Find an equivalent of an improper integral

Question

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?

Answer 1

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

Answer 2

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