \frac2{\left(m-n\right)\pi} & \text{if $m+n$ odd.} \end{cases}$$