How to prove
$$\left(\frac{c}{\pi}\right)^{\frac{1}{2}} \int_p^q e^{-c x^2} \mathrm{~d} x=\frac{1}{2}(\operatorname{erf}(q \sqrt{c})-\operatorname{erf}(p \sqrt{c})) .$$
I know$$ \operatorname{erf}(q \sqrt{c})=\frac{2}{\sqrt{\pi}} \int_0^{\sqrt{c} q} e^{-t^2} d t$$
Clearly the $e^{ct^2}$ isn't in the integrand.The "c" is missing.
Furthermore, How do you equate
$$\operatorname{erf}(q \sqrt{c})-\operatorname{erf}(p \sqrt{c})=\int_0^{\sqrt{c} p} \frac{2\left(-e^{-t^2}+\frac{e^{-\left(q^2 t^2\right) / p^2 q} q}{p}\right)}{\sqrt{\pi}} d t$$ with what we want to prove?
Is the answer anything related to the proof of$$ F(x)=\Phi\left(\frac{x-\mu}{\sigma}\right)=\frac{1}{2}\left[1+\operatorname{erf}\left(\frac{x-\mu}{\sigma \sqrt{2}}\right)\right]$$?
\begin{align} \sqrt{\frac{c}{\pi}}\int_{p}^{q}e^{cx^{2}} \ dx &= \sqrt{\frac{c}{\pi}}\left[\int_{0}^{q}e^{cx^{2}} \ dx - \int_{0}^{p}e^{cx^{2}} \ dx\right] \\ &= \sqrt{\frac{c}{\pi}}\left[\int_{0}^{q}e^{(\sqrt{c}x)^{2}} \ dx - \int_{0}^{p}e^{(\sqrt{c}x)^{2}} \ dx\right] \\ &= \sqrt{\frac{c}{\pi}}\left[\frac{1}{\sqrt{c}}\int_{0}^{\sqrt{c}q}e^{x^{2}} \ dx - \frac{1}{\sqrt{c}}\int_{0}^{\sqrt{c}p}e^{x^{2}} \ dx\right] \\ &= \frac{1}{2}\left[\frac{2}{\sqrt{\pi}}\int_{0}^{\sqrt{c}q}e^{x^{2}} \ dx - \frac{2}{\sqrt{\pi}}\int_{0}^{\sqrt{c}p}e^{x^{2}} \ dx\right] \\ &= \frac{1}{2}\left[\text{erf}(\sqrt{c}q) - \text{erf}(\sqrt{c}p)\right] \end{align}