Using FB=R = sqrt(d^2+r^2), and cos(alpha) = d/R, let the
angle BFD be e (for epsilon).

Then if the total flux emitted at F is 1, the flux through BD
is simply e/2pi, which is also the flux through BC. Now,
BC is BD/cos(alpha), and BD is Rtan(e), so the
average intensity of light passing through the segment BC is
therefore e/2pi * cos(alpha)/Rtan(e), which rearranges
quickly to

e/tan(e)  *  1/2pi  *  d/R^2

or
e/tan(e) * 1/2pi * d/(r^2+d^2)

and in the limit as e->0, this just gives

1/2pi * d/(r^2 + d^2)

for the flux density through the segment BC,
since e/tan(e)->1 as e->0.

I think that your approach, inasmuch as I can
work out what you're doing, is essentially assuming
that e is zero, so that e/tan(e) is already equal
to 1.

Check: the integral of this flux density over all values
of r should be 1/2 (NB, *not* 1, since only half
of the emitted light goes to the right of F, the rest
goes to the left).

1/2pi int_-infinity ^infinity d/(r^2+d^2)dr

=

1/2pi int_ -infinity ^infnity 1/(u^2+1) du

=

1/2pi (tan^{-1}(u)) _ -infinity ^infinity

=

1/2pi*(pi/2 - (-pi/2))

=

1/2pi * pi = 1/2.

Note that since your integral gives a value of 1 for
the total flux through the image plane, there is
presumably something wrong somewhere, since
not all the light emitted at F can pass through it.