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
 A: Make the substitutions $x=t^3(1-w)^2w$ and $y=t^2(1-w)w$, which turns $x^2-y^3$ in to $t^6(1-w)^3w^2(1-2w)$. This substitution has (usual) Jacobian $|t^4(1-w)^2w|$, which we'll need to square since we're integrating in the complex domain. This turns our integral in to $$\int \frac1{|t|^{12\lambda-8}|1-w|^{6\lambda-4}|w|^{4\lambda-2}|1-2w|^{2\lambda}}.$$
To study the convergence of this in the appropriate neighborhood, it suffices to determine when $\int |z|^{p}$ converges in a neighborhood of the origin. Converting to polar, the convergence is determined by whether $\int r^{p+1}dr$ converges, which happens exactly when $p+2>0$. In our case, this means we need $-(12\lambda-8)+2>0$, $-(6\lambda-4)+2>0$, $-(4\lambda-2)+2>0$, and $(2\lambda)+2>0$, or $\lambda<\frac56$, $\lambda < 1$, $\lambda < 1$, and $\lambda < 1$. So this integral converges iff $\lambda <\frac 56$, just like we expect.

But how did we get here? It turns out the easiest way to find the correct substitution actually comes from the algebraic geometry. If we can resolve the singularities of the pair $(\Bbb A^2,C)$ with what's called a strict resolution, then we'll actually exactly recover the change of variables we need here, and this procedure generalizes to any hypersurface.

Definition. Let $X$ be a scheme and $D\subset X$ a divisor. A strict resolution of $(X,D)$ is a proper birational morphism $\pi:Y\to X$ with $Y$ smooth and $\pi^{-1}(D)$ a strict normal crossings divisor - for any point in $\pi^{-1}(D)$, we require it to have a Zariski neighborhood $U\subset Y$ and an etale map $\alpha:U\to\Bbb A^n$ such that $\pi^{-1}(D)\cap U=\alpha^{-1}(D')$, where $D'\subset\Bbb A^n$ is a union of coordinate hyperplanes.

Let's produce one of these for the cuspidal cubic.

*

*Step 1: blow up the origin in $\Bbb A^2$, which on the relevant affine chart amounts to substituting $x=ty$ to get the equation $y^2(t^2-y)$, which gives us a divisor $2E_1+C_1$ which is a double line and a parabola intersecting non-transversely.

*Step 2: We blow up the origin again, which in the relevant affine chart amounts to substituting $y=tu$, which gives the equation $u^2t^3(t-u)$, which gives us a divisor $2E_1+3E_2+C_2$, which is three lines with multiplicities $2,3,1$ respectively intersecting in a point.

*Step 2.5: Before doing our final blowup to separate these lines, we'll make a linear change of variables to move one of the components of an exceptional divisor off an axis by replacing $t$ with $t-u$. This is not strictly necessary, but it makes our presentation nicer. Our equation becomes $u^2(t-u)^3(t-2u)$, and the relevant divisor is still $2E_1+3E_2+C_2$.

*Step 3: Blow up the origin again, which in the relevant affine chart amounts to substituting $u=tv$ to get the equation $t^6(v-1)^3v^2(1-2v)$, which is the divisor $2E_1+3E_2+6E_3+C_3$, and this is a strict normal crossings divisor because it's three nonintersecting lines with a fourth which meets all of them.

If you trace through these substitutions, you'll see that these are exactly the substitutions we used above. In fact, this is a general procedure: given hypersurface in $\Bbb A^n$ cut out by some $f\in \Bbb C[z_1,\cdots,z_n]$, we can calculate the log-canonical threshold of $V(f)$ by using the change of coordinates given by the strict resolution.
One very important property of a strict resolution $\pi:(Y,f^{-1}(D))\to (X,D)$ is that for every point, there's a neighborhood where both $\pi$ and the Jacobian divisor are locally monomial, that is, there are coordinates so that $f\circ \pi=u x_1^{a_1}\cdots x_n^{a_n}$ and $Jac(\pi)=vx_1^{b_1}\cdots x_n^{b_n}$ for invertible $u,v$ and integers $a_i,b_i$. So by using the map $\pi$ as a change of variables, we see that the finitness of $\int \frac1{|f|^{2\lambda}}$ in a neighborhood of zero is equivalent to the finiteness of the integral $\int \frac{|Jac(\pi)|^2}{|f\circ\pi|^{2\lambda}}$ in a neighborhood of every point $p\in\pi^{-1}(0)$. But this latter integral can be calculated in local coordinates as $$\int \frac 1{\prod_i |x_i|^{2\lambda a_i - 2b_i}},$$ which after applying the same trick of converting to polar, converges iff $2\lambda a_i - 2b_i -1 <1$, or $\lambda < \frac{b_i+1}{a_i}$. So computing log-canonical thresholds via integrals is equivalent to computing them via taking a log resolution and comparing the coefficients on the relative canonical divisor with the coefficients on the exceptional divisors in the preimage of $D$.

For a reference on where I learned all of this stuff the first time, Mustata's notes here were helpful, especially with connecting this representation in terms of integrals back to the more algebraic definitions, and Takumi Murayama's notes here from a minicourse taught by Harold Blum are also very relevant. Kollar's papers in this area are also invaluable, and this one is directly relevant and traces back some of the history through Atiyah's original observation of this fact, detailed in Resolutions of Singularities and Division of Distributions.
A: 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 $\lambda < 5/6$.
