Resonance production of charmonium

[gboff21]

The Question:

The cross section for the scattering of two particles with spins sa and sb via a resonance with
spin J is:
$\sigma(E)=\frac{\pi\lambda^{2}(2J+1)}{(2S_{a}+1)(2s_{b}+1)} \frac{\Gamma_{i}\Gamma_{j}}{(E-E_{0})^{2}+\frac{\Gamma^{2}}{4}}$
with $\lambda=1/p$, E is the centre-of-mass energy, E0 is the rest
mass energy of the resonance, $\Gamma$ is the total width of the resonance and $\Gamma_{i, f}$ are the partial widths for the decay of the resonance into the initial and final states, respectively.

The cross section for the production of the J/ψ resonance in e+e− collisions,
followed by its decay into e+e−, integrated over the centre-of-mass energy is:
$\int \sigma(E)dE=\frac{3\pi^{2}}{2}\lambda^{2}B^{2}_{J/\psi\rightarrow e^+e^-\Gamma}$
with
$B_{J/\psi\rightarrow e^+e^-}=\frac{\Gamma_{J/\psi\rightarrow e^+e^-}}{\Gamma}$
and we took the limits on the integral to be $\pm \infty$ and lambda is constant

The cross section for the production of (a) hadrons, (b) μ+μ− and (c)
e+e− is shown at http://imgur.com/94OYTaM. The measurements were made during a scan of the beam energies at the SPEAR
storage ring at SLAC using beams of e+ and e− circulating in opposite directions with the same
energy.
The observed width of the peak is due to the energy spread of the beams at each point in
the scan, the actual J/ width is much smaller than the observed width of the distributions.
However the relative centre-of-mass energies are known to about 1 part in 104.

At each scan point the beam energy spread produces a spread in the centre-of-mass energies
E′ distributed about the average centre-of-mass energy with a probability distribution
f (E − E′). Show that the measured area under the resonance peak is the same as the true
area under the peak, i.e.

$\int \sigma_{int}=\int \sigma_{measured}dE$My attempt:

Measured is wider than the actual resonance peak. So for the area to be the same, the actual peak has to be taller. All I can think of is that! How can I prove that mathematically?

[For some reason, the equations haven't come out well so please use this to view them http://www.codecogs.com/latex/eqneditor.php]

Thanks

 

[mfb]

The equations will be fine if you use correct tags: "[/itex]" instead of "[\itex]".

I think this can be solved with a simple manipulation of a double-integral (with E and E' as variables).

 

[gboff21]

Thanks for pointing the [/itex] out.

But how would you involve the f(E-E')?

 

[gboff21]

Is $\int\int \sigma f(E,E')dEdE' = \int\int \sigma dEdE'$ correct then?

 

[mfb]

σ depends on E, f(E,E') depends on the difference between those only, and I don't know what you mean with d.
In addition, some explanation or even steps how to get your integral would be useful.