Asymptotic expansion of an improper integral

Question

I'm having trouble proving this asymptotic expansion (with $0 \leq \delta<1/2$)

$$I(\delta) = \int_\delta ^{1-\delta} dx \frac{x (1-x)}{\sqrt{x^2-\delta^2}\sqrt{(1-x)^2-\delta^2}} \sim 1 - \delta^2 \ (\delta \rightarrow 0).$$

I tried naïve approaches like writing the Taylor expansion of the integrand before integrating it, but I didn't get a good result.

I also tried to calculate the derivatives $I^{(n)}(0)$, but I couldn't get anywhere.

Even for a simple integral like

$$J(\delta) = \int_\delta ^{1} dx \frac{x }{\sqrt{x^2-\delta^2}} = \sqrt{1-\delta^2} \sim 1 - \frac{\delta^2 }{2}\ (\delta \rightarrow 0) $$ I don't know how to find the asymptotic form without solving the integral directly first. Any advice would be greatly appreciated.

Answer 1

We have \begin{align*} I(\delta) &= \int_{\delta}^{1/2} + \int_{1/2}^{1-\delta} \\[6pt] &= 2\int_{\delta}^{1/2} \frac{x}{\sqrt{x^2-\delta^2}}\cdot \frac{1-x}{\sqrt{(1-x)^2-\delta^2}}\,\mathrm{d} x \\[6pt] &= 2\int_{\delta}^{1/2} \frac{x}{\sqrt{x^2-\delta^2}}\cdot \left(1 + \frac{1}{2(1-x)^2}\delta^2\right)\,\mathrm{d} x + o(\delta^2)\tag{1}\\[6pt] &= \frac{\delta^4\ln(1-2\delta^2 + \sqrt{1-\delta^2}\sqrt{1-4\delta^2})}{(1-\delta^2)^{3/2}} - \frac{\delta^4\ln \delta}{(1-\delta^2)^{3/2}} + \frac{\sqrt{1-4\delta^2}}{1-\delta^2} + o(\delta^2)\tag{2}\\[6pt] &= 1 - \delta^2 + o(\delta^2) \end{align*} where we use $\frac{1-x}{\sqrt{(1-x)^2-\delta^2}} = 1 + \frac{1}{2(1-x)^2} \delta^2 + o(\delta^2)$ and $2\int_{\delta}^{1/2} \frac{x}{\sqrt{x^2-\delta^2}}\,\mathrm{d} x = \sqrt{1 - 4t^2} \le 1$ in (1).
Note: We use Maple to obtain the closed form of the integral in (1) (the antiderivative admits a closed form as well). I think there is a way to obtain it by hand or there is another way to obtain $1 - \delta^2 + o(\delta^2)$ based on (1) not via (2) e.g. by IBP.

Answer 2

Here I will solve the integral in terms of complete elliptic integrals. Then the claim follows from the known asymptotic behaviour of these special functions.

Note that the integrand and the bounds are symmetric about $x = 1/2$. This becomes even clearer if we apply the shift $x = y + 1/2$. Writing $\delta = \varepsilon/2$ temporarily to simplify notation a bit, we find \begin{align} I (\varepsilon/2) &= \int \limits_{-(1-\varepsilon)/2}^{(1-\varepsilon)/2} \mathrm{d} y \, \frac{1 - 4 y^2}{\sqrt{[(1+\varepsilon)^2 - 4 y^2][(1-\varepsilon)^2 - 4 y^2]}} \\ &= 2 \int \limits_0^{(1-\varepsilon)/2} \mathrm{d} y \, \frac{1 - 4 y^2}{\sqrt{[(1+\varepsilon)^2 - 4 y^2][(1-\varepsilon)^2 - 4 y^2]}}\, . \end{align} Rescaling with $y = (1-\varepsilon)z/2$ and rearranging, we end up with \begin{align} I (\varepsilon/2) &= \int \limits_0^1 \mathrm{d} z \, \frac{1 - (1-\varepsilon)^2 z^2}{\sqrt{[(1+\varepsilon)^2 - (1-\varepsilon)^2 z^2][1-z^2]}} \\ &= \int \limits_0^1 \mathrm{d} z \, \left(\frac{(1+\varepsilon) \sqrt{1 - \left(\frac{1-\varepsilon}{1+\varepsilon}\right)^2 z^2}}{\sqrt{1-z^2}} - \frac{\varepsilon (2 + \varepsilon)}{(1+\varepsilon)\sqrt{1 - \left(\frac{1-\varepsilon}{1+\varepsilon}\right)^2 z^2}\sqrt{1-z^2}} \right)\, . \end{align} But these are just two complete elliptic integrals, which we can simplify using Landen transformations: \begin{align} I (\varepsilon/2) &= (1+\varepsilon) \operatorname{E} \left(\frac{1-\varepsilon}{1+\varepsilon}\right) - \frac{\varepsilon (2 + \varepsilon)}{1 + \varepsilon} \operatorname{K} \left(\frac{1-\varepsilon}{1+\varepsilon}\right) \\ &= \operatorname{E} \left(\sqrt{1-\varepsilon^2}\right) - \frac{\varepsilon^2}{2} \operatorname{K} \left(\sqrt{1-\varepsilon^2}\right) \, . \end{align}

Finally, their asymptotic expansions yield \begin{align} I(\delta) &= \operatorname{E} \left(\sqrt{1-4\delta^2}\right) - 2 \delta^2 \operatorname{K} \left(\sqrt{1-4\delta^2}\right) \\ &\sim 1 - \delta^2 - \left(\log(\delta/2) + \frac{5}{4}\right) \delta^4 + \operatorname{\mathcal{O}} \left(\log(\delta) \delta^6\right) \end{align} as $\delta \to 0^+$.