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 $$ C(s,t) = \mathbb{E}[\langle \phi(s-\cdot), W\rangle\langle \phi(t-\cdot), W\rangle] = \int \phi(s-x)\phi(t-x)dx $$ In the calculation below we want to move the $x$ into a position where we can easily integrate it out. Specifically, we try to collect the terms for the Gaussian density $$\begin{aligned} \phi(s-x)\phi(t-x) &= \exp\Bigl(-\frac{\|s-x\|^2 + \|t-x\|^2}{2\beta}\Bigr)\\ &= \exp\Bigl(-\frac{\|s\|^2 + \|t\|^2 - 2\langle t+s, x\rangle + 2\|x\|^2}{2\beta}\Bigr)\\ &= \exp\Bigl(-\frac{ \tfrac12\|s - t\|^2 +\tfrac12\|s+t\|^2 - 2\langle t+s, x\rangle + 2\|x\|^2 }{2\beta}\Bigr)\\ &= \exp\Bigl( -\frac{\tfrac12\|s - t\|^2 + \bigl\|\frac{s+t}{\sqrt{2}} - \sqrt{2} x\bigr\|^2 }{2\beta} \Bigr)\\ &= \exp\Bigl( -\frac{\tfrac12\|s - t\|^2}{2\beta}\Bigr) \exp\Bigl(-\frac{2\bigl\|\frac{s+t}{2} - x\bigr\|^2 }{2\beta} \Bigr) \end{aligned} $$ This finally implies for $s,t\in \mathbb R^d$ $$ C(s,t) = \exp\bigl(-\tfrac1{4\beta}\|s - t\|^2\bigr) \underbrace{\int\exp\Bigl(-\frac{\bigl\|\frac{s+t}{2} - x\bigr\|^2 }{2(\beta/2)}\Bigr) dx}_{=\sqrt{(\beta\pi)^d}} = (\beta \pi)^{d/2} \exp\bigl(-\tfrac1{4\beta}\|s - t\|^2\bigr) $$
To obtain the solution to the integral note that, as a density of the multivariate normal distribution integrates to $1$ by definition of a density, we have for any positive definite $\Sigma$ and any $\mu$ $$ (2\pi)^{-d/2}(\det(\Sigma))^{-1/2} \int \exp\Bigl( - \tfrac12 (x-\mu)^T \Sigma^{-1} (x-\mu)\Bigr)dx = 1 $$ Select the covariance matrix to be the scaled identity $\Sigma = \frac\beta2 \mathbb I$ and $\mu=\frac{s+t}{2}$ and the integral coincides with the one above, and due to $$ \det(\Sigma) = \det(\tfrac\beta2 \mathbb I) = (\tfrac\beta2)^d $$ the "$2$" cancels.