Let $E_{X_0(11)}$ be the elliptic curve (over $\bf Q$) of conductor $11$ defined by $$y^2+y=x^3-x^2-10x-20.$$
First, some theorems and formulas are introduced as follows.
The modularity theorem (Slow convergence):
\begin{align*} \displaystyle{}L(E_{X_{0}(11)},1)&=\sum_{n=1}^{+\infty}\dfrac{a(E_{X_{0}(11)})}{n}=\sum_{n=1}^{+\infty}\dfrac{a_n(f_{X_{0}(11)})}{n}\\ f(q)=f_{X_{0}(11)}(q)&=q\prod _{k=1}^{+\infty} \left(1-q^k\right)^2 \left(1-q^{11 k}\right)^2\\ f(q)=f_{X_{0}(11)}(q)&=\sum_{n=1}^{+\infty}a_n(f_{X_{0}(11)})q^n \end{align*}
The Manin-Drinfeld theorem+BSD formula: \begin{align*} L(E_{X_0(11)},1)&=\dfrac{\Omega(E_{X_0(11)})}{5}\\ &=\dfrac{1}{5}\cdot\dfrac{3+\operatorname{sgn}(\Delta(E_{X_0(11)}))}{2}\int_{E_{X_0(11)}(\mathbb{R})}\left|\omega\right|\\ &=\dfrac{1}{5}\int_{E_{X_0(11)}(\mathbb{R})}\dfrac{\mathrm{d}x}{2y+a_1x+a_3}\\ &=\dfrac{1}{5}\int_{E_{X_0(11)}(\mathbb{R})}\dfrac{\mathrm{d}x}{2y+1}\\ &=\dfrac{1}{5}\int_{\alpha}^{+\infty}\frac{\mathrm{d}x}{\sqrt{4x^3 - 4x^2 - 40x - 79}} \\ &\\ \alpha&=\tfrac{\small{2+(8-\sqrt{33})(19+3 \sqrt{33})^{1/3}+(8+\sqrt{33})(19-3 \sqrt{33})^{1/3}}}{6} \end{align*}
The Cohen formula of L-series (Fast convergence)[Cohen1993, Prop. 7.5.8, p. 405]:
\begin{align*} L(E_{X_0(11)},1) &=(1+\varepsilon(E_{X_0(11)}))\sum_{n=1}^{+\infty}\dfrac{a(E_{X_{0}(11)})}{n}e^{-\frac{2\pi{n}}{\sqrt{N}}}\\ &=2\sum_{n=1}^{+\infty}\dfrac{a(E_{X_{0}(11)})}{n}e^{-\frac{2\pi{n}}{\sqrt{11}}}\\ \end{align*}
The ODE satisfied by modular parameterization: \begin{align*} \frac{f(q)\mathrm{d}q}{q}&=\frac{\mathrm{d}x}{2y + a_1x + a_3} \\ \frac{f(q)\mathrm{d}q}{q}&=\frac{\mathrm{d}x}{2y + 1} \\ \frac{f(q)\mathrm{d}q}{q}&=\frac{\mathrm{d}x}{\sqrt{4x^3 - 4x^2 - 40x - 79}} \\ \end{align*}
\begin{align*} F=F(q)&=\frac{11 \left(1-24 \sum _{k=1}^{+\infty} \frac{k q^{11 k}}{1-q^{11 k}}\right)-\left(1-24 \sum _{k=1}^{+\infty} \frac{k q^k}{1-q^k}\right)}{10 q \prod _{k=1}^{+\infty} \left(1-q^k\right)^2 \left(1-q^{11 k}\right)^2}+\frac{8}{5}\\ G=G(q)&=\frac{\left(1+240 \sum _{k=1}^{+\infty} \frac{k^3 q^k}{1-q^k}\right)-121 \left(1+240 \sum _{k=1}^{+\infty} \frac{k^3 q^{11 k}}{1-q^{11 k}}\right)}{120 q^2 \prod _{k=1}^{+\infty} \left(1-q^k\right)^4 \left(1-q^{11 k}\right)^4}\\ \\ x=x(q)&=\dfrac{F^2 - 10 F - G + 10}{2}\\ y=y(q)&=\dfrac{10 - 28 F + 15 F^2 - F^3 - 5 G + FG}{2} \end{align*} Combining the codes of Mathematica and SageMath, you can verify the specific expression of modular parameterization
Column[Series[{(F^2 - 10*F - G + 10)/
2, (10 - 28*F + 15*F^2 - F^3 - 5*G + F*G)/2} /.
