Closed form for $\sum_{k=0}^{\infty}\int_0^1 e^{-(x+kL)^2}dx$ Is there a closed form for $\sum_{k=0}^{\infty}\int_0^1 e^{-(x+kL)^2}dx$ (where $L>0$ is a constant) ? Or is there a quick way to calculate this sum? This is a equation we came out to estimate the reliability of a system running a periodic schedule with period L. The brute force method simply requires too much time. If we can somehow bound the results in a narrow range, that would be very helpful. I like @marty cohen’s answer below, but not sure how to calculate $\sum_{k=1}^\infty e^{(-kL)^2}$ efficiently.
 A: I found an upper bound for $L>1$; I'm pretty sure with a bit more effort you might be even able to find the closed form.
Consider the integrand, $f(x) = e^{-(x+kL)^2}$. You can rewrite it as the density of a Gaussian rv:
\begin{align}
f(x) = \sqrt{\frac{2 \pi \frac{1}{2}}{2 \pi \frac{1}{2}}}e^{-\frac{(x+kL)^2}{2 \frac{1}{2}}} = \sqrt{\pi}\frac{1}{\sqrt{2 \pi \sigma^2}}e^{-\frac{(x+kL)^2}{2 \sigma^2}}
\end{align}
with $\sigma^2=\frac{1}{2}, \mu=-kL$. Now, convert this Normal rv to standard Normal:
\begin{align}
P(0<X<1) &= P\bigg(\frac{kL}{\frac{1}{\sqrt{2}}} < Z<\frac{1+kL}{\frac{1}{\sqrt{2}}}\bigg) = \Phi\bigg(\frac{1+kL}{\frac{1}{\sqrt{2}}}\bigg) - \Phi\bigg(\frac{kL}{\frac{1}{\sqrt{2}}}\bigg)\\
&< \Phi \bigg(\frac{(k+1)L}{\frac{1}{\sqrt{2}}}\bigg) - \Phi\bigg(\frac{kL}{\frac{1}{\sqrt{2}}}\bigg)
\end{align}
Here the first term is $\Phi(0)=0.5.$ Now, when you sum these terms on $k$, they cancel out (telescopic sum), and the limit $\lim_{n \to \infty} \Phi(\frac{(n+1)L}{\sigma}) = \Phi(\infty) = 1$, so the upper bound is $\sqrt{\pi}(1-\frac{1}{2})=\frac{\sqrt{\pi}}{2}$
A: Playing around,
probably not too useful.
$\begin{array}\\
f(L)
&=\sum_{k=0}^{\infty}\int_0^1 e^{-(x+kL)^2}dx\\
&=\sum_{k=0}^{\infty}\int_0^1 e^{-x^2-2xkL-k^2L^2}dx\\
&=\sum_{k=0}^{\infty}e^{-k^2L^2}\int_0^1 e^{-x^2-2xkL}dx\\
&=\sum_{k=0}^{\infty}e^{-k^2L^2}\dfrac{\sqrt{\pi}}{2}e^{k^2L^2}(erf(kL+1)-erf(kL))\\
&=\dfrac{\sqrt{\pi}}{2}\sum_{k=0}^{\infty}(erf(kL+1)-erf(kL))\\
&\text{more simply}\\
f(L)
&=\sum_{k=0}^{\infty}\int_0^1 e^{-(x+kL)^2}dx\\
&=\sum_{k=0}^{\infty}\int_{kL}^{kL+1} e^{-x^2}dx\\
&=\dfrac{\sqrt{\pi}}{2}\sum_{k=0}^{\infty}(erf(kL+1)-erf(kL))\\
&\text{simple bounds}\\
f(L)
&=\sum_{k=0}^{\infty}\int_{kL}^{kL+1} e^{-x^2}dx\\
&=\int_{0}^{1} e^{-x^2}dx+\sum_{k=1}^{\infty}\int_{kL}^{kL+1} e^{-x^2}dx\\
&\lt erf(1)+\sum_{k=1}^{\infty}e^{-(kL)^2}\\
f(L)
&=\sum_{k=0}^{\infty}\int_{kL}^{kL+1} e^{-x^2}dx\\
&=\int_{0}^{1} e^{-x^2}dx+\sum_{k=1}^{\infty}\int_{kL}^{kL+1} e^{-x^2}dx\\
&\gt erf(1)+\sum_{k=1}^{\infty}e^{-(kL+1)^2}\\
\end{array}
$
