Conjecture of Differential Geometry Line Element identity?

Question

Question

Lets say I have a line element $ds^2 = \sum_{\mu , \nu = 0}^n g^{\mu \nu} d \tilde x_\mu d \tilde x_\nu $ and

$1.$ It is of the form $ds^2 = \sum_{\mu=0}^n g^{\mu \mu} d \tilde x_\mu d \tilde x_\mu= \sum_{\mu=0}^n (d (\int \sqrt{g^{\mu \mu}} d\tilde x_\mu))^2 = \sum_{\mu=0}^n d x_\mu^2$. This form of the line element $\sum_{\mu=0}^n d x_\mu^2$ is achieved via coordinate transformation. If the square root is negative one can use $*$

$2.$ There exists a path whose path-length extends to infinity

We define $s = \int_{\lambda_0}^{\lambda_1} \sqrt{\sum_{\mu=0}^n \frac{ds_\mu}{d \lambda} \frac{d s^\mu}{d\lambda}} d \lambda $ with $s(0)= 0 = \lambda_0$

$3$. There is a unique one to one mapping between $s$ and its coordinates $x_i$ . For example, $s(x_{0c},x_{1c},\dots) = c$ then $s = c \implies (x_0,x_1,x_2,\dots) = (x_{0c},x_{1c},x_{2c},\dots)$

Further the path taken has if $s$ is expressed coordinates $x_i$

$4.$ Then also impose $d x_i \neq 0 $. For example if $ds^2 = dx^2 + dy^2$ then $s_\mu = (x,0)$ is not a valid path because $ds_1= 0 $ (counting $\mu$ from $0$). However, $s_\mu= (x,-y)$ is a valid path.

Then we have the following identity:

$$(\int_{s=0}^\infty e^{-\alpha s^2} ds )^2 = \sum_{i=0}^n (\int_{x_{i0}}^\infty e^{-\alpha s(x_i)^2} dx_i)^2 $$

$*$Note if $g^{j j}$ is negative then we use the same procedure with a negative sign. For example, $s_\mu = (t,x) = (\lambda, \lambda)$

$$ds^2 = - dt^2 + dx^2\implies (\int_{s=0}^\infty e^{-\alpha s^2} ds )^2 = -( \int_{t=0}^\infty e^{-\alpha s^2} dt )^2 + (\int_{x=0}^\infty e^{-\alpha s^2} dx )^2 =0 $$

Note: $ds^2$ is also $0$

Are there any counterexamples? Can I use this to find the mapping between $s$ and $x_i$? Does the formula still work if there isn't a unique mapping between $s$ and its coordinates?

Heuristic Rough Derivation and Example

The following proof is difficult to follow but is meant to give only a rough sketch.

We define $s = \int_{\lambda_1}^{\lambda_2} \sqrt{\sum_{\mu=0}^n \frac{ds_\mu}{d \lambda} \frac{d s^\mu}{d\lambda}} d \lambda $

For example, let's say I have the line element:

$$ ds^2 = dx^2 + dy^2 + dz^2$$

Now we multiply a factor of $\lambda_s e^{-\alpha s(x,y,z)}$ both sides. Note: $\lambda_s$ should not be thought as a function

$$ \lambda_s e^{-\alpha s(x,y,z)}ds^2 = \lambda_s e^{-\alpha s(x,y,z)} dx^2 +\lambda_s e^{-\alpha s(x,y,z)} dy^2 + \lambda_s e^{-\alpha s(x,y,z)}dz^2 $$

Now we assume the path takes is continuous and has a mapping such that $s(x,y,z) = c $ implies there exists a unique set of coordinates $(x_c,y_c,z_c)$ and $s(x_c ,y_c ,z_c) = c$. Further, the path chosen cannot be such that $dx = 0$ or $dy= 0$ or $dz= 0$. As an example let the path be $s(x,y,z)$. We then do a transformation based on this and cancel the common coefficient:

$$(\int_{s=0}^\infty e^{-\alpha s^2} ds )^2 = (\int_{x_{0}}^\infty e^{-\alpha s(x,y(x),z(x))^2} dx)^2 + (\int_{y_{0}}^\infty e^{-\alpha s(x(y),y,z(y))^2} dy)^2 + (\int_{z_{0}}^\infty e^{-\alpha s(x(z),y(z),z))^2} dz)^2 $$

$s = \sqrt{3} \lambda$ with $x=y=z=\lambda$ with $x_0 = y_0 =z_0 = \lambda_0 = 0$ when $s=0$. Thus,

$$ (\int_{s=0}^\infty e^{-\alpha s^2} ds )^2 = 3(\int_0^{\infty} e^{ - 3 \alpha x^2} dx )^2 = (\int_0^{\infty} e^{ - 3 \alpha x^2} d( \sqrt{ 3 }x) )^2 $$

Hence the identity holds.