# Showing that $\lim_{k\rightarrow 0}\int_0^1\frac{dx}{\sqrt{(1-x^2)(1-k^2x^2)}} = \int_0^1\frac{dx}{\sqrt{(1-x^2)}}$

So I'm trying to show that:

$$\lim_{k\rightarrow 0}\int_0^1\frac{dx}{\sqrt{(1-x^2)(1-k^2x^2)}} = \int_0^1\frac{dx}{\sqrt{(1-x^2)}}$$

I guess this boils down to a solid understanding of uniform convergence.

There's also the second issue that the theorem in Rudin's PMA which discusses the exchange of a definite integral and a limit says that the the sequence of integrands must converge to the limit integrand uniformly on a closed interval, in this case that would be $[0,1]$. But of course our function is only defined on $[0,1)$, thus should I be considering the interval $[0,1-\epsilon]$?

As far as proving uniform convergence, I was looking at the the sequence:

$$M_{a_n} = \sup_{x\in [0,1-\epsilon]}|\frac{1}{\sqrt{(1-x^2)(1-(a_n)^2x^2)}} - \frac{1}{\sqrt{(1-x^2)}}|$$

for any sequence $a_n\rightarrow 0$.

And trying to prove that the sequence $M_{a_n}$ goes to zero, but it doesn't seem to upon pulling out the factor $\frac{1}{\sqrt{1-x^2}}$.

To sum up, this problem with its complexities is just a bit beyond my level of understanding and comfort with analysis. Could someone help me sort through it? Thanks.

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What you want to show is that $$\frac{1}{\sqrt{(1-x^2)(1-k^2x^2)}}\to \frac{1}{\sqrt{1-x^2}}$$ on uniformly $[0,1-\epsilon]$ for all $\epsilon>0$ and that $$\lim\limits_{\epsilon\to 0}\int_0^{1-\epsilon}\frac{dx}{\sqrt{(1-x^2)(1-k^2x^2)}}\to \int_0^{1}\frac{dx}{\sqrt{(1-x^2)(1-k^2x^2)}}$$ uniformly in $k$. The first fact gives us $$\lim\limits_{k\to 0}\int_0^{1-\epsilon}\frac{dx}{\sqrt{(1-x^2)(1-k^2x^2)}}=\int_0^{1-\epsilon}\frac{dx}{\sqrt{1-x^2}}$$ for all $\epsilon>0$, and since integration is continuous $$\lim_{\epsilon\to 0}\int_0^{1-\epsilon}\frac{dx}{\sqrt{1-x^2}}=\int_0^{1}\frac{dx}{\sqrt{1-x^2}}$$ while adding the second fact gives $$\lim\limits_{\epsilon\to 0}\lim\limits_{k\to 0}\int_0^{1-\epsilon}\frac{dx}{\sqrt{(1-x^2)(1-k^2x^2)}}=\lim\limits_{k\to 0}\int_0^{1}\frac{dx}{\sqrt{(1-x^2)(1-k^2x^2)}}$$ and from the these three equalities we can conclude the desired result.
To prove the first fact, note that \begin{align} \sup_{x\in [0,1-\epsilon]}\left|\frac{1}{\sqrt{(1-x^2)(1-k^2x^2)}}-\frac{1}{\sqrt{1-x^2}}\right| &\le\left(\sup_{x\in [0,1-\epsilon]}\frac{1}{\sqrt{1-x^2}}\right)\sup_{x\in [0,1-\epsilon]}\left|\frac{1}{\sqrt{1-k^2x^2}}-1\right|\\ &\le\left(\sup_{x\in [0,1-\epsilon]}\frac{1}{\sqrt{1-x^2}}\right)\left|\frac{1}{\sqrt{1-k^2(1-\epsilon)^2}}-1\right|\to 0\\ \end{align} giving uniform convergence. For the second fact, note that $$\int_{1-\epsilon}^1\frac{dx}{\sqrt{(1-x^2)(1-k^2x^2)}}\le \frac{1}{\sqrt{1-k^2}}\int_{1-\epsilon}^1\frac{dx}{\sqrt{1-x^2}}=\frac{1}{\sqrt{1-k^2}}\int_{0}^1\chi_{[1-\epsilon,1]}(x)\frac{dx}{\sqrt{1-x^2}}$$ where $\chi_{[1-\epsilon,1]}$ is the indicator function for the interval $[1-\epsilon,1]$. The function $\chi_{[1-\epsilon,1]}(x)\frac{1}{\sqrt{1-x^2}}$ converges pointwise to $0$ except at $x=1$ and is dominated by $\frac{1}{\sqrt{1-x^2}}$ which has finite integral, thus by the Dominated Convergence Theorem we see that $\int_{1-\epsilon}^1\frac{dx}{\sqrt{1-x^2}}\to 0$. If we bound $k$ below $1$ (say $k\le 1/2$) we see that $$\frac{1}{\sqrt{1-k^2}}\int_{1-\epsilon}^1\frac{dx}{\sqrt{1-x^2}}\leq \frac{4}{3}\int_{1-\epsilon}^1\frac{dx}{\sqrt{1-x^2}}\to 0$$ which is clearly uniform in $k$.
Thanks for this Alex. I was hoping to avoid the Dominated Convergence Theorem since I'm just taking my first measure theory class right now. However I'm a bit confused by $\int_{1-\epsilon}^1\frac{dx}{\sqrt{(1-x^2)(1-k^2x^2)}}\leq \int_{1-\epsilon}^1 \frac{dx}{\sqrt{1-x^2}}$, shouldn't the inequality face the other way? –  cactuar Sep 26 '12 at 4:52