Asymptotic expansion of $\int_0^{\pi/2} dx \sqrt{1-(1-\epsilon^2)\sin^2x}$ as $\epsilon\to 0$

Question

I wish to find the asymptotic expansion of (at least the first few terms) \begin{equation} \int_0^{\pi/2} dx \sqrt{1-(1-\epsilon^2)\sin^2x} \end{equation} as $\epsilon\to 0$. This integral is closed related to the complete elliptic integral of the second kind, but I prefer not using any knowledge about that. I have no clue where to start.

The answer given by Mathematica is \begin{equation} 1-\frac{1}{2}\epsilon^2\ln\epsilon+O(\epsilon^2) \end{equation}

(PS: This is some integral I encountered while working on a problem of solid-state physics.)

Answer 1

As said in comments, using Mathematica notation,

$$I=\int_0^{\frac \pi 2} \sqrt{1-(1-\epsilon^2)\sin^2(x)}\,dx= E\left(1-\epsilon ^2\right)$$

Just type

Series[EllipticE[1-\[Epsilon]^2],{\[Epsilon],0,7}, Assumptions -> \[Epsilon] > 0]

and simplify (changing the $7$ to whatever will give more and more terms).

Answer 2

Too long for a comment

As was mentioned in the comment and the solution above, the systematic approach requires usage of the asymptotic of special functions. But it is always possible to find several first asymptotic terms directly. $$I(\epsilon)=\int_0^{\pi/2} \sqrt{1-(1-\epsilon^2)\sin^2x}\,\,dx\overset{\tan x=t}{=}\int_0^\infty\frac{\sqrt{1+\epsilon^2t^2}}{(1+t^2)^\frac32}dt$$ $$\overset{x=\frac1t}{=}\int_0^\infty\frac{\sqrt{x^2+\epsilon^2}}{(1+x^2)^\frac32}dx=\int_0^\infty\frac{\sqrt{x^2+\epsilon^2}-x+x}{(1+x^2)^\frac32}dx$$ $$=\int_0^\infty\frac x{(1+x^2)^\frac32}dx+\epsilon^2\frac{dx}{(1+x^2)^\frac32(x+\sqrt{x^2+\epsilon^2})}$$ $$\overset{x\to\epsilon x}{=}1+\epsilon^2\int_0^\infty\frac{dx}{(1+x^2\epsilon^2)^\frac32(x+\sqrt{1+x^2})}$$ $$=1+\epsilon^2\int_0^\infty\frac{dx}{(1+x^2\epsilon^2)^\frac32}\left(\frac1{x+\sqrt{1+x^2}}-\frac1{2\sqrt{1+x^2}}+\frac1{2\sqrt{1+x^2}}\right)$$ $$\overset{x=\frac t\epsilon}{=}1+\frac{\epsilon^2}2\int_0^\infty\frac{dx}{(1+x^2)^\frac32\sqrt{x^2+\epsilon^2}}+\frac{\epsilon^2}2\int_0^\infty\frac{dx}{(1+x^2)^\frac32}\frac{\sqrt{x^2+\epsilon^2}-x}{\sqrt{x^2+\epsilon^2}(x+\sqrt{x^2+\epsilon^2})}$$ Making the substitution $x=\epsilon t$ in the last integral $$=1+\frac{\epsilon^2}2\int_0^\infty\frac{dx}{(1+x^2)^\frac32\sqrt{x^2+\epsilon^2}}+\frac{\epsilon^2}2\int_0^\infty\frac{dt}{(1+t^2\epsilon^2)^\frac32\sqrt{1+t^2}(t+\sqrt{1+t^2}))^2}$$ We see that the last integral converges at $\epsilon=0$, so we drop $\epsilon$ inside the integral (i. e. dropping the terms $\sim\epsilon^4$). The first integral we integrate by part. $$I(\epsilon)=1+\frac{\epsilon^2}2\int_0^\infty\frac{dx}{\sqrt{1+x^2}(x+\sqrt{1+x^2})^2}+\frac{\epsilon^2}2\frac{\ln(x+\sqrt{x^2+\epsilon^2})}{(1+x^2)^\frac32}\,\bigg|_{x=0}^\infty+$$ $$+\frac{3\epsilon^2}2\int_0^\infty\frac{x\,\ln(x+\sqrt{x^2+\epsilon^2})}{(1+x^2)^\frac52}dx+O(\epsilon^4)$$ The last integral converges at $\epsilon=0$. Dropping $\epsilon$ inside the integral (dropping the terms $\sim\epsilon^4\ln\epsilon$) $$I(\epsilon)=1-\frac{\epsilon^2}2\ln\epsilon+\frac{\epsilon^2}2\int_0^\infty\frac{dx}{\sqrt{1+x^2}(x+\sqrt{1+x^2})^2}+\frac{3\epsilon^2}2\int_0^\infty\frac {x\,\ln(2x)}{(1+x^2)^\frac52}dx+O(\epsilon^4\ln\epsilon)$$ Using the substitution $x=\sinh t$ in the first integral $$\frac{\epsilon^2}2\int_0^\infty\frac{dx}{\sqrt{1+x^2}(x+\sqrt{1+x^2})^2}=\frac{\epsilon^2}2\int_0^\infty\frac{dt}{e^{2t}}=\frac{\epsilon^2}4$$ The last integral $$\frac{3\epsilon^2}2\int_0^\infty\frac {x\,\ln(2x)}{(1+x^2)^\frac52}dx=\frac{3\ln2\epsilon^2}2\int_0^\infty\frac{t\,dt}{(1+t^2)^\frac52}+\frac{3\epsilon^2}2\int_0^\infty\frac{t\,\ln t}{(1+t^2)^\frac52}dt$$ $$=\frac{\ln 2}2\epsilon^2+\frac{\ln2-1}2\epsilon^2=\left(\ln2-\frac12\right)\,\epsilon^2$$ Taking all together, $$I(\epsilon)=1-\frac{\epsilon^2\,\ln\epsilon}2+\left(\ln2-\frac14\right)\epsilon^2+O(\epsilon^4\ln\epsilon)$$ Numeric evaluation (WA) confirms the answer.