I am asking myself if there is any way of computing Gaussian integrals of the form $$\sum_{n=0}^\infty \int_{2n}^{2n+1} e^{-t^2} \, dt,$$ i.e. integrating over all of $\mathbb R^+$ but leaving out "half" the intervals. Maybe there is no precise value but a way of expressing it in terms of other functions?
I previously asked a question which was stated not in the way I actually intended it to be and thus led to a more trivial answer, see Gaussian integral over a union of intervals
I'm not sure you can obtain a closed form for that sum. But you can get an expression in terms of the Dawson function.
Let $$f(x)=\int_{|x|}^{+\infty}e^{-t^2}dt$$ We can rewrite the sum in question as $$\begin{split} \sum_{n=0}^\infty \int_{2n}^{2n+1} e^{-t^2} \, dt &= \sum_{n=0}^\infty\left[ f(2n)-f(2n+1)\right] \\ &=\frac 1 2\left(f(0)+\sum_{n\in\mathbb Z} f(2n)-\sum_{n\in\mathbb Z} f(2n+1)\right) \end{split}$$ By the Poisson summation formula, $$\left\{ \begin{split} \sum_{n\in\mathbb Z} f(2n)&=\frac 1 2 \sum_{n\in\mathbb Z}\hat f\left(\frac n 2\right)\\ \sum_{n\in\mathbb Z} f(2n+1)&=\frac 1 2 \sum_{n\in\mathbb Z}e^{i\pi n}\hat f\left(\frac n 2\right) \end{split} \right.$$ We obtain
$$\sum_{n=0}^\infty \int_{2n}^{2n+1} e^{-t^2} \, dt=\frac 1 4\left(2f(0)+\sum_{n\in\mathbb Z}\hat f\left(\frac n 2\right) -\sum_{n\in\mathbb Z}(-1)^n\hat f\left(\frac n 2\right) \right)=\frac 1 2\left(f(0)+\sum_{n\in\mathbb Z}\hat f\left(\frac {2n+1} 2\right) \right)$$
Now, for the Fourier transform of $f$, $$\begin{split} \hat f(\xi) &= \int_{\mathbb R}e^{-2i\pi\xi x}f(x)dx\\ &=2\int_0^{+\infty}\cos(2\pi \xi x)f(x)dx\\ &=2\int_0^{+\infty}\cos(2\pi \xi x)\int_x^{+\infty}e^{-t^2}dtdx\\ &=2\int_0^{+\infty}e^{-t^2}\int_0^{x}\cos(2\pi \xi x)dx dt\\ &= 2\int_0^{+\infty}e^{-t^2}\frac{\sin(2\pi\xi t)}{2\pi \xi}dt\\ &=\frac{D(\pi \xi)}{\pi \xi} \end{split}$$ where $D$ is the Dawson function. To recap, $$\boxed{\sum_{n=0}^\infty \int_{2n}^{2n+1} e^{-t^2} \, dt =\frac 1 2\left(\frac {\sqrt{\pi}}2 +\sum_{n\in\mathbb Z}\frac{D\left(\frac {2n+1}2 \pi\right)}{\frac{2n+1}{2}\pi}\right)}$$