Asymptotic expansion of an improper integral

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.

• for your simpler integral, perhaps putting $y=\frac{x-\delta}{1-\delta}$ to get $\delta$ out of the integration bounds can help? If this works then you can consider $y=\frac{x-\delta}{1-2\delta}$ for $I(\delta)$...? Commented Dec 22, 2023 at 17:34
• @CalvinKhor I gave it a shot but it didn't really help.
– Jean
Commented Dec 22, 2023 at 19:47
• This might be unimportant, but it looks like the parameter range should probably be $0\le\delta<\frac12$. Commented Dec 23, 2023 at 21:27
• @DavidH You are right this was a mistake I edited the question, thanks. But indeed the problem stays the same.
– Jean
Commented Dec 24, 2023 at 12:14

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 $$\delta \to 0^+$$.
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.