Question: Evaluate the first two terms as $\epsilon\to 0$ of $$\mathcal{N}(z)=\int_z^1\frac{dx}{\sqrt{x^3+\epsilon}}$$ where $0\le z <1$.
My approach: First, let us calculate $\mathcal{N}(z=0)$. It may be noted that when $x = \mathrm{ord}(\epsilon)$, the integrand is $\mathcal{O}(\epsilon^{-3/2})$, which implies that the integral is $\mathcal{O}(\epsilon^{-1/2})$. Now, when $x = \mathrm{ord}(1)$, the integrand is $\mathcal{O}(1)$, which implies that the integral is $\mathcal{O}(1)$. Therefore, clearly, the contribution to $\mathcal{N}(0)$ from the local region dominates that from the global region. So, we choose $\delta$ such that $\epsilon \ll \delta \ll \epsilon^{1/3}$, and split the integral into two regions: $$I_1=\int_0^{\delta}\frac{dx}{\sqrt{x^3+\epsilon}} \text{ and } I_2=\int_{\delta}^1\frac{dx}{\sqrt{x^3+\epsilon}}.$$
We first find a suitable asymptotic expansion of $I_1$ by substituting $x = \epsilon u$: $$I_1 = \epsilon^{-1/2}\int_0^{\delta/\epsilon}\frac{dx}{(u^3+\epsilon^{-2})^{1/2}} = \epsilon^{-1/2}\int_0^{\delta/\epsilon}\left(\epsilon -\frac{\epsilon^3}{2}u^3+\frac{3\epsilon^5}{8}u^6 - \frac{5\epsilon^7}{16}u^9+\cdots\right)dx.$$ The last equality is obtained by Taylor expanding $f(y) = y^{-1/2}$ about $\epsilon^{-2}$ with $h = u^3$ (difference value), since $0\le u^3\le \left(\frac{\delta}{\epsilon}\right)^3\ll \epsilon^{-2}$.
Similarly, in the global region, $\epsilon \ll x^3$, which implies that $I_2$ can be approximated as follows: $$I_2 = \int_{\delta}^1\left(x^{-3/2} + \frac{3\epsilon^2}{8}x^{-15/2}-\frac{5\epsilon^3}{16}x^{-21/2}+\cdots\right)dx.$$
I am not sure how to proceed after this. Can someone please help me out? Thanks in advance!
Using the Gaussian hypergeometric function $$\int\frac{dx}{\sqrt{x^3+\epsilon}}=\frac{x}{\sqrt{\epsilon }}\,\,\, _2F_1\left(\frac{1}{3},\frac{1}{2};\frac{4}{3};-\frac{x^3}{\epsilon }\right)$$ $$\mathcal{N}(z)=\int_z^1\frac{dx}{\sqrt{x^3+\epsilon}}=$$ $$\frac{1}{\sqrt{\epsilon }}\,\, _2F_1\left(\frac{1}{3},\frac{1}{2};\frac{4}{3};-\frac{1}{\epsilon }\right)- \Big(\frac{z}{\sqrt{\epsilon }}-\frac{z^4}{8 \epsilon ^{3/2}}+\frac{3 z^7}{56 \epsilon ^{5/2}}-\frac{z^{10}}{32 \epsilon ^{7/2}}+O\left(z^{13}\right) \Big)$$
Using the above for $\epsilon=\frac 1{100}$ and $z=\frac 1{10}$ leads to an absolute error of $1.96\times 10^{-6}$ (the value of the integral being $3.05523594895$).
Expanded as a series, the constant term write $$\frac{\Gamma \left(\frac{1}{6}\right) \Gamma \left(\frac{4}{3}\right)}{\sqrt{\pi } \, {\epsilon^{1/6 }}}-2\left(1-\frac{\epsilon }{14}+\frac{3 \epsilon ^2}{104}-\frac{5 \epsilon ^3}{304}+\frac{7 \epsilon ^4}{640}+O\left(\epsilon ^{5}\right) \right)$$