Can you prove the asymptotic formula of the following function?

Question

In the following equation,

$$P(x; a)= \frac{\gamma}{2\lambda L \eta} [\frac{1}{π^2N_F(a)\eta(1 - \frac{x}{a\eta})^2} + \frac{1}{π^2N_F(a)\eta(1 + \frac{x}{a\eta})^2} +\frac{2}{π^2N_F(a)\eta(1 - \frac{x^2}{a^2\eta^2})} [sin (\frac{π N_F(a)\eta(1 - \frac{x}{a\eta})^2}{2})sin (\frac{πN_F(a)\eta(1 + \frac{x}{a\eta})^2}{2}) + cos (\frac{πN_F(a)\eta(1 - \frac{x}{a\eta})^2}{2} )cos (\frac{πN_F(a)\eta(1 + \frac{x}{a\eta})^2}{2})]]$$

if $ 0<N_F(a) ≪ 1$ and $(x − aη)/aη ≫ 1/\sqrt{N_F (a)η}$

how do you arrive at

$$P(x; a) ≃ \frac{2γ}{π^2η^2} (\frac{a^2}{(\frac {x^2}{η^2} − a^2)^2} +\frac{1}{\frac{x^2}{η^2} − a^2} sin^2 (πN_F(a)\frac{x}{a}))$$

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Please see what I have done so far and check if I have errors in it.

Relevant Equations

$\alpha (x; a) = \sqrt{N_F(a)\eta } (1 - \frac{x}{a\eta})$

where $\eta = 1 + L/D$, $N_F(a) = \frac{2a^2}{\lambda L}$ and (additional definition) $\gamma = \eta - 1$

Starting with this:

$P(x; a)=\frac{1}{2\lambda(L+D)} ([C(α(x; a)) + C(α(x; −a))]^2 + [S(α(x; a)) + S(α(x; −a))]^2)$

$C[α(x; +a)] + C[α(x; −a)] ≃ \frac{1}{πα(x; a)} sin (\frac{πα(x; a)^2}{2} ) + \frac{1}{πα(x; -a)} sin (\frac{πα(x; -a)^2}{2} ) $

$S[α(x; +a)] + S[α(x; −a)] ≃ \frac{-1}{πα(x; a)} cos (\frac{πα(x; a)^2}{2} ) - \frac{1}{πα(x; -a)} cos (\frac{πα(x; -a)^2}{2} ) $

Here is what I have done so far: $(C[α(x; +a)] + C[α(x; −a)])^2 + (S[α(x; +a)] + S[α(x; −a)])^2 = \frac{1}{π^2α^2(x; a)} + \frac{1}{π^2α^2(x; -a)} + \frac{2}{π^2α(x; +a)α(x; −a)} [sin (\frac{πα(x; a)^2}{2})sin (\frac{πα(x; -a)^2}{2}) + cos (\frac{πα(x; a)^2}{2} )cos (\frac{πα(x; -a)^2}{2})]$

$=\frac{1}{π^2N_F(a)\eta(1 - \frac{x}{a\eta})^2} + \frac{1}{π^2N_F(a)\eta(1 + \frac{x}{a\eta})^2} +\frac{2}{π^2N_F(a)\eta(1 - \frac{x^2}{a^2\eta^2})} [sin (\frac{π N_F(a)\eta(1 - \frac{x}{a\eta})^2}{2})sin (\frac{πN_F(a)\eta(1 + \frac{x}{a\eta})^2}{2}) + cos (\frac{πN_F(a)\eta(1 - \frac{x}{a\eta})^2}{2} )cos (\frac{πN_F(a)\eta(1 + \frac{x}{a\eta})^2}{2})]$

And then the outer factor: $\frac{1}{2\lambda (L+D)} = \frac{\gamma}{2\lambda L \eta}$

So the new equation for P is: $P(x; a)= \frac{\gamma}{2\lambda L \eta} [\frac{1}{π^2N_F(a)\eta(1 - \frac{x}{a\eta})^2} + \frac{1}{π^2N_F(a)\eta(1 + \frac{x}{a\eta})^2} +\frac{2}{π^2N_F(a)\eta(1 - \frac{x^2}{a^2\eta^2})} [sin (\frac{π N_F(a)\eta(1 - \frac{x}{a\eta})^2}{2})sin (\frac{πN_F(a)\eta(1 + \frac{x}{a\eta})^2}{2}) + cos (\frac{πN_F(a)\eta(1 - \frac{x}{a\eta})^2}{2} )cos (\frac{πN_F(a)\eta(1 + \frac{x}{a\eta})^2}{2})]]$

But this is where I am not sure what to do anymore with $N_F (a) ≪ 1$ and if $(x − aη)/aη ≫ 1/\sqrt{N_F (a)η}$ to arrive at this equation.

