Simplifying a double integral to a single integral

Question

In a recent cross-validated post a comment was left by Dilip Sarwate that stated that the following double integral:

$$\int_0^L \int_0^L \rho(t-u)dudt$$

Could be simplified to the following single integral:

$$\int_{-L}^{L} \rho(s)(L-|s|)\,\mathrm ds$$

Where $\rho(t-u)$ is the correlation function of a stochastic process. Mathematica gives both integrals the same value for a correlation function of the form: $$\rho(t-u) = \exp\left(-\pi\left(\frac{t-u}{\theta}\right)^2\right) = \rho(s) = \exp\left(-\pi\left(\frac{s}{\theta}\right)^2\right)$$

Can you explain why this simplification is possible and how the double integral simplifies to the single integral.

Answer 1

Note that \begin{align*} \int_0^L\int_0^L\rho(t-u)dudt &= \int_0^L\int_{t-L}^t \rho(s) ds dt\\ &=\int_0^L\int_0^t \rho(s) ds dt + \int_0^L\int_{t-L}^0 \rho(s) ds dt\\ &=\int_0^L\int_s^L \rho(s) dt ds + \int_{-L}^0\int_0^{L+s} \rho(s) dt ds\\ &=\int_0^L\rho(s)(L-s) ds + \int_{-L}^0\rho(s) (L+s) ds\\ &=\int_0^L\rho(s)(L-|s|) ds + \int_{-L}^0\rho(s) (L-|s|) ds\\ &=\int_{-L}^L\rho(s)(L-|s|) ds. \end{align*}