The identity follows directly from the Fourier transforms given in <A HREF="http://www.ams.org/journals/proc/2001-129-03/S0002-9939-00-05821-4/S0002-9939-00-05821-4.pdf">
Orthogonal polynomials on the unit circle associated with the Laguerre polynomials.
</A> 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
$$\langle C\phi_m,S\phi_n\rangle=\frac{2}{\pi}\int_{0}^{\infty}[{\rm Re}\,(1-z)z^m][{\rm Im}\, (1-z)z^{n}]dk$$
$$\qquad = \frac{2i}{\pi}\int_{|z|=1,{\rm Im}\,z<0}[{\rm Re}\,(1-z)z^m][{\rm Im}\, (1-z)z^{n}]\frac{dz}{(1-z)^2}$$
$$\qquad = -\frac{2}{\pi}\int_{-\pi}^0[{\rm Re}\,(1-e^{i\phi})e^{im\phi}][{\rm Im}\, (1-e^{i\phi})e^{in\phi}]\frac{e^{i\phi}}{(1-e^{i\phi})^2}d\phi$$
$$\qquad=\frac{(2 m+1) +(-1)^{m+n+1}(2 n+1) }{(m-n) (m+n+1)\pi}$$
$$\qquad = \begin{cases} 
     \displaystyle
     \frac2{\left(m+n+1\right)\pi} & \text{if $m+n$ even,} \\
     \displaystyle
     \frac2{\left(m-n\right)\pi} & \text{if $m+n$ odd.}
 \end{cases}$$