I'm trying to solve the following integral: $$ \int_{x} \int_{y} \int_{z} \int_{w} \exp \left(-\frac{1}{2}\left\|\left(\begin{array}{l} y-x \\ z-y \\ y-z \end{array}\right)\right\|^{2}\right) d x d y d z d w $$
I tried writing it as $$ \int_{x} \int_{y} \int_{z} \int_{w} \exp \left(-\frac{1}{2}(y-x)^{2}-\frac{1}{2}(z-y)^{2}-\frac{1}{2}(w-z)^{2}\right) d x d y d z d w $$
and splitting this into $$ \int_{x} \int_{y} \exp \left(-\frac{1}{2}(y-x)^{2}\right) \int \exp \left(-\frac{1}{2}(z-y)^{2}\right) \int_{w} \exp \left(-\frac{1}{2}(w-z)^{2}\right) d w d z d y d x $$
And then moving out all of these as gaussians, but it seems that at the end I'm left with
$$ 2 \pi \int_{x} d x $$
So does this integral not converge? Or did I mess up somewhere?
As Matthew H. suggested and following this, let $u:=(w-z)/\sqrt{2}$, $t:=(z-y)/\sqrt{2}$ and $s:=(y-x)/\sqrt{2}$. Therefore, $$ \begin{aligned} \int_x \int_y \int_z \int_w &\exp \left[-\frac{1}{2}(y-x)^{2}\right] \exp \left[-\frac{1}{2}(z-y)^{2}\right] \exp \left[-\frac{1}{2}(w-z)^{2}\right] \mathrm{d}w \mathrm{d}z \mathrm{d}y \mathrm{d}x \\ &= 2^{3/2}\int_x\int_s\int_t\int_u \exp\left(-s^2\right) \exp\left(-t^2\right) \exp\left(-u^2\right) \mathrm{d}u \mathrm{d}t \mathrm{d}s \mathrm{d}x \\ &= x\left[\sqrt{\frac{\pi}{2}}\operatorname{erf}(s) + c_1\right]\left[\sqrt{\frac{\pi}{2}}\operatorname{erf}(t) + c_2\right]\left[\sqrt{\frac{\pi}{2}}\operatorname{erf}(u) + c_3\right] + c_4 \\ &= x\left[\sqrt{\frac{\pi}{2}}\operatorname{erf}\left(\frac{y-x}{\sqrt{2}}\right) + c_1\right]\left[\sqrt{\frac{\pi}{2}}\operatorname{erf}\left(\frac{z-y}{\sqrt{2}}\right) + c_2\right]\left[\sqrt{\frac{\pi}{2}}\operatorname{erf}\left(\frac{w-z}{\sqrt{2}}\right) + c_3\right] + c_4, \end{aligned} $$ where $\operatorname{erf}(x)$ is the error function. If you integrate over the reals, you will end up with $$ \begin{aligned} \int_{\mathbb{R}^4} &\exp \left[-\frac{1}{2}(y-x)^{2}\right] \exp \left[-\frac{1}{2}(z-y)^{2}\right] \exp \left[-\frac{1}{2}(w-z)^{2}\right] \mathrm{d}w \mathrm{d}z \mathrm{d}y \mathrm{d}x \\ &= 2^{3/2}\int_{\mathbb{R}^4} \exp\left(-s^2\right) \exp\left(-t^2\right) \exp\left(-u^2\right) \mathrm{d}u \mathrm{d}t \mathrm{d}s \mathrm{d}x \\ &= (2\pi)^{3/2}\int_\mathbb{R}\mathrm{d}x, \end{aligned} $$ whose integral clearly diverges.