How do I evaluate $\int \limits_{-\infty}^{a} e^{−t^2}dt$? I know that
$$I \equiv \int \limits_{-\infty}^\infty e^{−t^2} \, dt=\sqrt{\pi},\text{ and }\int \limits_{-\infty}^0 e^{−t^2} \, dt=\frac{\sqrt{\pi}}{2}.$$
However, I don't understand if (or how) I can find a similar solution for $\int \limits_{-\infty}^{a}e^{−t^2}dt, a \neq 0$, given that the error function actually does not yield closed form solutions.
Any help is greatly appreciated!

$$
\int_{-\infty}^a \exp(-t^2)\, \text dt = \\
\frac{\sqrt{\pi}}{2}+\int_0^a\exp(-t^2)\, \text dt=\\
\frac{\sqrt{\pi}}{2}+\int_0^a\sum_{n=0}^\infty \frac{(-1)^n t^{2n}}{n!} \, \text dt =\\
\frac{\sqrt{\pi}}{2}+\sum_{n=0}^\infty \frac{(-1)^n a^{2n+1}}{n!(2n+1)}
$$
You can use this to estimate the integral to sufficient accuracy (as GEdgar mentioned, however, this sum converges well when $a \approx 0$ but much slower when $a$ is large).  As was mentioned, no simple "closed form" for this integral exists.
For example, we may approximate $\int_{-\infty}^1 \exp(-t^2)\, \text dt$ with a truncated (6 terms) series:
$$\int_{-\infty}^1 \exp(-t^2)\, \text dt =1.63305\cdots \approx 
\frac{\sqrt{\pi}}{2}+\sum_{n=0}^6 \frac{(-1)^n}{n!(2n+1)}=1.66306\cdots$$
See also this paper for more strategies.
