Showing $\lim_{\Delta\rightarrow 0}\int_{-\Delta}^{\Delta}\frac{\mathrm{d}\omega}{(r+\omega)^2\sqrt{\Delta^2-\omega^2}}=\frac{\pi}{r^2}$ While I'm working on the proof of Bertrand's Theorem, I stuck at a limits calculation. I want to prove:
$$
\lim_{\Delta\rightarrow 0}\int_{-\Delta}^{\Delta}\frac{\mathrm{d}\omega}{(r+\omega)^2\sqrt{\Delta^2-\omega^2}}=\frac{\pi}{r^2}
$$
With the help of the software Mathematica, I got:
$$
\int_{-\Delta}^{\Delta}\dfrac{\mathrm{d}\omega}{(r+\omega)^2\sqrt{\Delta^2-\omega^2}}=\dfrac{\pi}{(1-\frac{\Delta^2}{r^2})^{\frac{3}{2}} \ r^2}
$$
but I have no clue how to formally calculate the limit or the integral. Thanks in advance if you can offer some help.
 A: Let
$\omega=\Delta \sin(x)$
$$I=\int\dfrac{d\omega}{(r+\omega)^2\sqrt{\Delta^2-\omega^2}}=\int\dfrac{dx}{(r+\Delta  \sin (x))^2}$$ Now, tanent half-angle substitution
$$I=\int \frac{2 \left(t^2+1\right)}{\left(r t^2+2 \Delta  t+r\right)^2}\,dt$$
Write
$$r t^2+2 \Delta  t+r=r (t-a)(t-b)$$ to make
$$I=\frac 2 {r^2} \int \frac{\left(t^2+1\right)}{(t-a)^2(t-b)^2}\,dt$$ Partial fracion decomposition
$$\frac{\left(t^2+1\right)}{(t-a)^2(t-b)^2}=\frac{a^2+1}{(a-b)^2 (t-a)^2}+\frac{b^2+1}{(a-b)^2 (t-b)^2}-\frac{2 (a
   b+1)}{(a-b)^3 (t-a)}+\frac{2 (a b+1)}{(a-b)^3 (t-b)}$$ Now, it is very simple.
When done, replace
$$a=\frac{\sqrt{\Delta ^2-r^2}-\Delta }{r} \qquad \text{and} \qquad b=\frac{-\sqrt{\Delta ^2-r^2}-\Delta }{r}$$
Edit
Since
$$I=\int_{-\Delta}^\Delta\dfrac{d\omega}{(r+\omega)^2\sqrt{\Delta^2-\omega^2}}=\frac{\pi  r^2 \sqrt{1-\frac{\Delta ^2}{r^2}}}{\left(r^2-\Delta ^2\right)^2}$$ the expansion around $\Delta=0$ is
$$\frac{\pi}{r^2} \left(1+\frac{3 \Delta ^2}{2 r^2}+\frac{15 \Delta ^4}{8 r^4}+\frac{35 \Delta ^6}{16
   r^6}+O\left(\Delta ^{8}\right) \right)$$
A: The substitution
$$\omega=\Delta\sin x\quad\left(-{\pi\over2}\leq x\leq{\pi\over2}\right),\qquad\sqrt{\Delta^2-\omega^2}=\Delta\cos x,\qquad d\omega=\Delta\cos x\>dx$$
gives
$$\int_{-\Delta}^\Delta{d\omega\over(r+\omega)^2\sqrt{\Delta^2-\omega^2}}=\int_{-\pi/2}^{\pi/2}{dx\over(r+\Delta\sin x)^2}\to{\pi\over r^2}\qquad(\Delta\to0)\ .$$
Since $r>0$ is fixed there is no question that we can do the limit $\Delta\to0$ under the integral sign.
A: Fix $r\neq 0$. Make the change of variables $\omega=\Delta x$. Then $d\omega = \Delta dx$ and$\sqrt{\Delta^2 - \omega^2} = \Delta \sqrt{1-x^2}$, so the factor of $\Delta$ cancels, leaving us with
$$ \int_{-1}^1 \frac{dx}{(r+\Delta x)^2\sqrt{1-x^2}}$$
now send $\Delta\to 0$; this is possible by Lebesgue dominated convergence and the fact that for $\Delta$ sufficiently small, $x<1<r/(2\Delta)$, so $$\frac1{(r+\Delta x)^2\sqrt{1-x^2}} \le \frac1{(r+\Delta x)^2\sqrt{1-x^2}} \le \frac4{r^2\sqrt{1-x^2}}\in L^1([-1,1]). $$
We are left with
$$ \frac1{r^2}\int_{-1}^1 \frac{dx}{\sqrt{1-x^2}}$$
This is well known, since $\frac1{\sqrt{1-x^2}} = \frac{d}{dx} \arcsin x$. The result follows.
