Consider the following definite integral:

$$ F(\alpha) = \int_{0}^{\frac{28}{15} \alpha^2} \left( \frac{1-u}{ 1-\frac{64}{9} u} \right)^{4/3} \frac{\mathrm{d} u}{\left( \alpha^2 - \frac{15}{28} u \right)^{1/2} }\, . $$

The goal is to find analytically the leading order term near the singularity point $\alpha_\mathrm{C}$ at which the integral diverges. Numerically, $\alpha_\mathrm{C} \approx 0.275$.

By evaluating the integral numerically, it can readily be noticed that $ F(\alpha-\alpha_\mathrm{C}) \sim \alpha-\alpha_\mathrm{C} \, . $

I am wondering whether this can be proven rigorously using e.g. asymptotic expansions around the singularity point. Any help is highly appreciated and desirable.




This is how I would get the leading-order term with the help of Mathematica and without worrying too much about rigorous justifications.

I'll assume $\alpha > 0$. First make the substitution $u = \frac{28}{15}\alpha^2 v$ to get

$$ F(\alpha) = \frac{28\alpha}{225 \sqrt[3]{15}}\int_0^1 (1-v)^{-1/2} (15-28\alpha^2 v)^{4/3} \left(1 - \frac{1792}{135}\alpha^2 v\right)^{-4/3}\,dv. $$

There are no issues coming from the factor $(1-v)^{-1/2} (15-28\alpha^2 v)^{4/3}$ since everything there is nice and integrable. The real trouble happens in $\left(1 - \frac{1792}{135}\alpha^2 v\right)^{-4/3}$: for $0 < \alpha < \frac{3}{16} \sqrt{\frac{15}{7}}$ this factor has no issues, but when $\alpha = \alpha_C := \frac{3}{16} \sqrt{\frac{15}{7}}$ we get a singularity of order $-4/3$ (plus $-1/2$ coming from the $(1-v)^{-1/2}$ factor), which is not integrable.

To get a first approximation we can approximate some of the factors which behave nicely when $\alpha \approx \alpha_C$. Since the singularity would be located at $v=1$, we can also approximate $(15-28\alpha^2v)^{4/3} \approx (15-28\alpha_C^2)^{4/3}$. All considered, we get

$$ \begin{align} F(\alpha) &\approx \frac{28\alpha_C}{225 \sqrt[3]{15}} (15-28\alpha_C^2)^{4/3} \int_0^1 (1-v)^{-1/2}\left(1 - \frac{1792}{135}\alpha^2 v\right)^{-4/3}\,dv \\ &= \frac{56\alpha_C}{225 \sqrt[3]{15}} (15-28\alpha_C^2)^{4/3} {}_2F_1\!\left(1,\frac{4}{3};\frac{3}{2};\frac{1792}{135}\alpha^2\right), \end{align} $$

where Mathematica gives the integral in terms of the hypergeometric function ${}_2F_1$. Also according to Mathematica,

$$ {}_2F_1\!\left(1,\frac{4}{3};\frac{3}{2};x\right) \sim \frac{\pi^{3/2}}{\operatorname{\Gamma}(1/6) \operatorname{\Gamma}(4/3)}(1-x)^{-5/6} \qquad \text{as } x \nearrow 1, $$

so get the approximation

$$ F(\alpha) \approx \frac{\pi^{3/2} 5^{5/6} 11^{4/3} \sqrt{21}}{512 \operatorname{\Gamma}(1/6) \operatorname{\Gamma}(4/3)} \left(1-\frac{1792}{135}\alpha^2\right)^{-5/6} \qquad \text{as } \alpha \nearrow \alpha_C, \tag{1} $$

where we've gone ahead and substituted $\alpha_C = \frac{3}{16} \sqrt{\frac{15}{7}}$.

Here is a plot of the numerical integration of $F(\alpha)$ in blue versus the approximation in $(1)$ in orange:

enter image description here

The coefficient in front of the approximation,

$$ \frac{\pi^{3/2} 5^{5/6} 11^{4/3} \sqrt{21}}{512 \operatorname{\Gamma}(1/6) \operatorname{\Gamma}(4/3)} \doteq 0.93789 \ 39338 \ 24880, $$

is pretty messy but I don't really see a way to simplify it.

  • $\begingroup$ Powerful answer, thank you for your effort! $\endgroup$ – Math Student Nov 21 '17 at 7:24

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