Let $r>0$. Show that $$\frac{1}{\sqrt{4\pi t}}\int\limits^{-r}_{-\infty} \exp \left(-\frac{x^{2}}{4t}\right) dx=\frac{1}{2}\left[1-\text{erf}\left(\frac{r}{\sqrt{4t}}\right)\right]$$ where $\text{erf}$ is the error function
I calculated this using software, but I can't prove .... somebody can help me please.
Since the integrand is even, we have $$\begin{align} \frac{1}{\sqrt{4\pi t}}\int\limits^{-r}_{-\infty} \exp \left(-\frac{x^{2}}{4t}\right) dx &= \frac{1}{\sqrt{4\pi t}}\int\limits^{\infty}_{r} \exp \left(-\frac{x^{2}}{4t}\right) dx. \end{align}$$ Then, making the $u$-substitution $u=x/(2\sqrt{t})$, we get $$\begin{align} \frac{1}{\sqrt{4\pi t}}\int\limits^{-r}_{-\infty} \exp \left(-\frac{x^{2}}{4t}\right) dx &= \frac{1}{\sqrt{\pi}}\int\limits^{\infty}_{r/(2\sqrt{t})} \exp \left(-u^2\right) du\\ &= \frac{1}{2}\left(\frac{2}{\sqrt{\pi}}\int\limits_0^\infty\exp \left(-u^2\right) du - \operatorname{erf}\left(\frac{r}{2\sqrt{t}}\right) \right)\\ &= \frac{1}{2}\left[1-\operatorname{erf}\left(\frac{r}{2\sqrt{t}}\right)\right]. \end{align}$$
