Determine the autocovariance $$ C(s,t) = \text{Cov}(X(s), X(t)) $$ of white noise $W$ convolved with a squared exponential (Gaussian) kernel $\phi$ $$ X(t) = (\phi* W) (t) = \int \phi(t-x) W(dx) $$ where the squared exponential kernel is given by $$ \phi(x) = \exp\bigl(-\frac{\|x\|^2}{2\beta}\bigr) $$ and white noise is defined as here.
The result was used here.
From the answer to the question Scalar product of a deterministic function with Gaussian white noise, we know that $$ C(s,t) = \mathbb{E}[\langle \phi(s-\cdot), W\rangle\langle \phi(t-\cdot), W\rangle] = \int \phi(s-x)\phi(t-x)dx $$ So we only need to calculate $$\begin{aligned} \phi(s-x)\phi(t-x) &= \exp\Bigl(-\frac{\|s-x\|^2 + \|t-x\|^2}{2\beta}\Bigr)\\ &= \exp\Bigl(-\frac{\|s\|^2 + \|t\|^2 - 2\langle t+s, x\rangle + 2\|x\|^2}{2\beta}\Bigr)\\ &= \exp\Bigl(-\frac{ \tfrac12\|s - t\|^2 +\tfrac12\|s+t\|^2 - 2\langle t+s, x\rangle + 2\|x\|^2 }{2\beta}\Bigr)\\ &= \exp\Bigl( -\frac{\tfrac12\|s - t\|^2 + \bigl\|\frac{s+t}{\sqrt{2}} - \sqrt{2} x\bigr\|^2 }{2\beta} \Bigr)\\ &= \exp\Bigl( -\frac{\tfrac12\|s - t\|^2}{2\beta}\Bigr) \exp\Bigl(-\frac{2\bigl\|\frac{s+t}{2} - x\bigr\|^2 }{2\beta} \Bigr) \end{aligned} $$ This finally implies $$ C(s,t) = \exp\bigl(-\tfrac1{4\beta}\|s - t\|^2\bigr) \underbrace{\int\exp\Bigl(-\frac{\bigl\|\frac{s+t}{2} - x\bigr\|^2 }{2(\beta/2)}\Bigr) dx}_{=\sqrt{(\beta\pi)^d}} = (\beta \pi)^{d/2} \exp\bigl(-\tfrac1{4\beta}\|s - t\|^2\bigr) $$