{F -> (11*(1 -
24*Sum[(k*q^(11*k))/(1 - q^(11*k)), {k, 1, 100}]) - (1 -
24*Sum[(k*q^k)/(1 - q^k), {k, 1, 100}]))/
(10*q*
Product[(1 - q^k)^2*(1 - q^(11*k))^2, {k, 1, 100}]) + 8/5,
G -> ((1 + 240*Sum[(k^3*q^k)/(1 - q^k), {k, 1, 100}]) -
121*(1 + 240*Sum[(k^3*q^(11*k))/(1 - q^(11*k)), {k, 1, 100}]))/
(120*q^2*
Product[(1 - q^k)^4*(1 - q^(11*k))^4, {k, 1, 100}])}, {q, 0,
18}]]
E = EllipticCurve('11a1')
phi = E.modular_parametrization()
x,y = phi.power_series(prec=20)
for i in [x, y]:
print(i)
The following SageMath code can verify this ODE:$\frac{f(q)\mathrm{d}q}{q}=\frac{\mathrm{d}x}{2y + a_1x + a_3}$
E = EllipticCurve('11a1')
phi = E.modular_parametrization()
x,y = phi.power_series(prec=20)
a1,_,a3,_,_ = E.a_invariants()
f = E.q_expansion(21)
q = f.parent().gen()
for i in [f/q, (X.derivative()/(2*y+a1*x+a3))]:
print(i)
For this elliptic curve, the module parameterization satisfies the following ODE \begin{align*} \frac{f(q)\mathrm{d}q}{q}=\frac{\mathrm{d}x}{2y+1}\tag{e.g.1} \end{align*} In addition, according to the long Weierstrass equation of the elliptic curve, we can get the expression of 2y+1: \begin{align*} y^2+y&=x^3-x^2-10x-20\\ &\Downarrow\\ \quad{}2y+1&=\sqrt{4x^3 - 4x^2 - 40x - 79}\tag{e.g.2}\\ \end{align*} According to (e.g.1) and (e.g.2), we can definitely get \begin{align*} \dfrac{f(q)}{q}\mathrm{d}q&=\frac{\mathrm{d}x}{\sqrt{4x^3 - 4x^2 - 40x - 79}} \end{align*} The integral on the left hand side of the equal sign is convergent, while the integral on the right hand side of the equal sign, the x value obtained according to the modular parameterization is singular. Can other modular parameterizations be constructed so that both sides of the following integral relation are regular? \begin{align*} \int_{q_1\to0}^{q_2\to1}\dfrac{f(q)}{q}\mathrm{d}q&=\int_{x_1\in\mathbb{R}}^{x_2\in\mathbb{R}}\dfrac{\mathrm{d}x}{\sqrt{4x^3 - 4x^2 - 40x - 79}}\quad(\text{Expected situation 1})\\ \int_{q_1\to0}^{q_2\to1}\dfrac{f(q)}{q}\mathrm{d}q&=\int_{x_1\in\mathbb{C}}^{x_2\in\mathbb{C}}\dfrac{\mathrm{d}x}{\sqrt{4x^3 - 4x^2 - 40x - 79}}\quad(\text{Expected situation 2})\\ \end{align*} If the integral is changed to a complex integral, is there an analytical continuation method to obtain the corresponding x value, so that the integral on the right side of the equal sign can be calculated, and the x value corresponds to the modular parameter.
Question 1: Can the value of the L function of the elliptic curve $E_{X_0(11)}$ at $s=1$ be equal to an integral, and the coefficient in front of the integral is equal to one? According to the modularity theorem and the differential equation satisfied by modular parameterization, how to determine the integration limit of the above integral?
\begin{align*} \displaystyle{}L(E_{X_{0}(11)},1)&=\sum_{n=1}^{+\infty}\dfrac{a_n(f_{X_{0}(11)})}{n}\overset{?}{=}\int_0^1\dfrac{f(q)}{q}\mathrm{d}q\\ \int_0^1\dfrac{f(q)}{q}\mathrm{d}q&={\color{red}{1}}\int_{?}^{?}\frac{\mathrm{d}x}{\sqrt{4x^3 - 4x^2 - 40x - 79}} \end{align*}
Question 2: Can the order of integration and summation here be exchanged? If possible, is this quadruple integral series equal to the value of the $L$ function of the elliptic curve $E_{X_0(11)}$ at $s=1$?
\begin{align*} L(E_{X_0(11)},1)\overset{?}{=}\dfrac{1}{2}\sum _{(j,k,m,n)\in\mathbb{Z}^4}\dfrac{(-1)^{k+n}}{\scriptsize{(2 j-1)^2+(2 j-1) k+3 k^2+(2 m-1)^2+(2 m-1) n+3 n^2}} \\ \end{align*}
Since the convergence speed of this quadruple series is also extremely slow, it is not certain whether they are equal even if numerical verification is used. Fortunately, we can show that this infinite product can be expressed as a quadruple series.
\begin{align*} f(q)&=\dfrac{1}{4}\sum _{(j,k,m,n)\in\mathbb{Z}^4}(-1)^{k+n} q^{\frac{(2 j-1)^2+(2 j-1) k+3 k^2+(2 m-1)^2+(2 m-1) n+3 n^2}{2}}\\ &=q\prod _{k=1}^{+\infty} \left(1-q^k\right)^2 \left(1-q^{11 k}\right)^2 \end{align*}
Ref:
[Cohen1993] H. Cohen. A course in computational algebraic number theory. Springer-Verlag, Berlin, 1993.
[Weston2001] Tom Weston. Modular curves X0(11) and X1(11).
https://swc-math.github.io/notes/files/01Weston1.pdf
Arizona Winter School 2001 Course Descriptions and Notes