Consider the following integral:
$$ I(t) = \int_{0}^{t} e^{\alpha(\tau+\beta)^2} \ d\tau. $$
I want to compute this integral analytically, which I believe can be done through the use of error functions.
The error function is defined as
$$ \textrm{erf}(z) := \frac{2}{\sqrt{\pi}}\int_{0}^{z} e^{-t^2} \ dt. $$
Changing variables $z \mapsto iz$ gives
$$ \textrm{erf}(iz) = \frac{2}{\sqrt{\pi}}\int_{0}^{iz} e^{-t^2} \ dt = \frac{2i}{\sqrt{\pi}}\int_{0}^{z} e^{y^2} \ dy, $$ where in the last equality, I used the change of variables $t = iy$. Next, let $y = ax$ for some $a\in\mathbb{R}$. This transformation gives
$$ \textrm{erf}(iz) = \frac{2ai}{\sqrt{\pi}}\int_{0}^{z/a} e^{a^2x^2}\ dx. $$
Next, change variables $z \mapsto az$ which gives
$$ \textrm{erf}(iaz) = \frac{2ai}{\sqrt{\pi}}\int_{0}^{z} e^{a^2x^2} \ dx. $$
This is very close to $I(t)$ which is the goal integral to compute, however I am stuck on how to deal with the extra 'bias' $\beta$. Any help would be appreciated!
Notice that: $$I(t) = \int_{0}^{t} e^{\alpha(\tau+\beta)^2} \ d\tau = \int_{\beta}^{t+\beta} e^{\alpha u^2} \ du = \int_{0}^{t+\beta} e^{\alpha u^2} \ du - \int_{0}^{\beta} e^{\alpha v^2} \ dv$$