High-order antiderivatives of the Weierstrass P-function

The Weierstrass P-function, $\wp(z;g_2,g_3)$ and its derivative $\wp'(z;g_2,g_3)$ are the building blocks of all elliptic functions of a given period lattice. The differential equation $$\wp''=6\wp^2-\frac{1}{2} g_2$$ implies that all derivatives $\wp^{(n)}$ can be expressed as polynomials in $\wp,\wp'$. Recently, I've tried exploring antiderivatives.

The Mathematica command

Integrate[WeierstrassP[z,{g2,g3}],z]

shows that $$\int \wp(z;g_2,g_3) \mathrm{d} z=-\zeta(z;g_2,g_3)+A$$ where $A$ is an arbitrary constant, and $\zeta$ is the Weierstrass Zeta function. Moreover, the command

Integrate[WeierstrassP[z,{g2,g3}],z,z]

shows that

$$\int - \zeta(z; g_2,g_3) \mathrm{d} z=- \log \sigma(z;g_2,g_3)+B$$ where $B$ is an arbitrary constant, and $\sigma$ is the Weierstrass Sigma function. Trying a third time, the command

Integrate[WeierstrassP[z,{g2,g3}],z,z,z]

returns the unevaluated integral $$-\int \log \sigma(z;g_2,g_3) \mathrm{d} z.$$

My questions are: is it really not possible to express $\wp^{(-n)}$ in terms of other known functions for $n \geq 3$? Are there any other known expressions for $\wp^{(-n)}$ with $n \geq 3$?

Thank you!

There could be...the integral$$\int x^2\wp(x+z)\,dx\tag1$$has two periods as a function of $z$, and using IBP twice, will produce$$-x^2\zeta(x+z)+2x\ln\sigma(x+z)-2\int \ln\sigma(x+z)\,dx$$
But the integrand in $(1)$ has second-order poles; thus its integral can be evaluated as a series of residues, implying it can be rewritten as a sum of Weierstrass zeta functions (in $z$). Finally, set $z=0$.
Higher-order antiderivatives can be handled similarly by increasing the exponent in $(1)$.
• The second anti-derivative of $\wp_(z) = \frac{1}{z^2}+\sum_{n,m}' \frac{1}{(z-n\tau-m)^2}-\frac{1}{(n\tau+m)^2}$ is $A+Cz-\log(z)-\sum_{n,m} (\log(1+\frac{z}{n\tau-m})-\frac{z}{n\tau-m}+\frac{z^2}{2(n\tau-m)^2})$ Nov 23 '17 at 8:16
• What happens when you add $1$ or $\tau$ to $z$? I think the sum can be regularized...I'll look into it soon. Try treating it formally! Nov 24 '17 at 1:54
• Oh, OK. The summand can be rewritten as $\sum_{k\ge3}\tfrac{(-1)^{k-1}z^k}{k(n\tau-m)^k}$, but I don't know if that makes the series any easier to evaluate. Nov 24 '17 at 3:03
• And this is the logarithm of Weierstrass $\sigma$ function, which is how he obtained doubly periodic functions and his Weierstrass product for arbitrary entire functions Nov 24 '17 at 3:12