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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 $$\langle C\phi_m,S\phi_n\rangle=\frac{2}{\pi}\int_{0}^{\infty}[{\rm Re}\,(1-z)z^m][{\rm Im}\, (1-z)z^{n}]dk$$ $$\qquad = \frac{2i}{\pi}\int_{|z|=1,{\rm Im}\,z<0}[{\rm Re}\,(1-z)z^m][{\rm Im}\, (1-z)z^{n}]\frac{dz}{(1-z)^2}$$ $$\qquad = -\frac{2}{\pi}\int_{-\pi}^0[{\rm Re}\,(1-e^{i\phi})e^{im\phi}][{\rm Im}\, (1-e^{i\phi})e^{in\phi}]\frac{e^{i\phi}}{(1-e^{i\phi})^2}d\phi$$ $$\qquad=\frac{(2 m+1) +(-1)^{m+n+1}(2 n+1) }{(m-n) (m+n+1)\pi}$$ $$\qquad = \begin{cases} \displaystyle \frac2{\left(m+n+1\right)\pi} & \text{if $m+n$ even,} \\ \displaystyle \frac2{\left(m-n\right)\pi} & \text{if $m+n$ odd.} \end{cases}$$