It's an example given in my book after monotone convergence theorem and dominated convergence theorem (without explanation) :

Find an equivalent of $$\int_0^{\pi/2}\dfrac {dx} {\sqrt{\sin^2(x)+\epsilon \cos^2(x)}}$$

when $\epsilon\to 0^{+}$.

Inspired of the theorems, I naturally think of the sequence $(\epsilon_n)$ that converges to $0$, and it is monotone. However, the limit (1/sin(x)) doesn't converge in the integral (the other examples converge to a finite number...). Would someone give a hint about how to deal with the divergent case?

  • $\begingroup$ What do you mean with "Find an equivalent of"? $\endgroup$ – Robert Z Sep 23 '17 at 7:01
  • $\begingroup$ Denote the integral by $I$. We look for a closed form $f(\epsilon)$ s.t. $f(\epsilon)\sim I$... $\endgroup$ – pqros Sep 23 '17 at 7:18
  • $\begingroup$ I was typing my answer when came Robert Z's one ! He won the race about elliptic integrals ! $\endgroup$ – Claude Leibovici Sep 23 '17 at 9:00

At least for the time being, this is probaly not an answer.

Sooner or later, you will learn about elliptic integrals and, assuming $\epsilon >0$, $$\int\dfrac {dx} {\sqrt{\sin^2(x)+\epsilon \cos^2(x)}}=\frac{1}{\sqrt{\epsilon }}F\left(x\left|\frac{\epsilon -1}{\epsilon }\right.\right)$$ where appears the elliptic integral of the first kind. This makes $$I=\int_0^{\pi/2}\dfrac {dx} {\sqrt{\sin^2(x)+\epsilon \cos^2(x)}}=\frac{1}{\sqrt{\epsilon }}K\left(\frac{\epsilon -1}{\epsilon }\right)$$ where appears the complete elliptic integral of the first kind.

Now, you could use the expansion $$K\left(\frac{\epsilon -1}{\epsilon }\right)=\sqrt{\epsilon } \left(2 \log (2)-\frac{\log (\epsilon )}{2}\right)+\frac{1}{8} \epsilon ^{3/2} (4 \log (2)-\log (\epsilon )-2)+O\left(\epsilon ^{5/2}\right)$$ which makes $$I=\left(2 \log (2)-\frac{\log (\epsilon )}{2}\right)+\frac{1}{8} \epsilon (4 \log (2)-\log (\epsilon )-2)+O\left(\epsilon ^2\right)$$ which clearly shows the limit and how it is approached.

Let $\epsilon=10^{-k}$ and compare the exact value with the above approximation $$\left( \begin{array}{ccc} k & \text{exact} & \text{approximation} \\ 0 & 1.570796327 & 1.482867951 \\ 1 & 2.578092113 & 2.576026580 \\ 2 & 3.695637363 & 3.695601653 \\ 3 & 4.841132561 & 4.841132044 \\ 4 & 5.991589341 & 5.991589334 \\ 5 & 7.142772451 & 7.142772450 \\ 6 & 8.294051464 & 8.294051464 \end{array} \right)$$

  • $\begingroup$ Is it $K\left(\sqrt{\frac{\epsilon -1}{\epsilon }}\right)$? $\endgroup$ – Robert Z Sep 23 '17 at 9:04
  • $\begingroup$ @RobertZ. This is the famous problem of notations in elliptic integrals ! $\endgroup$ – Claude Leibovici Sep 23 '17 at 9:05
  • $\begingroup$ Thanks a lot. I learnt something more than the question itself. (That's why I like this forum so much :) $\endgroup$ – pqros Sep 23 '17 at 9:24
  • $\begingroup$ @pqros. Same for me ! I learn every single day here. By the way, you are welcome. $\endgroup$ – Claude Leibovici Sep 23 '17 at 9:31

Let us consider the complete elliptic integral of the first kind $$K(k) = \int_0^{\pi/2} \frac{d x}{\sqrt{1 - k^2 \sin^2 x}}.$$ Then $$I_{\epsilon}:=\int_0^{\pi/2}\dfrac {dx} {\sqrt{\sin^2(x)+\epsilon \cos^2(x)}} =\int_0^{\pi/2}\dfrac {dx} {\sqrt{1-(1-\epsilon)\cos^2(x)}}= K\left(\sqrt{1-\epsilon}\right)$$ Now by Asymptotic expansion of the complete elliptic integral of the first kind, we have that the asymptotic expansion of $K(k)$ at $1^-$ is $$K(k) = -\frac12 \ln(1-k^2) + O(1)\implies \lim_{\epsilon\to 0^{+}}\frac{I_{\epsilon}}{\ln(\epsilon)}=-\frac{1}{2}.$$

  • $\begingroup$ I am typing almost the same ! I shall continue with numerical aspects. Cheers. $\endgroup$ – Claude Leibovici Sep 23 '17 at 8:46
  • $\begingroup$ This happens quite often. You are too fast for the old man. What do you think about the approximations I put in my answer ? By the way, where are you in Europe ? $\endgroup$ – Claude Leibovici Sep 23 '17 at 9:01
  • $\begingroup$ Your approximations are fine (+1)! P.S. This user prefers to keep an air of mystery about him. ;-) $\endgroup$ – Robert Z Sep 23 '17 at 9:09
  • $\begingroup$ Is Europe $O(x)$ ? $\endgroup$ – Claude Leibovici Sep 23 '17 at 9:11
  • $\begingroup$ No, it is $O(1)$... Apparently we are more than 1000km away. $\endgroup$ – Robert Z Sep 23 '17 at 9:18

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