I got the following cubic elliptic curve from some physical problem $$E_c(\mathbb{C}): w^2=4 z^3-zG_2-G_3,$$ where $G_2=3 \alpha ^2+\gamma$ and $G_3=\alpha ^3-\alpha \gamma -\beta ^2$ for known real parameters $\alpha,\beta,\gamma$. In fact, $\alpha,\beta,\gamma$ are such that the roots of $E_c(\mathbb{C})$ are real.
Under the birational change of variables $$ x\to \frac{w-\beta }{2 (z-\alpha )} \quad \text{and} \quad y\to \frac{(w-\beta )^2}{4 (z-\alpha )^2}-2 z-\alpha $$ with inverse $$ z\to \frac{1}{2}(x^2-y-\alpha) \quad \text{and} \quad w\to x^3-xy-3x\alpha+\beta ,$$ we get a quartic curve $$E_q(\mathbb{C}): y^2=x^4-6 \alpha x^2+4 \beta x+\gamma.$$
I am interested in writing the integral $$\xi(z):=\int_\infty^z \frac{dz}{w},$$
defined on $E_c(\mathbb{C})$ in terms of the variables on $E_q(\mathbb{C})$. Using the above birational map and working out the Jacobian, I find $$\int_\infty^{z(x)} \frac{dz}{w}=\int_\infty^{x(z)=\frac{w(z)-\beta }{2
(z-\alpha )}}\frac{dx}{y}.$$
I tested this relation using NIntegrate[] in Mathematica and it seems the two integrals agree to arbitrary precision far from $z=\alpha$. However, for $z\sim \alpha$, the LHS integral gives a finite number (up to periods) and the RHS integral gives zero (up to periods). I could have expected this difference, since $z=\alpha$ gives $x=\infty$.
I am wondering about what I could do to 'regulate' or 'blow up' this singularity locally and make the above two expressions equal globally. The point $z=\alpha$ is of physical relevance in my problem so understanding its image onto $E_q(\mathbb{C})$ is important. Any suggestion?
Also, is my problem related to the so-called 'points at infinity'? I am asking because one of the comment by Michael Stoll says "The "points at infinity" then are the points one has to add to the affine curve to obtain the smooth projective model." This seems to resonate with my problem here.
