1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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 $$\left( \begin{array}{ccc} k & \text{exact} & \text{approximation} \\ 0 & 1.570796327 & 1.482867951 \\ 1 & 2.578092113 & 2.576026580 \\ 2 & 3.695637363 & 3.695601653 \\ 3 & 4.841132561 & 4.841132044 \\ 4 & 5.991589341 & 5.991589334 \\ 5 & 7.142772451 & 7.142772450 \\ 6 & 8.294051464 & 8.294051464 \end{array} \right)$$

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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 $$K(k) = -\frac12 \ln(1-k^2) + O(1)\implies \lim_{\epsilon\to 0^{+}}\frac{I_{\epsilon}}{\ln(\epsilon)}=-\frac{1}{2}.$$

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.