Determine the $\lambda \in \mathbb{R}$ for which this integral converges Consider the cuspoidal cubic given by $x^2 - y^3 =0$ in $\mathbb{C}^2$. The log-canonical threshold of the cuspoidal cubic is determined by finding the largest value of $\lambda \in \mathbb{R}$ for which the integral $$\int \frac{1}{| x^2 - y^3|^{2\lambda}}$$ converges in a neighborhood of $0$.
There is an algebraic way of showing that $\lambda = \frac{5}{6}$, but I'm curious as to whether we can deduce this from the convergence of the above integral.
Does anyone know how to show that the above integral converges in a neighborhood of $0$ only if $\lambda =\frac{5}{6}$?
Additional Remark: From what I understand, to compute the lct, the integral needs to converge in a neighborhood of $0$ in $\mathbb{C}^2$. Regardless, I do not know how to integrate the function if $x$ and $y$ are real variables. Both settings may be of interest. Thank you for your interest in this problem

Here is another approach that I have learned from Donaldson:
For a non-negative integer $r$, consider the annular regions $$\Omega_r = \{ (z,w) \in \mathbb{C}^2 : 2^{-3(r+1)} \leq | x | \leq 2^{-3r}, \ 2^{-2(r+1)} \leq | y | \leq 2^{-2r} \}.$$
Let $I_r = \int_{\Omega_r} | x^2 - y^3 |^{-2 \lambda}$. The substitution $z = 2^3 x$ and $w = 2^2 y$ shows that $$I_{r+1} = 2^{12\lambda -10} I_r.$$ Hence, $\sum_r I_r$ is finite if $\lambda < 5/6$.