$P(x; a) ≃ \frac{2γ}{π^2η^2} (\frac{a^2}{(\frac {x^2}{η^2} − a^2)^2} +\frac{1}{\frac{x^2}{η^2} − a^2} sin^2 (πN_F(a)\frac{x}{a}))$

For more info about these derivations, please visit Page 14 of https://arxiv.org/pdf/1110.2346.pdf Picture of the final result in the paper

Answer 1

$P(x; a)= \frac{\gamma}{2\lambda L \eta} [\frac{1}{π^2N_F(a)\eta(1 - \frac{x}{a\eta})^2} + \frac{1}{π^2N_F(a)\eta(1 + \frac{x}{a\eta})^2} +\frac{2}{π^2N_F(a)\eta(1 - \frac{x^2}{a^2\eta^2})} [cos (\frac{πN_F(a)\eta(1 - \frac{x}{a\eta})^2}{2} - \frac{πN_F(a)\eta(1 + \frac{x}{a\eta})^2}{2})]]$

$P(x; a)= \frac{\gamma}{2\lambda L \eta} [\frac{2(1+\frac{x^2}{a^2\eta^2})}{π^2N_F(a)\eta(1 - \frac{x^2}{a^2\eta^2})^2} +\frac{2}{π^2N_F(a)\eta(1 - \frac{x^2}{a^2\eta^2})} [cos (2πN_F(a)\frac{x}{a})]]$

$P(x; a)= \frac{\gamma}{2\lambda L \eta} [\frac{2}{π^2N_F(a)\eta(1 - \frac{x^2}{a^2\eta^2})} [\frac{1 + \frac{x^2}{a^2\eta^2}}{1 - \frac{x^2}{a^2\eta^2}}+cos (2πN_F(a)\frac{x}{a})]] $

$P(x; a)= \frac{\gamma}{2\lambda L \eta} [\frac{2}{π^2N_F(a)\eta( \frac{x^2}{a^2\eta^2}-1)} [\frac{ \frac{x^2}{a^2\eta^2}}{ \frac{x^2}{a^2\eta^2}-1}+\frac{1 }{\frac{x^2}{a^2\eta^2}-1}-cos (2πN_F(a)\frac{x}{a})]] $

$P(x; a)≃ \frac{\gamma}{2\lambda L \eta} [\frac{2}{π^2N_F(a)\eta( \frac{x^2}{a^2\eta^2}-1)} [\frac{1 }{\frac{x^2}{a^2\eta^2}-1} +1-cos (2πN_F(a)\frac{x}{a})]] $

$P(x; a)≃ \frac{\gamma}{2\lambda L \eta} [\frac{2}{π^2N_F(a)\eta( \frac{x^2}{a^2\eta^2}-1)} [\frac{1 }{\frac{x^2}{a^2\eta^2}-1} +2 sin^2 (πN_F(a)\frac{x}{a})]] $

$P(x; a)≃ \frac{\gamma}{ \eta} \frac{1}{π^2a^2\eta( \frac{x^2}{a^2\eta^2}-1)} [\frac{2 }{\frac{x^2}{a^2\eta^2}-1} +2 sin^2 (πN_F(a)\frac{x}{a})] $

$P(x; a)≃ \frac{2\gamma}{π^2 \eta^2} (\frac{1}{a^2( \frac{x^2}{a^2\eta^2}-1)} [\frac{1 }{\frac{x^2}{a^2\eta^2}-1} + sin^2 (πN_F(a)\frac{x}{a})]) $

$P(x; a)≃ \frac{2\gamma}{π^2 \eta^2} (\frac{1}{( \frac{x^2}{\eta^2}-a^2)} [\frac{a^2 }{\frac{x^2}{\eta^2}-a^2} + sin^2 (πN_F(a)\frac{x}{a})]) $

$P(x; a)≃ \frac{2\gamma}{π^2 \eta^2} ( \frac{a^2 }{(\frac{x^2}{\eta^2}-a^2)^2} + \frac{1}{( \frac{x^2}{\eta^2}-a^2) }sin^2 (πN_F(a)\frac{x}{a})) $