In the notes "Introduction to Stochastic PDE's" from Martin Hairer, one reads:
2.3.2 A formal calculation
Define the covariance for the solution to the stochastic heat eqaution by $$C(s,t,xy)=\mathbf{E}u(s,x)u(t,y),\tag{2.7}$$ where $u$ is given by $(2.6)$.
$\ \ $ By (statistical) translation invariance, it is clear that $C(s,t,x,y)=C(s,t,0,x-y)$. Using $(2.6)$ and the expression for the covariance of $\xi$, one has $$\begin{align}C(s,t,& 0,x)\\ &=\frac 1{(4\pi)^n}\mathbf{E}\int_0^t\int_0^s\int_{\mathbf{R}^n}\int_{\mathbf{R}^n}\frac1{|s-r'|^{n/2}|t-r|^{n/2}}e^{-\frac{|x-y|^2}{4(t-r)}-\frac{|y'|^2}{4(s-r')}}\xi(r,y)\xi(r',y')\,dy\,dy'\,dr'\,dr \\ & = \frac1{(4\pi)^n}\int_0^{s\wedge t}\int_{\mathbf{R}^n}\frac1{|s-r|^{n/2}|t-r|^{n/2}}e^{-\frac{|x-y|^2}{4(t-r)}-\frac{|y'|^2}{4(s-r')}}\,dy\,dr \\ &=\frac1{(4\pi)^n}\int_0^{s\wedge t}\int_{\mathbf{R}^n}\frac1{|s-r|^{n/2}|t-r|^{n/2}}\\ &\qquad\times \exp\left(-\frac{|x|^2}{4(t-r)}-\frac{\langle x,y\rangle}{2(t-r)}-\frac{|y|^2}{4(s-r)}-\frac{|y|^2}{4(t-r)}\right)\,dy\,dr.\end{align}$$ The integral over $y$ can be performed explicitly by 'completing the square' and one obtains $$\begin{align}C(s,t,0,x)&=2^{-n}\int_0^{s\wedge t}(s+t-2r)^{-n/2}\exp\left(-\frac{|x|^2}{4(s+t-2r)}\right)\,dr\\ &=2^{-(n+1)}\int_{|s-t|}^{s+t}\ell^{-n/2}\exp\left(-\frac{|x|^2}{4\ell}\right)\,d\ell.\tag{2.8} \end{align}$$
I am trying to compute the covariance $C(s,t,0,x)$ I understand up to $$C(s,t,0,x) = \frac{1}{(4\pi)^n}\int_0^{s\wedge t} \int_{\Bbb{R}^n}\frac{1}{|s-r|^{n/2}|t-r|^{n/2}} e^{-\frac{|x-y|^2}{4(t-r)}}e^{-\frac{|y|^2}{4(s-r)}}dy dr$$ Now we write $|x-y|^2 = |x|^2 - 2\langle x,y\rangle + |y|^2$ then we should obtain
$$ C(s,t,0,x) = \frac{1}{(4\pi)^n}\int_0^{s\wedge t} \int_{\Bbb{R}^n}\frac{1}{|s-r|^{n/2}|t-r|^{n/2}} \\ \times \exp\bigg(-\frac{|x|^2}{4(t-r)} + \frac{ \langle x,y\rangle}{2(t-r)}-\frac{|y|^2}{4(s-r)}-\frac{ |y|^2}{4(t-r)}\bigg)dy dr$$
So you see a sign difference here (Is this a typo?)
Moving on to completing the square we obtain
$$\bigg( \frac{ \langle x,y\rangle}{2(t-r)}-\frac{|y|^2}{4(s-r)}-\frac{ |y|^2}{4(t-r)}\bigg) \\ =\bigg(-\frac{(t+s-2r)}{4(s-r)(t-r)}|y|^2 + \frac{ \langle x,y\rangle}{2(t-r)}\bigg) \\ =\bigg(-\frac{(t+s-2r)}{4(s-r)(t-r)}\bigg[|y|^2 + \frac{2(s-r) \langle x,y\rangle }{(t + s -2r)}\bigg]\bigg) \\ =\bigg(-\frac{(t+s-2r)}{4(s-r)(t-r)}\bigg[\bigg|y - \frac{(s-r) x }{(t + s -2r)}\bigg|^2 - \frac{(s-r)^2 |x|^2 }{(t + s -2r)^2}\bigg]\bigg) \\ =\exp\bigg(\frac{(s-r) |x|^2 }{4(t + s -2r) (t-r)}\bigg) \exp\bigg( -\frac{(t+s-2r)}{4(s-r)(t-r)}\bigg|y - \frac{(s-r) x }{(t + s -2r)}\bigg|^2\bigg) $$
Combining this with the term $\frac{|x|^2}{4(t-r)}$ and ignoring the other terms in the integral we obtain $$ \exp\bigg(\frac{(s-r) |x|^2 }{4(t + s -2r) (t-r)}-\frac{|x|^2}{4(t-r)}\bigg)\\ \exp\bigg(\frac{(s-r) |x|^2 -|x|^2 (t + s -2r)}{4(t + s -2r) (t-r)}\bigg)\\ \exp\bigg(\frac{ -|x|^2 (t -r)}{4(t + s -2r) (t-r)}\bigg)\\ \exp\bigg(\frac{ -|x|^2 }{4(t + s -2r)}\bigg)\\ $$ It remains to compute the integral
$$\int_{\Bbb{R}^n}\exp\bigg(-\frac{(t+s-2r)}{4(s-r)(t-r)} \bigg|y - \frac{(s-r) x }{(t + s -2r)}\bigg|^2\bigg)\, dy \\ =\int_{\Bbb{R}^n}\exp\bigg( -\frac{(t+s-2r)}{4(s-r)(t-r)}|y|^2\bigg)\, dy\\ =\bigg(\frac{4\pi(s-r)(t-r)}{(t+s-2r)}\bigg)^{-n/2}$$
Putting this together we arrive at
$$C(s,t,0,x) = \frac{1}{(4\pi)^n} (4\pi)^{-n/2} \int_0^{s\wedge t} \bigg(\frac{1}{(t+s-2r)}\bigg)^{-n/2}\exp\bigg(\frac{-|x|^2}{4(t + s -2r)}\bigg) dr \\ = \frac{1}{(\pi)^{n/2}|} \frac{1}{2^{n+1}}\int_{|s-t|}^{s+t} \bigg(\frac{1}{\ell}\bigg)^{-n/2}\exp\bigg(\frac{ -|x|^2 }{4\ell}\bigg) d\ell$$
So there is a constant $\pi^{-n/2}$ that appeared in my computation, but not in the notes.
Is this a typo?
Also, one reads
By (statistical) translation invariance, it is clear that $C(s,t,x,y)= C(s,t,0,x-y)$.
Instead, shouldn't it be $C(s,t,x,y)= C(s,t,0,y-x)$?