Given four random lines in $\mathbb{R}P^3$, how many lines intersect all of those lines? In the recent paper Probabilistic Schubert Calculus, Peter Bürgisser and Antonio Lerario discuss this question and much more general versions of it.

In **Proposition 6.7**, they determine the expected number of lines meeting four random lines in $\mathbb{R}P^3$ to be:

$$\operatorname{edeg}G(2,4) =\\ 2^{-13}\int_{[0,2\pi]^6}\left| \det\begin{pmatrix} \sin{t_1}\sin{s_1}&\sin{t_2}\sin{s_2}&\sin{t_3}\sin{s_3}\\ \cos{t_1}\sin{s_1}&\cos{t_2}\sin{s_2}&\cos{t_3}\sin{s_3}\\ \sin{t_1}\cos{s_1}&\sin{t_2}\cos{s_2}&\sin{t_3}\cos{s_3} \end{pmatrix}\right|dt_1dt_2dt_3ds_1ds_2ds_3$$

The integrand can be expanded to $$\left|\cos(s_2)\sin(s_1)\sin(s_3)\sin(t_2)\sin(t_1 - t_3)- \sin(s_2)\big(\cos(s_1)\sin(s_3)\sin(t_1)\sin(t_2 - t_3) + \cos(s_3) \sin(s_1)\sin(t_3)\sin(t_1 - t_2)\big)\right|$$

The absolute value in the integrand and the integration over six variables seem to make it difficult to evaluate this integral to high precision. I am interested in the exact value of this number. Bürgisser and Lerario computed it to be $1.72$ rounded to two digits; I ran $10^{11}$ Monte Carlo evaluations to obtain $1.726225\dots$ with an estimated error of $7.3\cdot 10^{-6}$

My questions are:

- Can the exact value of $\operatorname{edeg}G(2,4)$ be determined; is there perhaps a relation with other constants; is it, for example, algebraic over $\mathbb{Q}[\pi]$?
- If this is too difficult, what are ways of evaluating the above integral to higher accuracy numerically?

**Update**: In AdamP.Goucher's great answer, he provides more digits ($1.726230876$) and also uses a reformulation described by Matt F. in a comment to get an integrand without any absolute values:

$$G(2,4) = 2^{-6} \int \dfrac{h\left((y-u)^2+(v-x)^2\right)^2}{\left((1+(u + (y-u)c - (v-x)h)^2)(1+(x + (v-x)c + (y-u)h)^2)\prod_{\alpha\in\{u,y,x,v\}}(1+\alpha^2)\right)^{3/2}}$$ where we integrate over $\mathbb{R}^5\times \mathbb{R}^+$ (and $h$ is the positive variable).

Perhaps this form could come in handy for evaluating this integral to higher accuracy?!

**Update 2** The new formula from L.Mathis and Antonio Lerario is very useful for calculating digits! The following mpmath code can returns in less than $3$ minutes

$1.7262312489219034885256331685361697650475579915479447$

(the last few digits might not be accurate) I expect to make that even much faster when solving the two integrals $F$ and $G$ symbolically first.

```
import functools
from mpmath import mp
@functools.lru_cache(maxsize=1000)
def F(u):
return mp.quad(lambda phi: (u*mp.sin(phi)**2)/(mp.sqrt(mp.cos(phi)**2 + u**2*mp.sin(phi)**2)), [0, mp.pi/2])
@functools.lru_cache(maxsize=1000)
def G(u):
return mp.quad(lambda phi: (mp.sin(phi)**2)/(mp.sqrt(mp.sin(phi)**2 + u**2*mp.cos(phi)**2)), [0, mp.pi/2])
def H(u):
return F(u)/G(u)
def L(u):
return F(u)*G(u)
def integrand3(u):
return L(u)**2*(1/H(u) - H(u))*mp.diff(H, u)/H(u)
dps = 50
mp.dps = dps
%time z = 3*mp.quad(integrand, [0,1]); z
```

3more comments