If I want to calculate the following integral in terms of the Error function, is this correct?

$$\frac{1}{\sqrt{2\pi}}\int_{f(x)}^{-\infty}e^{-p^2}\mathrm{d}p = \mathrm{Erf}(-\infty) - \mathrm{Erf}(f(x))$$

  • 1
    $\begingroup$ Yes, other than the fact that $$\frac1{\sqrt{2\pi}}\int_{f(x)}^{-\infty}e^{-p^2}dp=-\operatorname{Erf}(f(x))+\lim_{L\to-\infty}\operatorname{Erf}(L)$$ $\endgroup$ – clathratus Oct 26 '18 at 16:42

You can check that $$\frac{1}{\sqrt{2\pi}}\int_{f(x)}^{\infty}e^{-t^2}dt = \frac{1}{2\sqrt{2}}\big(1-\mathrm{erf}(f(x))\big).$$

Hint: Use the definition of $\mathrm{erf}(x)$:

$$\mathrm{erf}(x)=\frac{2}{\sqrt{\pi}}\int_{0}^{x}e^{-t^2}dt$$ and the identity: $$\int_{-\infty}^{\infty}e^{-t^2}dt = \sqrt{\pi}.$$


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